CommeUnJeu · L1 PCSI

Polynomials

⌚ ~147 min ▢ 18 blocks ✓ 77 exercises ➣ Prerequisites : Real Polynomials, Algebraic computations

A polynomial in $\mathbb{K}[X]$ is a formal expression $P = a_n X^n + \dots + a_1 X + a_0$ where $X$ is an indeterminate --- not a number. Addition and multiplication are defined formally on the coefficients; only later, by evaluation, do we apply $P$ to a specific element $a$ to obtain $P(a)$. This separation between the formal polynomial $P$ and its associated polynomial function $\tilde{P} : x \mapsto P(x)$ is what makes the algebraic theory work cleanly: it is the only way to honestly state and prove the result « $P = 0 \iff \tilde{P} = 0$ », which would otherwise be a circular tautology. Throughout, $\mathbb{K}$ denotes $\mathbb{R}$ or $\mathbb{C}$.
The chapter proceeds in seven movements. We first set up the ring $\mathbb{K}[X]$, the degree, and the operations of addition, multiplication, and composition; the key invariant is the degree, which controls everything that follows. We next introduce divisibility and prove the central computational tool of the chapter: the Euclidean division of $A$ by $B \neq 0$, which produces a unique pair $(Q, R)$ with $\deg R < \deg B$. We then bring in polynomial functions, roots, multiplicity, and the relations between coefficients and roots of a split polynomial; the key fact is the rigidity theorem (a non-zero polynomial has at most $\deg P$ roots). We introduce the formal derivative and the polynomial Taylor formula, and use them to characterize multiplicity by successive vanishing of derivatives. We state the d'Alembert--Gauss theorem (admis), classify the irreducible polynomials of $\mathbb{C}[X]$ and $\mathbb{R}[X]$, and prove the unique-factorization theorems in both. We then treat Lagrange interpolation: $n$ data points with distinct abscissas determine a unique polynomial of degree at most $n - 1$. We close with the partial-fraction decomposition of rational functions in simple situations --- the practical tool that turns a quotient $A / B$ into a sum of simple pieces, ready for integration, differentiation, and summation.
Two limits are set from the start. The construction of $\mathbb{K}[X]$ is beyond our scope: we take $\mathbb{K}[X]$ as a commutative ring with unit and reason from there. The deeper arithmetic of $\mathbb{K}[X]$ --- the gcd of two polynomials, Bézout's identity, the systematic use of common factors --- is likewise beyond our scope; whenever we need to express the absence of a common factor we say so directly, in terms of common divisors or common roots. The proof of d'Alembert--Gauss is admitted, and we use the theorem as a black box for the factorization and decomposition sections below.

I Ring $\mathbb{K}[X]$$\virgule$ degree$\virgule$ composition

We work over $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$ throughout the chapter (the program restricts itself to these two fields). A polynomial is a formal expression $\sum_{k \geq 0} a_k X^k$ with finite support. The construction of $\mathbb{K}[X]$ is hors programme; we announce it axiomatically as a commutative ring with unit and reason from there. The crucial operational tool is the degree: it controls the size of $P + Q$, controls (additively) the size of $PQ$, forces integrality, and reappears in every theorem of the chapter.

Definition — Polynomial$\virgule$ coefficient

A polynomial (with one indeterminate, with coefficients in $\mathbb{K}$) is a formal expression $$ P = \sum_{k \geq 0} a_k X^k = a_0 + a_1 X + a_2 X^2 + \dots $$ where $a_k \in \mathbb{K}$ for every $k \in \mathbb{N}$ and $a_k = 0$ for $k$ large enough (the sum has only finitely many non-zero terms). The scalar $a_k$ is the coefficient of degree $k$ of $P$. The set of all polynomials with coefficients in $\mathbb{K}$ is denoted $\mathbb{K}[X]$.
Two polynomials $P = \sum a_k X^k$ and $Q = \sum b_k X^k$ are equal if and only if $a_k = b_k$ for every $k \in \mathbb{N}$ (identification of coefficients).
Admis (the rigorous construction is hors programme): $\mathbb{K}[X]$, equipped with the addition $(P + Q)_k = P_k + Q_k$ and the Cauchy product $(PQ)_k = \sum_{i + j = k} P_i Q_j$, is a commutative ring with unit (the additive neutral is the polynomial $0$, the multiplicative neutral is the constant polynomial $1$). The integrality of this ring is proved below from the degree formula for the product.

Example

$7 X^3 + 2 X^2 - X + 5$ is a polynomial of $\mathbb{R}[X]$. Its non-zero coefficients are $a_0 = 5$, $a_1 = -1$, $a_2 = 2$, $a_3 = 7$, with $a_k = 0$ for $k \geq 4$. The constant polynomials $0$ and $1$, and the indeterminate $X$ itself (with $a_1 = 1$ and all others zero), are in $\mathbb{K}[X]$.

Definition — Degree$\virgule$ dominant coefficient$\virgule$ monic polynomial

Let $P \in \mathbb{K}[X]$ with $P \neq 0$. The degree of $P$ is $$ \deg P = \max\{k \in \mathbb{N} \,\mid\, a_k \neq 0\}. $$ The coefficient $a_{\deg P}$ is the dominant coefficient of $P$. We say $P$ is monic (« unitaire ») if its dominant coefficient equals $1$.
By convention, the degree of the zero polynomial is $\deg 0 = -\infty$. The zero polynomial has no dominant coefficient.

Definition — $\mathbb{K}_n\lbrack X\rbrack$

For $n \in \mathbb{N}$, we denote $$ \mathbb{K}_n[X] = \{P \in \mathbb{K}[X] \,\mid\, \deg P \leq n\}, $$ the set of polynomials of degree at most $n$. The zero polynomial belongs to $\mathbb{K}_n[X]$ for every $n$ (since $-\infty \leq n$).

Proposition — Degree of a sum

For all $P, Q \in \mathbb{K}[X]$, $$ \deg(P + Q) \leq \max(\deg P, \deg Q), $$ with equality when $\deg P \neq \deg Q$.

Proof

The result is immediate if $P = 0$ or $Q = 0$. Suppose $P \neq 0$ and $Q \neq 0$, with $P = \sum a_k X^k$ and $Q = \sum b_k X^k$, and set $m = \max(\deg P, \deg Q)$. For every $k > m$, both $a_k = 0$ and $b_k = 0$, so $a_k + b_k = 0$, hence $\deg(P + Q) \leq m$.
If $\deg P \neq \deg Q$, say $\deg P > \deg Q$, the coefficient of $X^{\deg P}$ in $P + Q$ is $a_{\deg P} + 0 = a_{\deg P} \neq 0$, hence $\deg(P + Q) = \deg P = \max(\deg P, \deg Q)$.

Proposition — Degree of a product

For all $P, Q \in \mathbb{K}[X]$ with $P \neq 0$ and $Q \neq 0$, the product $PQ$ is non-zero and $$ \deg(PQ) = \deg P + \deg Q. $$ The dominant coefficient of $PQ$ equals the product of the dominant coefficients of $P$ and $Q$.

Proof

Set $p = \deg P$, $q = \deg Q$, with dominant coefficients $a_p \neq 0$ and $b_q \neq 0$. The coefficient of degree $k$ of $PQ$ is $\displaystyle c_k = \sum_{i + j = k} a_i b_j$.
For $k > p + q$: every pair $(i, j)$ with $i + j = k$ has $i > p$ or $j > q$, so $a_i = 0$ or $b_j = 0$, hence $c_k = 0$. Thus $\deg(PQ) \leq p + q$.
For $k = p + q$: in the sum $c_{p + q}$, the only $(i, j)$ with $i \leq p$ and $j \leq q$ is $(p, q)$ (any other would require $i > p$ or $j > q$). Hence $c_{p + q} = a_p b_q \neq 0$. This shows simultaneously $PQ \neq 0$, $\deg(PQ) = p + q$, and that the dominant coefficient of $PQ$ is $a_p b_q$.

Proposition — Integrality of $\mathbb{K}\lbrack X\rbrack$

The ring $\mathbb{K}[X]$ is an integral domain: for all $P, Q \in \mathbb{K}[X]$, $$ PQ = 0 \implies P = 0 \text{ or } Q = 0. $$

Proof

Contrapositive of the degree-of-product Proposition. If $P \neq 0$ and $Q \neq 0$, then $\deg(PQ) = \deg P + \deg Q \geq 0$, in particular $PQ \neq 0$. By contraposition, $PQ = 0$ implies $P = 0$ or $Q = 0$.

Definition — Composition

For $P = \sum a_k X^k \in \mathbb{K}[X]$ and $Q \in \mathbb{K}[X]$, the composition $P \circ Q$ is the polynomial $$ P \circ Q = \sum_{k \geq 0} a_k \, Q^k = a_0 + a_1 Q + a_2 Q^2 + \dots $$ (the sum has finitely many non-zero terms since $P$ has finite support).

Proposition — Degree of a composition

Let $P, Q \in \mathbb{K}[X]$ with $P \neq 0$ and $Q$ non-constant (i.e. $\deg Q \geq 1$). Set $p = \deg P$ and $q = \deg Q$. Then $$ \deg(P \circ Q) = p \cdot q. $$ Edge cases. If $P = 0$, $P \circ Q = 0$. If $\deg P = 0$, $P \circ Q = P$ has degree $0$. If $Q = c$ is constant, $P \circ Q$ is the constant polynomial $P(c)$ (possibly the zero polynomial); the formula $p q$ does not apply.

Proof

With $P = \sum_{k=0}^{p} a_k X^k$ ($a_p \neq 0$) and $\deg Q = q \geq 1$, the degree-of-product Proposition gives $\deg(Q^k) = k q$ for every $k \in \mathbb{N}$ (by induction on $k$). Hence:

$\deg(a_p Q^p) = p q$ with dominant coefficient $a_p \cdot (\text{dominant coeff. of } Q)^p$, which is non-zero.
For every $k < p$, $\deg(a_k Q^k) \leq k q < p q$.

The leading term $a_p Q^p$ dominates strictly, so $\deg(P \circ Q) = p q$.

Method — Computing degrees of $P + Q$$\virgule$ $PQ$$\virgule$ $P \circ Q$$\virgule$ $P^n$

The four formulas to memorize, valid when all polynomials involved are non-zero (and $Q$ non-constant for composition): $$ \textcolor{colorprop}{\begin{aligned} \deg(P + Q) &\leq \max(\deg P, \deg Q) \quad \text{(equality if degrees differ)} \\ \deg(P Q) &= \deg P + \deg Q \\ \deg(P \circ Q) &= \deg P \cdot \deg Q \quad \text{(if } \deg Q \geq 1 \text{)} \\ \deg(P^n) &= n \cdot \deg P \quad \text{(} n \in \mathbb{N} \text{)} \end{aligned}} $$ The last identity is iteration of the second. Always check non-zero hypotheses before applying these --- the zero polynomial has $\deg 0 = -\infty$, which makes the formulas degenerate.

Example

Compute $\deg\big((X^3 + X + 1)(X^2 - 1)\big)$ and $\deg\big((X^2 + 1) \circ (X - 2)\big)$.

Answer

Both factors of the product are non-zero of degrees $3$ and $2$, hence $$ \deg\big((X^3 + X + 1)(X^2 - 1)\big) = 3 + 2 = 5. $$ For the composition: $P = X^2 + 1$ has degree $2$, $Q = X - 2$ has degree $1$ (non-constant), so $\deg(P \circ Q) = 2 \cdot 1 = 2$. Direct check: $(X - 2)^2 + 1 = X^2 - 4X + 5$.

Skills to practice

Computing degrees and operating on polynomials

II Divisibility and Euclidean division

Just as for integers, divisibility in $\mathbb{K}[X]$ is the relation « $A = B Q$ for some $Q$ ». The decisive fact is that $\mathbb{K}[X]$ admits a Euclidean division: $A$ divided by $B \neq 0$ produces a unique pair $(Q, R)$ with $\deg R < \deg B$. The role played by absolute value in the integer Euclidean division is played here by the degree. We treat divisibility lightly --- enough to formulate the theorem and recognise associated polynomials --- and leave the deeper arithmetic (PGCD, Bézout, Gauss) aside, as it is beyond the scope of this course.

Definition — Divisibility

Let $A, B \in \mathbb{K}[X]$. We say that $B$ divides $A$, written $B \mid A$, if there exists $Q \in \mathbb{K}[X]$ such that $$ A = B Q. $$ We also say $B$ is a divisor of $A$, that $A$ is divisible by $B$, or that $A$ is a multiple of $B$.

Proposition — Basic properties of divisibility

The relation $\mid$ on $\mathbb{K}[X]$ satisfies, for all $A, B, C, D \in \mathbb{K}[X]$:

Reflexivity: $A \mid A$.
Transitivity: $A \mid B$ and $B \mid C$ imply $A \mid C$.
Linearity: if $D \mid A$ and $D \mid B$, then $D \mid (\lambda A + \mu B)$ for all $\lambda, \mu \in \mathbb{K}$.
Product: if $A \mid B$ and $C \mid D$, then $A C \mid B D$.
$A \mid 0$ for every $A$; $0 \mid A$ if and only if $A = 0$.

Proof

All five points are direct from the definition.

Reflexivity: $A = A \cdot 1$.
Transitivity: $B = A Q_1$ and $C = B Q_2$ give $C = A (Q_1 Q_2)$.
Linearity: $A = D U_1$ and $B = D U_2$ give $\lambda A + \mu B = D (\lambda U_1 + \mu U_2)$.
Product: $B = A U$ and $D = C V$ give $B D = (A C)(U V)$.
For $A \mid 0$: $0 = A \cdot 0$. For $0 \mid A \implies A = 0$: $0 \mid A$ means $A = 0 \cdot Q = 0$.

Definition — Associated polynomials

Two polynomials $A, B \in \mathbb{K}[X]$ are associated if $A \mid B$ and $B \mid A$.

Proposition — Characterization of associated polynomials

For all $A, B \in \mathbb{K}[X]$, the following are equivalent:

$A$ and $B$ are associated.
There exists $\lambda \in \mathbb{K}^*$ such that $B = \lambda A$.

Proof

$(\Leftarrow)$ If $B = \lambda A$ with $\lambda \in \mathbb{K}^*$, then $A \mid B$ (with quotient $\lambda$) and $B \mid A$ (with quotient $\lambda^{-1}$, since $\lambda \neq 0$).
$(\Rightarrow)$ Suppose $A \mid B$ and $B \mid A$. Then $B = A Q_1$ and $A = B Q_2$ for some $Q_1, Q_2 \in \mathbb{K}[X]$, so $A = A Q_1 Q_2$, i.e. $A (1 - Q_1 Q_2) = 0$.
- Case $A = 0$. Then $B = 0 \cdot Q_1 = 0$, so $B = 1 \cdot A$ with $\lambda = 1 \in \mathbb{K}^*$.
- Case $A \neq 0$. By integrality of $\mathbb{K}[X]$, $1 - Q_1 Q_2 = 0$, i.e. $Q_1 Q_2 = 1$. Then $\deg Q_1 + \deg Q_2 = \deg 1 = 0$ with $\deg Q_1, \deg Q_2 \in \mathbb{N}$, hence $\deg Q_1 = \deg Q_2 = 0$. So $Q_1$ is a non-zero constant $\lambda \in \mathbb{K}^*$, and $B = \lambda A$.

Theorem — Euclidean division in $\mathbb{K}\lbrack X\rbrack$

For all $A, B \in \mathbb{K}[X]$ with $B \neq 0$, there exists a unique pair $(Q, R) \in \mathbb{K}[X] \times \mathbb{K}[X]$ such that $$ A = B Q + R \quad \text{and} \quad \deg R < \deg B. $$ $Q$ is the quotient and $R$ the remainder of the Euclidean division of $A$ by $B$.

Proof

Set $b = \deg B \geq 0$ and let $\beta$ be the dominant coefficient of $B$ ($\beta \neq 0$).
Existence (by strong induction on $\deg A$). We prove: for every $A \in \mathbb{K}[X]$, there exist $Q, R$ with $A = B Q + R$ and $\deg R < b$.
Base. If $\deg A < b$ (in particular if $A = 0$), take $Q = 0$ and $R = A$.
Inductive step. Assume the result for all polynomials of degree $< n$, and let $A$ be of degree $n \geq b$. Write $A = a_n X^n + (\text{lower-degree terms})$ with $a_n \neq 0$. Set $$ A' = A - \frac{a_n}{\beta} X^{n - b} \, B. $$ The product $\frac{a_n}{\beta} X^{n - b} B$ has degree $(n - b) + b = n$ and dominant coefficient $\frac{a_n}{\beta} \cdot \beta = a_n$, which equals the dominant coefficient of $A$. So the leading $X^n$ terms cancel in $A'$, giving $\deg A' \leq n - 1 < n$. By the inductive hypothesis, $A' = B Q' + R$ with $\deg R < b$. Hence $$ A = A' + \frac{a_n}{\beta} X^{n - b} B = B \left( Q' + \frac{a_n}{\beta} X^{n - b} \right) + R, $$ which is the desired writing of $A$ with quotient $Q = Q' + \frac{a_n}{\beta} X^{n - b}$ and remainder $R$.
Uniqueness. Suppose $(Q_1, R_1)$ and $(Q_2, R_2)$ both satisfy $A = B Q_i + R_i$ and $\deg R_i < b$. Subtracting: $$ B (Q_1 - Q_2) = R_2 - R_1. $$ The right-hand side has degree $\leq \max(\deg R_1, \deg R_2) < b$. If $Q_1 \neq Q_2$, the left-hand side has degree $b + \deg(Q_1 - Q_2) \geq b$ (since $B \neq 0$), contradiction. Hence $Q_1 = Q_2$, and then $R_1 = A - B Q_1 = A - B Q_2 = R_2$.

Proposition — Divisibility via remainder

For $A, B \in \mathbb{K}[X]$ with $B \neq 0$, $$ B \mid A \iff \text{the remainder of the Euclidean division of } A \text{ by } B \text{ is } 0. $$

Proof

$(\Leftarrow)$ If $A = B Q + 0 = B Q$, then $B \mid A$ by definition.
$(\Rightarrow)$ Conversely, suppose $B \mid A$, i.e. $A = B Q_0$ for some $Q_0 \in \mathbb{K}[X]$. Then $(Q_0, 0)$ satisfies $A = B Q_0 + 0$ with $\deg 0 = -\infty < \deg B$. By uniqueness of the Euclidean division, the remainder is $0$.

Method — Euclidean division --- long-division algorithm

To divide $A$ by $B \neq 0$, iterate:

If $\deg A < \deg B$, stop: $Q = 0$, $R = A$.
Otherwise, divide the leading term of $A$ by the leading term of $B$: this gives a monomial $\frac{a_n}{\beta} X^{n - b}$, the next term of $Q$.
Subtract $\frac{a_n}{\beta} X^{n - b} \cdot B$ from $A$: this kills the leading term, producing a new $A'$ with $\deg A' < \deg A$.
Replace $A$ by $A'$ and go back to step 1.

The algorithm terminates because $\deg A$ strictly decreases at each step. The successive monomials $\frac{a_n}{\beta} X^{n - b}$ are the successive terms of the quotient $Q$, and the final remainder is $R$.

Example

Compute the Euclidean division of $A = X^4 + X^3 - 2 X^2 + 1$ by $B = X^2 - X + 1$.

Answer

Apply the long-division algorithm: $$ \begin{aligned} A &= X^4 + X^3 - 2 X^2 + 0 \cdot X + 1 \\ &\quad - X^2 \cdot B = X^4 - X^3 + X^2 \quad \text{(first step: } X^4 / X^2 = X^2 \text{)} \\ A_1 &= 2 X^3 - 3 X^2 + 0 \cdot X + 1 \\ &\quad - 2X \cdot B = 2 X^3 - 2 X^2 + 2 X \quad \text{(second step: } 2 X^3 / X^2 = 2 X \text{)} \\ A_2 &= -X^2 - 2 X + 1 \\ &\quad - (-1) \cdot B = -X^2 + X - 1 \quad \text{(third step: } -X^2 / X^2 = -1 \text{)} \\ A_3 &= -3 X + 2. \end{aligned} $$ Now $\deg A_3 = 1 < \deg B = 2$, so we stop. The quotient is $Q = X^2 + 2 X - 1$ and the remainder is $R = -3 X + 2$.
Verification. $B Q + R = (X^2 - X + 1)(X^2 + 2X - 1) + (-3 X + 2) = X^4 + X^3 - 2 X^2 + 1 = A$. $\checkmark$

Skills to practice

Performing Euclidean division and using it

III Polynomial functions$\virgule$ roots$\virgule$ Viète relations

A polynomial $P$ and the function it computes (« evaluate $P$ at $x$ ») are conceptually distinct objects. The bridge is the evaluation map $a \mapsto P(a)$. Once we have evaluation, we can ask: when does $P(a) = 0$? Such an $a$ is a root of $P$, and the algebraic content of the chapter is encoded in the equivalence root $\iff$ $(X - a) \mid P$. Multiplicity, factorization, rigidity, and Viète's relations all follow from this single observation.

Definition — Polynomial function$\virgule$ evaluation

For $P = \sum a_k X^k \in \mathbb{K}[X]$ and $a \in \mathbb{K}$, the evaluation of $P$ at $a$ is the scalar $$ P(a) = \sum_{k \geq 0} a_k a^k \in \mathbb{K}. $$ The map $\tilde{P} : \mathbb{K} \to \mathbb{K}$, $x \mapsto P(x)$, is the polynomial function associated to $P$.

Proposition — Evaluation is a ring morphism

For $a \in \mathbb{K}$ fixed, the evaluation map $\mathrm{ev}_a : P \mapsto P(a)$ from $\mathbb{K}[X]$ to $\mathbb{K}$ satisfies $$ \textcolor{colorprop}{\begin{aligned} (P + Q)(a) &= P(a) + Q(a), \\ (P Q)(a) &= P(a) \cdot Q(a), \\ 1(a) &= 1. \end{aligned}} $$

Proof

Direct from the coefficient definitions. For the sum: $(P + Q)(a) = \sum (P_k + Q_k) a^k = \sum P_k a^k + \sum Q_k a^k = P(a) + Q(a)$. For the product, the Cauchy product: $$ (PQ)(a) = \sum_k \left( \sum_{i + j = k} P_i Q_j \right) a^k = \sum_{i, j} P_i Q_j a^{i + j} = \left( \sum_i P_i a^i \right) \left( \sum_j Q_j a^j \right) = P(a) Q(a). $$ The constant polynomial $1$ has $a_0 = 1$ and all other coefficients zero, so $1(a) = 1$.

Definition — Root of a polynomial

Let $P \in \mathbb{K}[X]$ and $a \in \mathbb{K}$. We say that $a$ is a root (or zero) of $P$ in $\mathbb{K}$ if $P(a) = 0$.

Proposition — Root $\iff$ linear factor

For $P \in \mathbb{K}[X]$ and $a \in \mathbb{K}$: $$ a \text{ is a root of } P \iff (X - a) \mid P. $$

Proof

Perform the Euclidean division of $P$ by $X - a$ (which has degree $1$): there exist $Q \in \mathbb{K}[X]$ and a constant $r \in \mathbb{K}$ (since $\deg R < 1$) such that $$ P = (X - a) Q + r. $$ Evaluate at $a$: the evaluation morphism gives $P(a) = (a - a) Q(a) + r = 0 + r = r$. Hence $r = P(a)$, and $$ (X - a) \mid P \iff r = 0 \iff P(a) = 0 \iff a \text{ is a root of } P. $$

Method — Horner's scheme

To evaluate $P = a_n X^n + a_{n-1} X^{n-1} + \dots + a_0$ at a scalar $\alpha$, rewrite in the nested form $$ P(\alpha) = \big(\dots \big((a_n \alpha + a_{n-1}) \alpha + a_{n-2}\big) \alpha + \dots\big) \alpha + a_0. $$ This computes $P(\alpha)$ in exactly $n$ multiplications and $n$ additions, without computing the powers $\alpha^k$ separately.
Bonus property. The intermediate values $b_{n-1} = a_n$, $b_{n-2} = b_{n-1} \alpha + a_{n-1}$, \dots, $b_0 = b_1 \alpha + a_1$ are exactly the coefficients of the quotient $Q = b_{n-1} X^{n-1} + \dots + b_0$ in the Euclidean division of $P$ by $X - \alpha$, and the final number $P(\alpha) = b_0 \alpha + a_0$ is the remainder. Justification: writing $P = (X - \alpha) Q + r$ with $Q = \sum_{k=0}^{n-1} b_k X^k$ and matching coefficients gives $a_k = b_{k-1} - \alpha b_k$ (with $b_{n-1} = a_n$ and $b_{-1} := r$), rearranging to $b_{k-1} = b_k \alpha + a_k$ --- Horner's recurrence. Thus Horner gives both $P(\alpha)$ and the quotient in one pass.

Definition — Multiplicity of a root

Let $P \in \mathbb{K}[X]$ with $P \neq 0$ and $a \in \mathbb{K}$. We say that $a$ is a root of multiplicity $m \geq 1$ of $P$ if $$ (X - a)^m \mid P \quad \text{and} \quad (X - a)^{m+1} \nmid P. $$ This makes sense: since $P \neq 0$, any $m$ with $(X - a)^m \mid P$ satisfies $m \leq \deg P$ (a divisor has degree at most that of the dividend), so the set of such $m$ is bounded, and the multiplicity --- the largest $m$ with $(X - a)^m \mid P$ --- is a well-defined integer. By convention, if $a$ is not a root of $P$, its multiplicity is $0$. A root of multiplicity $1$ is called simple, of multiplicity $2$ double, and so on. (For $P = 0$, every $a$ is a root and finite multiplicity is not defined --- the definition deliberately excludes this case.)

Proposition — Number of roots vs degree

Let $P \in \mathbb{K}[X]$ with $P \neq 0$ of degree $n$, and let $a_1, \dots, a_r \in \mathbb{K}$ be its distinct roots with multiplicities $m_1, \dots, m_r \geq 1$. Then $$ (X - a_1)^{m_1} (X - a_2)^{m_2} \cdots (X - a_r)^{m_r} \mid P, $$ and in particular $$ m_1 + m_2 + \dots + m_r \leq n. $$ The sum of multiplicities of roots of $P$ is at most $\deg P$. A non-zero polynomial of degree $n$ has at most $n$ roots counted without multiplicity.

Proof

We use strong induction on $\deg P = n$.
Base $n = 0$. A non-zero constant polynomial has no roots, hence the sum of multiplicities is $0 \leq 0$.
Inductive step. Suppose the result holds for every non-zero polynomial of degree $< n$, and let $P$ have degree $n \geq 1$.

If $P$ has no root in $\mathbb{K}$: the sum $\sum m_i = 0 \leq n$, and the divisibility statement is the empty product « $1 \mid P$ », trivially true.
If $P$ has at least one root in $\mathbb{K}$: pick $a_1$ a root with multiplicity $m_1 \geq 1$. By definition of multiplicity, $P = (X - a_1)^{m_1} \cdot A$ with $A(a_1) \neq 0$ (otherwise $(X - a_1)^{m_1 + 1} \mid P$, contradicting the maximality of $m_1$). The polynomial $A$ has degree $n - m_1 < n$ and is non-zero. The roots of $P$ other than $a_1$ are exactly the roots of $A$: indeed, for $a \neq a_1$, the factor $(X - a_1)^{m_1}$ evaluated at $a$ is $(a - a_1)^{m_1} \neq 0$, hence $P(a) = 0 \iff A(a) = 0$. Multiplicity preservation lemma. For $a \neq a_1$, the multiplicity of $a$ as a root of $P$ equals its multiplicity as a root of $A$. Proof: write $A = (X - a)^{m'} B$ with $B(a) \neq 0$ (so $m'$ is the multiplicity of $a$ in $A$). Then $P = (X - a_1)^{m_1} (X - a)^{m'} B$, and evaluating $(X - a_1)^{m_1} B$ at $a$ gives $(a - a_1)^{m_1} B(a) \neq 0$. Hence $(X - a)^{m'} \mid P$ but $(X - a)^{m' + 1} \nmid P$, i.e.\ $a$ has multiplicity exactly $m'$ in $P$. By the inductive hypothesis applied to $A$, $$ (X - a_2)^{m_2} \cdots (X - a_r)^{m_r} \mid A \quad \text{and} \quad m_2 + \dots + m_r \leq \deg A = n - m_1. $$ Multiplying by $(X - a_1)^{m_1}$: $$ (X - a_1)^{m_1} (X - a_2)^{m_2} \cdots (X - a_r)^{m_r} \mid P, $$ and adding $m_1$ to both sides of the degree inequality gives $\sum_{i=1}^{r} m_i \leq n = \deg P$.

Proposition — Identification polynomial $\iff$ function

For $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$, the map $P \mapsto \tilde{P}$ from $\mathbb{K}[X]$ to the space of functions $\mathbb{K} \to \mathbb{K}$ is injective: for all $P, Q \in \mathbb{K}[X]$, $$ \tilde{P} = \tilde{Q} \iff P = Q. $$ In particular, $P = 0 \iff \tilde{P} = 0$.

Proof

The map $P \mapsto \tilde{P}$ is a ring morphism (Proposition above), so $\tilde{P} = \tilde{Q} \iff \widetilde{P - Q} = 0$. It thus suffices to show that $\tilde{P} = 0$ implies $P = 0$.
Suppose $\tilde{P} = 0$, i.e. $P(a) = 0$ for every $a \in \mathbb{K}$. Then every $a \in \mathbb{K}$ is a root of $P$. Since $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$ is infinite, $P$ has infinitely many roots. By the previous Proposition (number of roots $\leq$ degree), if $P \neq 0$ then $P$ has at most $\deg P$ roots, a finite number. Contradiction. Hence $P = 0$.

Definition — Split polynomial

Let $P \in \mathbb{K}[X]$ with $\deg P = n \geq 1$. We say that $P$ is split over $\mathbb{K}$ (« polynôme scindé sur $\mathbb{K}$ ») if $P$ has $n$ roots in $\mathbb{K}$ counted with multiplicity, i.e. if $P$ can be written $$ P = \alpha \prod_{i=1}^{n} (X - x_i), $$ with $\alpha \in \mathbb{K}^*$ the dominant coefficient of $P$ and $x_1, \dots, x_n \in \mathbb{K}$ the roots (not necessarily distinct).

Proposition — Viète relations$\virgule$ sum and product

Let $P = \alpha \prod_{i=1}^{n} (X - x_i) = a_n X^n + a_{n-1} X^{n-1} + \dots + a_0$ be a split polynomial with $n \geq 1$ roots $x_1, \dots, x_n$ counted with multiplicity. Then $$ \textcolor{colorprop}{\begin{aligned} \sum_{i=1}^{n} x_i &= -\frac{a_{n-1}}{a_n}, \\ \prod_{i=1}^{n} x_i &= (-1)^n \frac{a_0}{a_n}. \end{aligned}} $$

Proof

Expand $P = a_n \prod_{i=1}^{n} (X - x_i)$. The coefficient of $X^n$ in the product $\prod (X - x_i)$ equals $1$, so $a_n = \alpha$ confirms $\alpha$ is the dominant coefficient.
Coefficient of $X^{n-1}$. In the expansion of $\prod_{i=1}^{n} (X - x_i)$, the term $X^{n-1}$ is obtained by choosing $X$ from $n - 1$ factors and $-x_i$ from the remaining one. Summing over the choice: $$ \text{coeff of } X^{n-1} \text{ in } \prod (X - x_i) = -(x_1 + x_2 + \dots + x_n) = -\sum x_i. $$ Hence $a_{n-1} = a_n \cdot (- \sum x_i)$, giving $\sum x_i = -a_{n-1}/a_n$.
Coefficient of $X^0$. Setting $X = 0$ in $\prod (X - x_i)$ gives $\prod (- x_i) = (-1)^n \prod x_i$. So $a_0 = a_n \cdot (-1)^n \prod x_i$, giving $\prod x_i = (-1)^n a_0 / a_n$.

Method — Reading sum and product of roots without computing them

Given a split polynomial $a_n X^n + a_{n-1} X^{n-1} + \dots + a_0$, the sum of the roots reads off the ratio $-a_{n-1}/a_n$ and the product reads off $(-1)^n a_0 / a_n$. The general elementary-symmetric relations $$ \sigma_k = \sum_{i_1 < i_2 < \dots < i_k} x_{i_1} x_{i_2} \cdots x_{i_k} = (-1)^k \frac{a_{n-k}}{a_n}, \quad k \in \{1, \dots, n\}, $$ must be re-derivable quickly (program: « doivent être retrouvées rapidement »). The proof follows the same expansion of $\prod (X - x_i)$.

Example

Admitting the fundamental theorem of algebra (d'Alembert--Gauss, proved later in this chapter), $P = X^3 - 6 X^2 + 11 X - 6$ is split over $\mathbb{C}$ and has three complex roots counted with multiplicity. Find their sum and product without computing them, then guess the roots by inspection.

Answer

Apply Viète with $a_3 = 1$, $a_2 = -6$, $a_0 = -6$, $n = 3$: $$ \sum x_i = -\frac{a_2}{a_3} = 6, \qquad \prod x_i = (-1)^3 \frac{a_0}{a_3} = -(-6) = 6. $$ Inspection: $1 + 2 + 3 = 6$ and $1 \cdot 2 \cdot 3 = 6$ --- try $\{1, 2, 3\}$. Verification: $P(1) = 1 - 6 + 11 - 6 = 0$, $P(2) = 8 - 24 + 22 - 6 = 0$, $P(3) = 27 - 54 + 33 - 6 = 0$. $\checkmark$ Hence $P = (X - 1)(X - 2)(X - 3)$.

Skills to practice

Finding roots$\virgule$ factoring$\virgule$ computing multiplicity
Using Viète's relations

IV Formal derivative and Taylor formula

The formal derivative of a polynomial is defined coefficient by coefficient --- purely algebraic, no analysis required. It enjoys the same algebraic properties as the analytic derivative (linearity, product rule, Leibniz formula). The polynomial Taylor formula is exact (no remainder term) and re-expresses $P$ as a finite combination of $(X - a)^k$. Together, these tools give a clean characterization of multiplicity: $a$ is a root of multiplicity $m$ of $P$ if and only if $P(a) = P'(a) = \dots = P^{(m-1)}(a) = 0$ and $P^{(m)}(a) \neq 0$.

Definition — Formal derivative

For $P = \sum_{k \geq 0} a_k X^k \in \mathbb{K}[X]$, the formal derivative of $P$ is $$ P' = \sum_{k \geq 1} k \, a_k X^{k - 1}. $$ The successive derivatives are defined inductively: $P^{(0)} = P$ and $P^{(n+1)} = (P^{(n)})'$ for $n \in \mathbb{N}$.

Proposition — Degree of the derivative

For $P \in \mathbb{K}[X]$:

If $\deg P = n \geq 1$, then $\deg P' = n - 1$.
If $\deg P \leq 0$ (i.e. $P$ is constant), then $P' = 0$.

Proof

If $P = a_0$ is constant, the formula $P' = \sum_{k \geq 1} k a_k X^{k-1}$ gives $P' = 0$.
If $\deg P = n \geq 1$ with dominant coefficient $a_n \neq 0$, the formula gives $P' = \sum_{k=1}^{n} k a_k X^{k-1}$. The leading term is $n a_n X^{n-1}$ with coefficient $n a_n$. Since we work over $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$ (characteristic $0$), $n a_n \neq 0$. Hence $\deg P' = n - 1$.

Proposition — Linearity$\virgule$ product rule$\virgule$ Leibniz formula

For all $P, Q \in \mathbb{K}[X]$, $\lambda, \mu \in \mathbb{K}$, and $n \in \mathbb{N}$: $$ \textcolor{colorprop}{\begin{aligned} (\lambda P + \mu Q)' &= \lambda P' + \mu Q' \quad &&\text{(linearity)} \\ (P Q)' &= P' Q + P Q' \quad &&\text{(product rule)} \\ (P Q)^{(n)} &= \sum_{k=0}^{n} \binom{n}{k} P^{(k)} Q^{(n - k)} \quad &&\text{(Leibniz)} \end{aligned}} $$

Proof

Linearity. Direct from the formula: $(\lambda P + \mu Q)' = \sum_{k \geq 1} k (\lambda a_k + \mu b_k) X^{k-1} = \lambda P' + \mu Q'$.
Product rule. It is enough to prove $(PQ)' = P' Q + P Q'$ for monomials $P = X^p$ and $Q = X^q$, then extend by bilinearity (the formula is bilinear in $(P, Q)$). If $p = 0$ or $q = 0$, one factor is a constant $c$, whose derivative is $0$: say $P = c$, then $(PQ)' = c Q' = P Q' = P' Q + P Q'$ since $P' = 0$; the formula holds trivially (symmetrically if $q = 0$). For $p, q \geq 1$, $$ (X^p X^q)' = (X^{p + q})' = (p + q) X^{p + q - 1}, $$ while $$ (X^p)' X^q + X^p (X^q)' = p X^{p - 1} X^q + q X^p X^{q - 1} = (p + q) X^{p + q - 1}. $$ The two sides agree. By bilinearity, $(PQ)' = P' Q + P Q'$ holds for all $P, Q \in \mathbb{K}[X]$.
Leibniz. By induction on $n$. The case $n = 0$ is $(PQ)^{(0)} = PQ = \binom{0}{0} P Q$. Assume the formula holds for some $n$. Differentiate: $$ (PQ)^{(n+1)} = \big((PQ)^{(n)}\big)' = \sum_{k=0}^{n} \binom{n}{k} \big( P^{(k+1)} Q^{(n-k)} + P^{(k)} Q^{(n-k+1)} \big). $$ Re-index and use Pascal's identity $\binom{n}{k-1} + \binom{n}{k} = \binom{n+1}{k}$: $$ (PQ)^{(n+1)} = \sum_{k=0}^{n+1} \binom{n+1}{k} P^{(k)} Q^{(n+1-k)}. $$

Theorem — Polynomial Taylor formula

For $P \in \mathbb{K}[X]$ and $a \in \mathbb{K}$: $$ P(X) = \sum_{k \geq 0} \frac{P^{(k)}(a)}{k!} (X - a)^k, $$ where the sum is finite (the terms vanish for $k > \deg P$).

Proof

Set $Q(Y) = P(a + Y) \in \mathbb{K}[Y]$. The substitution $X \mapsto a + Y$ is a polynomial composition, so $Q$ is a polynomial in $Y$ of degree $\leq \deg P$. Write $$ Q(Y) = \sum_{k=0}^{\deg P} b_k Y^k. $$ Differentiate $Q$ formally $k$ times with respect to $Y$ and evaluate at $0$: $$ Q^{(k)}(0) = k! \, b_k, \quad \text{hence} \quad b_k = \frac{Q^{(k)}(0)}{k!}. $$ It remains to justify that $Q^{(k)}(0) = P^{(k)}(a)$. By linearity, it is enough to check this for $P(X) = X^m$. Then $Q(Y) = (a + Y)^m = \sum_{\ell = 0}^{m} \binom{m}{\ell} a^{m - \ell} Y^\ell$, so $Q^{(k)}(0) = k! \, b_k$ where $b_k$ is the coefficient of $Y^k$ in $Q$. If $k \leq m$, $$ Q^{(k)}(0) = k! \binom{m}{k} a^{m - k} = m (m - 1) \cdots (m - k + 1) \, a^{m - k} = P^{(k)}(a), $$ where the last equality uses $P^{(k)}(X) = m (m-1) \cdots (m-k+1) X^{m-k}$ and evaluation at $a$. If $k > m$, both sides vanish: $Q$ has degree $m$ so its coefficient of $Y^k$ is $0$, giving $Q^{(k)}(0) = 0$; likewise $P^{(k)} = 0$ since $\deg P = m < k$, so $P^{(k)}(a) = 0$ (consistently, $\binom{m}{k} = 0$ for $k > m$). In both cases $Q^{(k)}(0) = P^{(k)}(a)$. By linearity, the identity $Q^{(k)}(0) = P^{(k)}(a)$ extends to every polynomial $P$. Therefore $b_k = P^{(k)}(a)/k!$, and $$ P(a + Y) = Q(Y) = \sum_{k \geq 0} \frac{P^{(k)}(a)}{k!} Y^k. $$ Substitute $Y = X - a$: $$ P(X) = \sum_{k \geq 0} \frac{P^{(k)}(a)}{k!} (X - a)^k. $$

Method — Taylor expansion of a polynomial at a point

To expand $P$ as a sum of the powers $(X - a)^k$ at a chosen $a \in \mathbb{K}$, compute the successive derivatives $P, P', P'', \dots$, evaluate each at $a$, and divide by the corresponding factorial: $$ P(X) = P(a) + P'(a)(X - a) + \frac{P''(a)}{2!}(X - a)^2 + \frac{P^{(3)}(a)}{3!}(X - a)^3 + \dots $$ The expansion stops at $k = \deg P$ (higher derivatives vanish).

Example

Compute the Taylor expansion of $P = X^3 + 2 X^2 - X + 1$ at $a = 1$.

Answer

Compute successive derivatives and evaluate at $1$: $$ \begin{aligned} P(1) &= 1 + 2 - 1 + 1 = 3, \\ P'(X) &= 3 X^2 + 4 X - 1, & P'(1) &= 3 + 4 - 1 = 6, \\ P''(X) &= 6 X + 4, & P''(1) &= 10, \\ P^{(3)}(X) &= 6, & P^{(3)}(1) &= 6. \end{aligned} $$ The Taylor expansion is $$ P(X) = 3 + 6 (X - 1) + \frac{10}{2!}(X - 1)^2 + \frac{6}{3!}(X - 1)^3 = 3 + 6(X - 1) + 5(X - 1)^2 + (X - 1)^3. $$

Theorem — Multiplicity by successive derivatives

Let $P \in \mathbb{K}[X]$ with $P \neq 0$, $a \in \mathbb{K}$, and $m \in \mathbb{N}^*$. Then $a$ is a root of multiplicity exactly $m$ of $P$ if and only if $$ P(a) = P'(a) = \dots = P^{(m-1)}(a) = 0 \quad \text{and} \quad P^{(m)}(a) \neq 0. $$

Proof

By the polynomial Taylor formula, $$ P(X) = \sum_{k = 0}^{\deg P} \frac{P^{(k)}(a)}{k!} (X - a)^k. $$ Let $j$ be the smallest index such that $P^{(j)}(a) \neq 0$ (this exists since $P \neq 0$ and $P^{(\deg P)}(a) = (\deg P)! \cdot \text{dom. coeff} \neq 0$). Then $$ P(X) = (X - a)^j \cdot \underbrace{\sum_{k \geq 0} \frac{P^{(j+k)}(a)}{(j+k)!} (X - a)^k}_{=: Q(X)}, $$ with $Q(a) = P^{(j)}(a)/j! \neq 0$.
Hence $(X - a)^j \mid P$ but $(X - a)^{j + 1}$ does not divide $P$ (otherwise $(X - a)^{j+1} \mid P$ would force $(X - a) \mid Q$, i.e. $Q(a) = 0$, contradicting $Q(a) \neq 0$). By the definition of multiplicity, $j$ is the multiplicity of $a$.
Therefore: $a$ has multiplicity exactly $m$ $\iff$ $j = m$ $\iff$ $P^{(0)}(a) = \dots = P^{(m-1)}(a) = 0$ and $P^{(m)}(a) \neq 0$.

Skills to practice

Computing derivatives and Leibniz formula
Using Taylor formula and multiplicity test

V Factorization in $\mathbb{C}\lbrack X\rbrack$ and $\mathbb{R}\lbrack X\rbrack$

The chapter culminates with the factorization theorems. The d'Alembert--Gauss theorem (the fundamental theorem of algebra) asserts that every non-constant polynomial of $\mathbb{C}[X]$ has a complex root --- proof hors programme, admis. From this single fact follows: every non-constant polynomial of $\mathbb{C}[X]$ factors completely into linear terms; the irreducibles of $\mathbb{C}[X]$ are exactly the degree-1 polynomials; the irreducibles of $\mathbb{R}[X]$ are degree-1 plus degree-2-with-$\Delta < 0$; every $P \in \mathbb{R}[X]$ factors as a product of these. The factorization of $X^n - 1$ via the $n$-th roots of unity is the canonical worked example and bridges to Complex numbers.

Theorem — d'Alembert--Gauss

Every non-constant polynomial $P \in \mathbb{C}[X]$ has at least one root in $\mathbb{C}$.

Admis

The proof is hors programme --- we admit it. The classical proofs use either complex analysis (Liouville's theorem on bounded entire functions) or topology (a degree argument on $|P|$, which attains its minimum on a large enough disk). Here, we use the theorem as a black box.

Definition — Irreducible polynomial

Let $P \in \mathbb{K}[X]$. We say that $P$ is irreducible (in $\mathbb{K}[X]$) if $\deg P \geq 1$ and the only divisors of $P$ in $\mathbb{K}[X]$ are the non-zero constants and the polynomials associated to $P$ (i.e. $\lambda P$ with $\lambda \in \mathbb{K}^*$).

Proposition — Irreducibles of $\mathbb{C}\lbrack X\rbrack$

The irreducible polynomials of $\mathbb{C}[X]$ are exactly the polynomials of degree $1$, i.e.\ the $a X + b$ with $a \in \mathbb{C}^*$ and $b \in \mathbb{C}$.

Proof

Degree-1 polynomials are irreducible. A divisor $D$ of $a X + b$ has $\deg D \in \{0, 1\}$. If $\deg D = 0$, $D$ is a non-zero constant. If $\deg D = 1$, then $a X + b = D \cdot Q$ with $\deg Q = 0$, so $Q$ is a non-zero constant $\mu$ and $a X + b = \mu D$, i.e. $D$ is associated to $a X + b$.
Degree-$\geq 2$ polynomials are reducible. Let $P \in \mathbb{C}[X]$ with $\deg P \geq 2$. By d'Alembert--Gauss, $P$ has a root $z \in \mathbb{C}$, so $(X - z) \mid P$ (the « Root $\iff$ linear factor » Proposition above). The divisor $X - z$ has $\deg(X - z) = 1$, which is neither $0$ (not a constant) nor $\deg P \geq 2$ (so $X - z$ is not associated to $P$). Hence $P$ admits a non-trivial divisor and is not irreducible.

Theorem — Factorization in $\mathbb{C}\lbrack X\rbrack$

Every non-constant $P \in \mathbb{C}[X]$ of degree $n \geq 1$ is split over $\mathbb{C}$ and factors uniquely (up to permutation of the roots) as $$ P = a_n \prod_{i=1}^{r} (X - z_i)^{m_i}, $$ where $a_n$ is the dominant coefficient of $P$, $z_1, \dots, z_r \in \mathbb{C}$ are the distinct roots of $P$, and $m_1, \dots, m_r \geq 1$ are their multiplicities, with $\sum_{i=1}^{r} m_i = n$.

Proof

Existence by induction on $n = \deg P$.
Base $n = 1$. $P = a_1 X + a_0 = a_1 (X - (-a_0/a_1))$ --- already in the announced form with $r = 1$, $z_1 = -a_0/a_1$, $m_1 = 1$.
Inductive step. Let $\deg P = n \geq 2$. By d'Alembert--Gauss, $P$ has a root $z \in \mathbb{C}$, hence $(X - z) \mid P$, so $P = (X - z) \cdot P_1$ with $\deg P_1 = n - 1 \geq 1$. The dominant coefficient of $P_1$ equals that of $P$ (since $X - z$ is monic). By induction, $P_1 = a_n \prod_{j=1}^{s} (X - w_j)^{n_j}$ with $\sum n_j = n - 1$. Therefore $P = a_n (X - z) \prod_{j=1}^{s} (X - w_j)^{n_j}$. Group the factor $(X - z)$ with any equal $(X - w_j)$ to get the announced form with $\sum m_i = n$.
Uniqueness. The set of distinct roots and the multiplicity of each root are intrinsic to $P$ (defined directly from $P$, not from a particular factorization). Hence two factorizations $P = a_n \prod (X - z_i)^{m_i} = b_n \prod (X - w_j)^{n_j}$ have the same dominant coefficient ($a_n = b_n$, both equal to the leading coefficient of $P$), the same set of distinct roots ($\{z_i\} = \{w_j\}$), and the same multiplicities. The factorizations agree up to permutation.

Method — Factor $X^n - 1$ and $X^n + 1$ in $\mathbb{C}\lbrack X\rbrack$

For $n \in \mathbb{N}^*$:

Roots of $X^n - 1$: the $n$-th roots of unity $\omega_k = e^{2 i k \pi / n}$ for $k \in \{0, 1, \dots, n - 1\}$. Hence $$ X^n - 1 = \prod_{k=0}^{n-1} (X - \omega_k). $$
Roots of $X^n + 1$: the $n$-th roots of $-1$, $\zeta_k = e^{i (2 k + 1) \pi / n}$ for $k \in \{0, 1, \dots, n - 1\}$. Hence $$ X^n + 1 = \prod_{k=0}^{n-1} (X - \zeta_k). $$

For the corresponding real factorizations in $\mathbb{R}[X]$, pair each non-real root with its conjugate: $(X - \omega)(X - \overline{\omega}) = X^2 - 2 \mathrm{Re}(\omega) X + |\omega|^2 \in \mathbb{R}[X]$.

Example

Factor $X^4 - 1$ in $\mathbb{C}[X]$, then in $\mathbb{R}[X]$.

Answer

The four 4th roots of unity are $\omega_k = e^{2 i k \pi / 4} = e^{i k \pi / 2}$ for $k = 0, 1, 2, 3$, i.e.\ $1, i, -1, -i$. So $$ X^4 - 1 = (X - 1)(X - i)(X + 1)(X + i) \quad \text{in } \mathbb{C}[X]. $$ For the real factorization, pair the conjugate roots $i$ and $-i$: $$ (X - i)(X + i) = X^2 - i^2 = X^2 + 1. $$ Hence $$ X^4 - 1 = (X - 1)(X + 1)(X^2 + 1) \quad \text{in } \mathbb{R}[X]. $$

Proposition — Conjugate roots come in pairs

Let $P \in \mathbb{R}[X]$ (real coefficients) and $z \in \mathbb{C}$. Then $P(\overline{z}) = \overline{P(z)}$. In particular, if $z$ is a root of $P$, so is $\overline{z}$, and they have the same multiplicity.

Proof

Write $P = \sum a_k X^k$ with $a_k \in \mathbb{R}$, hence $\overline{a_k} = a_k$. Conjugation $\bar{\cdot}$ is a ring morphism on $\mathbb{C}$: $$ P(\overline{z}) = \sum a_k \overline{z}^k = \sum \overline{a_k} \cdot \overline{z^k} = \overline{\sum a_k z^k} = \overline{P(z)}. $$ For multiplicity: the polynomial $P^{(j)}$ also has real coefficients (each derivation preserves « real-coefficient »), so by the same argument, $P^{(j)}(\overline{z}) = \overline{P^{(j)}(z)}$ for every $j \in \mathbb{N}$. In particular, $$ P^{(j)}(z) = 0 \iff \overline{P^{(j)}(z)} = 0 \iff P^{(j)}(\overline{z}) = 0. $$ By the Theorem of multiplicity by successive derivatives above, the multiplicity of $z$ and the multiplicity of $\overline{z}$ are determined by the same vanishing conditions, hence are equal.

Proposition — Irreducibles of $\mathbb{R}\lbrack X\rbrack$

The irreducible polynomials of $\mathbb{R}[X]$ are exactly:

the degree-1 polynomials $a X + b$ with $a \in \mathbb{R}^*$;
the degree-2 polynomials $a X^2 + b X + c$ with $a \in \mathbb{R}^*$ and $\Delta = b^2 - 4 a c < 0$.

Proof

Degree-1 is irreducible. Same argument as for $\mathbb{C}[X]$ (the field plays no role here).
Degree-2 with $\Delta < 0$ is irreducible. Let $P = a X^2 + b X + c$ with $\Delta < 0$. A non-trivial divisor $D$ would have $\deg D = 1$, hence $D = \alpha (X - r)$ with $r \in \mathbb{R}$, forcing $r$ to be a real root of $P$. But $\Delta < 0$ means $P$ has no real root --- contradiction.
Degree-2 with $\Delta \geq 0$ is reducible. If $\Delta \geq 0$, the quadratic $P$ has at least one real root $r$. Hence $(X - r) \mid P$. Since $\deg P = 2$ and $\deg(X - r) = 1$, this gives a non-trivial factorization in $\mathbb{R}[X]$, so $P$ is not irreducible.
Degree $\geq 3$ is reducible. Let $P \in \mathbb{R}[X]$ with $\deg P \geq 3$. By d'Alembert--Gauss applied to $P$ viewed as an element of $\mathbb{C}[X]$, $P$ has a complex root $z$. Case $z \in \mathbb{R}$. Then $(X - z) \mid P$ in $\mathbb{R}[X]$ (the Euclidean division of $P$ by $X - z$ in $\mathbb{R}[X]$ has remainder $P(z) = 0$). Since $\deg(X - z) = 1 < \deg P \geq 3$, $X - z$ is a non-trivial divisor. Case $z \notin \mathbb{R}$. Set $Q = (X - z)(X - \overline{z}) = X^2 - 2 \mathrm{Re}(z) X + |z|^2 \in \mathbb{R}[X]$ (degree $2$). We show $Q \mid P$ in $\mathbb{R}[X]$. Perform the Euclidean division of $P$ by $Q$ in $\mathbb{R}[X]$: $P = A Q + R$ with $A, R \in \mathbb{R}[X]$ and $\deg R < 2$. Evaluating in $\mathbb{C}$ at $z$: $0 = P(z) = A(z) Q(z) + R(z) = R(z)$. Hence $z$ is a root of $R$. But $R$ has real coefficients and $\deg R \leq 1$ --- a polynomial of $\mathbb{R}[X]$ of degree $\leq 1$ that has a non-real root must be the zero polynomial. Hence $R = 0$, i.e. $Q \mid P$. Since $\deg Q = 2 < \deg P$, $Q$ is a non-trivial divisor.

Conclusion. An irreducible polynomial of $\mathbb{R}[X]$ has degree $\leq 2$ (and degree $\geq 1$ by definition). Among degree-1 it is always irreducible; among degree-2 only those with $\Delta < 0$.

Theorem — Factorization in $\mathbb{R}\lbrack X\rbrack$

Every $P \in \mathbb{R}[X]$ of degree $n \geq 1$ factors uniquely (up to order) as $$ P = a_n \prod_{i} (X - x_i)^{m_i} \prod_{j} (X^2 + b_j X + c_j)^{n_j}, $$ where $a_n$ is the dominant coefficient, the $x_i \in \mathbb{R}$ are the distinct real roots of $P$ with multiplicities $m_i$, and the $X^2 + b_j X + c_j$ are the distinct irreducible-quadratic factors (i.e.\ $b_j^2 - 4 c_j < 0$) with multiplicities $n_j$, satisfying $$ \sum_i m_i + 2 \sum_j n_j = n. $$

Proof

View $P$ as an element of $\mathbb{C}[X]$ and apply the $\mathbb{C}$-factorization theorem: $$ P = a_n \prod_{k} (X - z_k)^{\mu_k} $$ where the $z_k$ are the distinct complex roots of $P$ with multiplicities $\mu_k$.
Group the roots: real roots stay as linear factors $(X - x_i)^{m_i}$. By the conjugate-pair Proposition, non-real roots come in conjugate pairs $z, \overline{z}$ with the same multiplicity; group each such pair into the real quadratic $$ (X - z)^{\mu}(X - \overline{z})^{\mu} = \big(X^2 - 2 \mathrm{Re}(z) X + |z|^2\big)^{\mu}. $$ The result is the announced real factorization. The dominant coefficient $a_n$ is preserved.
The degree count $\sum m_i + 2 \sum n_j = n$ follows by reading degrees: each real linear factor contributes $1$, each real quadratic contributes $2$.
Uniqueness comes from the uniqueness in $\mathbb{C}[X]$ (the distinct roots and their multiplicities are intrinsic) combined with the fact that two complex conjugate roots determine a unique real quadratic.

Example

Factor $X^4 + 1$ in $\mathbb{R}[X]$.

Answer

The four roots of $X^4 + 1$ are the 4-th roots of $-1$: $\zeta_k = e^{i (2k + 1) \pi / 4}$ for $k = 0, 1, 2, 3$. Explicitly: $$ \zeta_0 = e^{i \pi / 4}, \quad \zeta_1 = e^{i 3 \pi / 4}, \quad \zeta_2 = e^{i 5 \pi / 4} = \overline{\zeta_1}, \quad \zeta_3 = e^{i 7 \pi / 4} = \overline{\zeta_0}. $$ None is real. Pair conjugates: $$ (X - \zeta_0)(X - \overline{\zeta_0}) = X^2 - 2 \mathrm{Re}(\zeta_0) X + |\zeta_0|^2 = X^2 - 2 \cos(\pi/4) X + 1 = X^2 - \sqrt{2} X + 1, $$ $$ (X - \zeta_1)(X - \overline{\zeta_1}) = X^2 - 2 \cos(3 \pi / 4) X + 1 = X^2 + \sqrt{2} X + 1. $$ Therefore $$ X^4 + 1 = (X^2 - \sqrt{2} X + 1)(X^2 + \sqrt{2} X + 1) \quad \text{in } \mathbb{R}[X]. $$ Both quadratic factors have $\Delta = 2 - 4 = -2 < 0$, confirming irreducibility.

Skills to practice

Factoring $X^n - 1$ and $X^n + 1$
Using conjugate roots and identifying irreducibles

VI Lagrange interpolation

Given $n$ data points $(x_1, y_1), \dots, (x_n, y_n)$ with pairwise distinct abscissas $x_i$, can we find a polynomial passing through all of them? Yes, and it is unique once we fix the maximum allowed degree to $n - 1$. The proof produces the explicit polynomial via the Lagrange polynomials $L_i$. This gives a third way of specifying a polynomial --- by $n$ values at $n$ distinct abscissas, in addition to the coefficient and root viewpoints used earlier in the chapter.

Definition — Lagrange polynomials

Let $x_1, \dots, x_n \in \mathbb{K}$ be pairwise distinct ($n \geq 1$). For each $i \in \{1, \dots, n\}$, the $i$-th Lagrange polynomial is $$ L_i(X) = \prod_{\substack{j = 1 \\ j \neq i}}^{n} \frac{X - x_j}{x_i - x_j}. $$ Each $L_i \in \mathbb{K}_{n - 1}[X]$, and the $L_i$ satisfy the Kronecker property: $$ L_i(x_j) = \delta_{i j} = \begin{cases} 1 & \text{if } j = i, \\ 0 & \text{if } j \neq i. \end{cases} $$

Theorem — Lagrange interpolation

Let $x_1, \dots, x_n \in \mathbb{K}$ be pairwise distinct and $y_1, \dots, y_n \in \mathbb{K}$ arbitrary. There exists a unique polynomial $P \in \mathbb{K}_{n - 1}[X]$ such that $$ P(x_i) = y_i \quad \text{for every } i \in \{1, \dots, n\}. $$ It is given explicitly by $$ P(X) = \sum_{i = 1}^{n} y_i \, L_i(X). $$

Proof

Existence. Set $P = \sum_{i=1}^{n} y_i L_i$. Each $L_i$ has degree $\leq n - 1$, so $P \in \mathbb{K}_{n-1}[X]$. By the Kronecker property, for $j \in \{1, \dots, n\}$: $$ P(x_j) = \sum_{i=1}^{n} y_i L_i(x_j) = \sum_{i=1}^{n} y_i \delta_{ij} = y_j. $$ Uniqueness. Suppose $P_1, P_2 \in \mathbb{K}_{n-1}[X]$ both satisfy $P_k(x_i) = y_i$ for every $i$. The difference $D = P_1 - P_2 \in \mathbb{K}_{n-1}[X]$ satisfies $D(x_i) = 0$ for every $i$, i.e.\ $D$ has at least $n$ distinct roots. By the « Number of roots vs degree » Proposition above, if $D \neq 0$ then $D$ has at most $\deg D \leq n - 1 < n$ roots --- contradiction. Hence $D = 0$, i.e.\ $P_1 = P_2$.

Method — Computing the Lagrange interpolation polynomial

Given data $(x_1, y_1), \dots, (x_n, y_n)$ with the $x_i$ pairwise distinct:

Form the $n$ Lagrange polynomials $L_1, \dots, L_n$, each a product of $n - 1$ linear factors normalized so that $L_i(x_i) = 1$.
The interpolation polynomial is the linear combination $P = y_1 L_1 + y_2 L_2 + \dots + y_n L_n$.

Always check the answer by evaluating $P$ at each $x_i$ --- a quick way to detect arithmetic errors.

Example

Compute the Lagrange polynomial $P \in \mathbb{R}_2[X]$ passing through $(0, 1), (1, 0), (2, 1)$.

Answer

With $n = 3$ and $(x_1, x_2, x_3) = (0, 1, 2)$, $(y_1, y_2, y_3) = (1, 0, 1)$, compute: $$ \begin{aligned} L_1(X) &= \frac{(X - 1)(X - 2)}{(0 - 1)(0 - 2)} = \frac{(X-1)(X-2)}{2}, \\ L_2(X) &= \frac{(X - 0)(X - 2)}{(1 - 0)(1 - 2)} = -X(X - 2), \\ L_3(X) &= \frac{(X - 0)(X - 1)}{(2 - 0)(2 - 1)} = \frac{X(X - 1)}{2}. \end{aligned} $$ The interpolation polynomial is $$ P = 1 \cdot L_1 + 0 \cdot L_2 + 1 \cdot L_3 = \frac{(X-1)(X-2)}{2} + \frac{X(X-1)}{2} = \frac{(X-1)\big[(X-2) + X\big]}{2} = \frac{(X-1)(2X-2)}{2} = (X - 1)^2. $$ Verification: $P(0) = 1$, $P(1) = 0$, $P(2) = 1$. $\checkmark$

Proposition — Description of all $Q$ with $Q(x_i) \equal y_i$

Let $x_1, \dots, x_n \in \mathbb{K}$ be pairwise distinct, $y_1, \dots, y_n \in \mathbb{K}$, and $P$ the Lagrange interpolation polynomial in $\mathbb{K}_{n-1}[X]$. The polynomials $Q \in \mathbb{K}[X]$ such that $Q(x_i) = y_i$ for every $i$ are exactly the polynomials of the form $$ Q = P + \prod_{i=1}^{n} (X - x_i) \cdot S, \quad S \in \mathbb{K}[X]. $$

Proof

$(\Leftarrow)$ For $Q = P + \prod_{i} (X - x_i) \cdot S$ and any $j \in \{1, \dots, n\}$: $$ Q(x_j) = P(x_j) + \prod_{i} (x_j - x_i) \cdot S(x_j) = y_j + 0 \cdot S(x_j) = y_j. $$ $(\Rightarrow)$ Suppose $Q(x_i) = y_i$ for every $i$. If $Q - P = 0$, then $Q = P$, which is of the announced form with $S = 0$. Otherwise $Q - P \neq 0$: the difference $Q - P$ satisfies $(Q - P)(x_i) = 0$ for every $i$, i.e.\ $Q - P$ has $n$ pairwise distinct roots. By the « Number of roots vs degree » Proposition above (applied to the non-zero polynomial $Q - P$ with the $r = n$ distinct roots $x_1, \dots, x_n$, each of multiplicity $m_i \geq 1$), we get $$ \prod_{i=1}^{n} (X - x_i)^{m_i} \mid (Q - P), \quad \text{a fortiori} \quad \prod_{i=1}^{n} (X - x_i) \mid (Q - P). $$ Hence $Q - P = \prod_i (X - x_i) \cdot S$ for some $S \in \mathbb{K}[X]$.

Skills to practice

Computing the Lagrange interpolation polynomial
Using Lagrange in proofs

VII Partial-fraction decomposition

A rational function is a quotient $R = A / B$ of two polynomials of $\mathbb{K}[X]$ with $B \neq 0$; the set of all such quotients is written $\mathbb{K}(X)$ (so $\mathbb{R}(X)$ over $\mathbb{R}$ and $\mathbb{C}(X)$ over $\mathbb{C}$). We always assume the quotient written in irreducible form, i.e.\ $A$ and $B$ share no common complex root --- which we take here as the working definition (it is the same as having no common factor, a fact we admit at this level). For $R \neq 0$, the degree of $R = A / B$ is set by the convention $\deg R = \deg A - \deg B$ (and $\deg 0 = -\infty$). A pole of $R$ is a root of the denominator $B$, and its order is its multiplicity as a root of $B$. By the Euclidean division of $A$ by $B$ (Section « Divisibility and Euclidean division »), one writes $A = E B + A_1$ with $\deg A_1 < \deg B$, so $R = E + A_1 / B$ with $E \in \mathbb{K}[X]$; this $E$ is the polynomial part (French: partie entière) of $R$, and $A_1 / B$ is a proper fraction. A fraction $A / B$ is proper exactly when $\deg A < \deg B$, equivalently $\deg R \leq -1$ (no polynomial part). The goal of partial-fraction decomposition is to write the proper part $A_1 / B$ as a sum of simple pieces $a / (X - \lambda)^j$ (and, over $\mathbb{R}$, of pieces $(\alpha X + \beta) / (X^2 + p X + q)^j$), one family per pole. This turns a single quotient into a sum that is easy to integrate, differentiate, or sum --- which is exactly why the tool is introduced here. We treat only simple situations: the existence and uniqueness theorem is admitted, and when a pole is multiple or a denominator carries an irreducible degree-$2$ factor, the form of the decomposition is given to us and only the coefficients are computed.

Suppose $R \in \mathbb{C}(X)$ has poles $\lambda_1, \ldots, \lambda_r$ of respective orders $m_1, \ldots, m_r$. The strategy of partial fractions is to split $R$ into a polynomial part $E$ plus a sum of simple polar pieces $a / (X - \lambda_i)^j$, one per pole and per order of vanishing. The existence and uniqueness of this decomposition is a theorem (admis: the proof is hors programme); two computational tools are required --- the formula $a = A(\lambda) / B'(\lambda)$ for the coefficient at a simple pole, and the decomposition of $P' / P$ when $P$ splits.

Definition — Simple element over $\mathbb{C}$

A simple element (French: élément simple) of $\mathbb{C}(X)$ is a rational fraction of the form $$ \frac{a}{(X - \lambda)^k}, $$ with $a \in \mathbb{C}$, $\lambda \in \mathbb{C}$, $k \in \mathbb{N}^*$.

Theorem — Decomposition over $\mathbb{C}$

Let $R \in \mathbb{C}(X) \setminus \{0\}$, written in irreducible form $R = A / B$, with $B = c \prod_{i = 1}^{r} (X - \lambda_i)^{m_i}$ ($c \in \mathbb{C}^*$, the $\lambda_i$ pairwise distinct). Then $R$ admits a unique decomposition $$ \textcolor{colorprop}{R = E + \sum_{i = 1}^{r} \sum_{j = 1}^{m_i} \frac{a_{i, j}}{(X - \lambda_i)^j},} $$ with $E \in \mathbb{C}[X]$ the partie entière of $R$ and $a_{i, j} \in \mathbb{C}$. The sub-sum $\displaystyle\sum_{j = 1}^{m_i} \frac{a_{i, j}}{(X - \lambda_i)^j}$ is called the partie polaire of $R$ at $\lambda_i$. The zero fraction has the trivial decomposition $0 = 0$.

Proof

Admitted (hors programme).

Proposition — Coefficient at a simple pole

Let $R = A / B \in \mathbb{C}(X)$ in irreducible form, and let $\lambda \in \mathbb{C}$ be a simple pole of $R$ (i.e. a simple root of $B$, with $A(\lambda) \neq 0$). Write $B = (X - \lambda) B_1$ with $B_1(\lambda) \neq 0$. Then the coefficient of $\dfrac{1}{X - \lambda}$ in the partial-fraction decomposition of $R$ is $$ \textcolor{colorprop}{a = \frac{A(\lambda)}{B'(\lambda)} = \frac{A(\lambda)}{B_1(\lambda)}.} $$

Proof

Write the decomposition with partie entière $E$ and partial parts at every pole. Isolate the pole at $\lambda$: $$ R(X) = \frac{a}{X - \lambda} + S(X), $$ where $S$ regroups all the other terms of the decomposition --- $S$ is regular at $\lambda$, i.e. $S(\lambda) \in \mathbb{C}$ is finite.
Multiply both sides by $(X - \lambda)$ and simplify the simple-pole term: $$ \begin{aligned} (X - \lambda) R(X) &= (X - \lambda) \cdot \frac{a}{X - \lambda} + (X - \lambda) S(X) && \text{(distribute)} \\ &= a + (X - \lambda) S(X) && \text{(simplify the simple-pole factor).} \end{aligned} $$ Now take the limit $X \to \lambda$ in $\mathbb{C}$. On the right side, $S$ regular at $\lambda$ gives $(X - \lambda) S(X) \to 0$, hence the right side $\to a$. On the left side, using $R = A / B$ in irreducible form and $B = (X - \lambda) B_1$: $$ \begin{aligned} (X - \lambda) R(X) &= (X - \lambda) \cdot \frac{A(X)}{(X - \lambda) B_1(X)} && \text{(substitute } R\text{)} \\ &= \frac{A(X)}{B_1(X)} && \text{(simplify } X - \lambda\text{)} \\ &\xrightarrow[X \to \lambda]{} \frac{A(\lambda)}{B_1(\lambda)} && \text{(} A, B_1 \text{ continuous, } B_1(\lambda) \neq 0\text{).} \end{aligned} $$ Identifying the two limits yields $a = A(\lambda) / B_1(\lambda)$.
For the alternate form $a = A(\lambda) / B'(\lambda)$: differentiate $B = (X - \lambda) B_1$ via the product rule: $$ B'(X) = B_1(X) + (X - \lambda) B_1'(X), \quad \text{so} \quad B'(\lambda) = B_1(\lambda). $$ Substituting gives $a = A(\lambda) / B'(\lambda)$.

Method — Coefficient at a simple pole --- the $A / B'$ formula

At a simple pole $\lambda$ of $R = A / B$ in irreducible form, the coefficient of $1 / (X - \lambda)$ is $$ a = \frac{A(\lambda)}{B'(\lambda)}. $$ This formula is preferred when $B$ is given as an explicit polynomial whose derivative is easy to evaluate. The equivalent form $a = A(\lambda) / B_1(\lambda)$ (with $B = (X - \lambda) B_1$) is preferred when $B$ is already factored.

Method — Coefficient at a double pole

Let $\lambda$ be a pole of order $2$ of $R = A / B$, i.e. $B = (X - \lambda)^2 B_2$ with $B_2(\lambda) \neq 0$ and $A(\lambda) \neq 0$. The partie polaire at $\lambda$ has the form $$ \frac{a_2}{(X - \lambda)^2} + \frac{a_1}{X - \lambda}. $$

Multiply $R$ by $(X - \lambda)^2$: $$ (X - \lambda)^2 R(X) = a_2 + a_1 (X - \lambda) + (X - \lambda)^2 \cdot (\text{regular at } \lambda). $$ Evaluate at $X = \lambda$: the right side reduces to $a_2$, and the left side equals $A(\lambda) / B_2(\lambda)$. So $a_2 = A(\lambda) / B_2(\lambda)$.
Recover $a_1$: subtract the known $a_2 / (X - \lambda)^2$ term from $R$, and use any cheap method --- evaluation at one or two convenient values of $X$, identification of coefficients of like powers, or comparison of limits at infinity.

Program note. For $m \geq 3$ poles, the present cours does not give a general derivative-based closed formula --- the program excludes technical examples (« toute technicité dans les exemples est exclue ») and the calculation is done case-by-case in the exo file.

Method — Constraint from the limit at infinity

Suppose $R \in \mathbb{C}(X)$ has $\deg R \leq -1$ (no polynomial part, the fraction is proper) and decomposes as $$ R = \sum_i \frac{a_{i, 1}}{X - \lambda_i} + (\text{terms involving higher powers } 1 / (X - \lambda_i)^j \text{ with } j \geq 2). $$ Note that $a_{i, 1}$ is the coefficient of $1 / (X - \lambda_i)$ for every pole $\lambda_i$, regardless of its order. Then $$ X R(X) \xrightarrow[X \to \infty]{} \sum_i a_{i, 1}. $$ This gives a single linear constraint among the $a_{i, 1}$ coefficients. It is useful as a consistency check after the other coefficients have been computed, or as a linear constraint that, combined with already-known coefficients, can save one substitution step --- when only one $a_{i, 1}$ remains unknown, the constraint suffices to extract it (cf. the calculation of $a_1$ in the next worked example). Used alone, it gives only the sum $\sum_i a_{i, 1}$, never an individual coefficient.

Example

Decompose $R = \dfrac{1}{X^2 - 1}$ as a sum of simple elements over $\mathbb{C}$.

Answer

The denominator factors over $\mathbb{C}$ as $X^2 - 1 = (X - 1)(X + 1)$ --- two simple poles, $1$ and $-1$. The fraction is irreducible (the numerator is the constant $1$). The decomposition has the form $$ R = \frac{a}{X - 1} + \frac{b}{X + 1}. $$ Coefficient at $\lambda = 1$. Apply the $A / B'$ formula with $A = 1$ and $B = X^2 - 1$, so $B' = 2 X$: $$ a = \frac{A(1)}{B'(1)} = \frac{1}{2}. $$ Coefficient at $\lambda = -1$. Same formula: $$ b = \frac{A(-1)}{B'(-1)} = \frac{1}{-2} = -\frac{1}{2}. $$ Conclusion. $$ \frac{1}{X^2 - 1} = \frac{1 / 2}{X - 1} - \frac{1 / 2}{X + 1}. $$ Quick check. The right side equals $\frac{1}{2} \cdot \dfrac{(X + 1) - (X - 1)}{(X - 1)(X + 1)} = \frac{1}{2} \cdot \dfrac{2}{X^2 - 1} = \dfrac{1}{X^2 - 1}$.

Example

Decompose $R = \dfrac{X^2 + 3 X + 1}{(X - 1)^2 (X - 2)}$ as a sum of simple elements over $\mathbb{C}$.

Answer

The denominator factors as $(X - 1)^2 (X - 2)$, with poles $1$ (order $2$) and $2$ (simple). The numerator $X^2 + 3 X + 1$ does not vanish at $1$ (gives $5$) or at $2$ (gives $11$), so the fraction is irreducible. The decomposition has the form $$ R = \frac{a_2}{(X - 1)^2} + \frac{a_1}{X - 1} + \frac{c}{X - 2}. $$ Coefficient $c$ (simple pole at $2$). Multiply $R$ by $(X - 2)$ and evaluate at $X = 2$ (or use $A / B'$): $$ c = \left[ (X - 2) R(X) \right]_{X = 2} = \left[ \frac{X^2 + 3 X + 1}{(X - 1)^2} \right]_{X = 2} = \frac{4 + 6 + 1}{1} = 11. $$ Coefficient $a_2$ (highest at the double pole). Multiply $R$ by $(X - 1)^2$ and evaluate at $X = 1$: $$ a_2 = \left[ (X - 1)^2 R(X) \right]_{X = 1} = \left[ \frac{X^2 + 3 X + 1}{X - 2} \right]_{X = 1} = \frac{1 + 3 + 1}{-1} = -5. $$ Coefficient $a_1$ (remaining). Use the limit-at-infinity constraint: $\deg R = 2 - 3 = -1$, so $\deg(X R) = 0$ and $X R(X)$ tends to a finite limit at infinity. Compute that limit directly: $$ \begin{aligned} X R(X) &= X \cdot \frac{X^2 + 3 X + 1}{(X - 1)^2 (X - 2)} = \frac{X^3 + 3 X^2 + X}{X^3 - 4 X^2 + 5 X - 2} && \text{(expand denominator)} \\ &\xrightarrow[X \to \infty]{} \frac{1}{1} = 1 && \text{(ratio of leading coefficients).} \end{aligned} $$ The limit-at-infinity constraint then gives $\lim_{X \to \infty} X R(X) = a_1 + c$ (sum of simple-pole coefficients). With $c = 11$ already known, only $a_1$ remains unknown, so the constraint extracts it: $$ a_1 + c = 1, \qquad a_1 = 1 - c = 1 - 11 = -10. $$ Conclusion. $$ \frac{X^2 + 3 X + 1}{(X - 1)^2 (X - 2)} = \frac{-5}{(X - 1)^2} + \frac{-10}{X - 1} + \frac{11}{X - 2}. $$

Proposition — Decomposition of $P' / P$

Let $P \in \mathbb{C}[X]$ be a non-constant polynomial that splits over $\mathbb{C}$: $$ P = c \prod_{i = 1}^{r} (X - a_i)^{m_i}, \qquad c \in \mathbb{C}^*, \ a_i \text{ pairwise distinct}, \ m_i \in \mathbb{N}^*. $$ Then $$ \textcolor{colorprop}{\frac{P'}{P} = \sum_{i = 1}^{r} \frac{m_i}{X - a_i}.} $$ This is a partial-fraction decomposition with simple poles only --- one per root of $P$, with coefficient equal to the multiplicity of that root.

Proof

Differentiate the product $P = c \prod_i (X - a_i)^{m_i}$ via the product rule: $$ P' = c \sum_{i = 1}^{r} m_i (X - a_i)^{m_i - 1} \prod_{j \neq i} (X - a_j)^{m_j}. $$ Divide both sides by $P = c \prod_i (X - a_i)^{m_i}$: $$ \begin{aligned} \frac{P'}{P} &= \sum_{i = 1}^{r} \frac{m_i (X - a_i)^{m_i - 1} \prod_{j \neq i} (X - a_j)^{m_j}}{\prod_{j} (X - a_j)^{m_j}} && \text{(divide each term)} \\ &= \sum_{i = 1}^{r} \frac{m_i}{X - a_i} && \text{(simplify factor by factor).} \end{aligned} $$ The simplification works because in the $i$-th term, the numerator's $(X - a_i)^{m_i - 1}$ matches the denominator's $(X - a_i)^{m_i}$, leaving $(X - a_i)^{-1}$, and the other factors $(X - a_j)^{m_j}$ for $j \neq i$ cancel exactly.
Remark. Formally, the identity is the « logarithmic derivative » of the factorization (one would write $\ln P = \ln c + \sum_i m_i \ln (X - a_i)$ and differentiate); we obtain it without invoking a formal logarithm.

Example

Apply the $P' / P$ formula to $P = (X - 1)^3 (X + 2)^2$.

Answer

$P$ has two roots: $1$ (multiplicity $3$) and $-2$ (multiplicity $2$). The proposition gives directly $$ \frac{P'}{P} = \frac{3}{X - 1} + \frac{2}{X + 2}. $$ This bypasses computing $P'$ explicitly. Sanity check: $P$ has degree $5$, so $P'$ has degree $4$, hence $P' / P$ has degree $-1$. On the right-hand side, $$ \frac{3}{X - 1} + \frac{2}{X + 2} \sim_{X \to \infty} \frac{5}{X}, $$ which also has degree $-1$ --- consistent.

Over $\mathbb{R}$, the irreducible polynomials are of two kinds: degree-$1$ factors $(X - \lambda)$, and irreducible degree-$2$ factors $(X^2 + p X + q)$ with discriminant $p^2 - 4 q < 0$ (recall from Polynomials). Accordingly, the partial-fraction decomposition over $\mathbb{R}$ uses two species of simple elements: 1st species $a / (X - \lambda)^k$ and 2nd species $(\alpha X + \beta) / (X^2 + p X + q)^k$. Three computational tools cover most cases: a direct method that writes the expected real form and identifies coefficients, the grouping of complex conjugate poles starting from the $\mathbb{C}$ decomposition, and the use of parity to halve the number of unknowns.

Definition — Simple element of the first species over $\mathbb{R}$

A simple element of the first species of $\mathbb{R}(X)$ is a rational fraction of the form $$ \frac{a}{(X - \lambda)^k}, \qquad a, \lambda \in \mathbb{R}, \ k \in \mathbb{N}^*. $$

Definition — Simple element of the second species over $\mathbb{R}$

A simple element of the second species of $\mathbb{R}(X)$ is a rational fraction of the form $$ \frac{\alpha X + \beta}{(X^2 + p X + q)^k}, \qquad \alpha, \beta, p, q \in \mathbb{R}, \ k \in \mathbb{N}^*, \ p^2 - 4 q < 0. $$ The numerator is a polynomial of degree at most $1$; the denominator is a power of an irreducible degree-$2$ polynomial of $\mathbb{R}[X]$.

Theorem — Decomposition over $\mathbb{R}$

Let $R \in \mathbb{R}(X) \setminus \{0\}$, written in irreducible form $R = A / B$, with denominator factored over $\mathbb{R}$: $$ B = c \prod_{i = 1}^{r} (X - \lambda_i)^{m_i} \prod_{j = 1}^{s} (X^2 + p_j X + q_j)^{n_j}, $$ where $c \in \mathbb{R}^*$, the $\lambda_i$ are pairwise distinct, the irreducible quadratic factors $X^2 + p_j X + q_j$ are pairwise distinct, and $p_j^2 - 4 q_j < 0$. Then $R$ admits a unique decomposition $$ \textcolor{colorprop}{R = E + \sum_{i = 1}^{r} \sum_{u = 1}^{m_i} \frac{a_{i, u}}{(X - \lambda_i)^u} + \sum_{j = 1}^{s} \sum_{v = 1}^{n_j} \frac{\alpha_{j, v} X + \beta_{j, v}}{(X^2 + p_j X + q_j)^v},} $$ with $E \in \mathbb{R}[X]$ the partie entière of $R$ and all coefficients in $\mathbb{R}$. The zero fraction has the trivial decomposition $0 = 0$.

Proof

Admitted (hors programme).

Method — Direct real method

Factor the denominator over $\mathbb{R}$ to identify the real poles and the irreducible quadratic factors.
Write the expected real form, with unknown coefficients $a_{i, u}, \alpha_{j, v}, \beta_{j, v}$.
Multiply both sides by the denominator to clear fractions; obtain a polynomial identity.
Determine the coefficients by either:
- identifying coefficients of like powers of $X$, or
- evaluating at convenient real values (e.g. $X = 0$, $X = $ a root of the numerator, etc.).

This method does not require detouring through $\mathbb{C}$ --- valuable when the irreducible quadratic factors do not lend themselves to easy complex manipulations.

Method — From $\mathbb{C}$ decomposition: group conjugate poles

Suppose $R \in \mathbb{R}(X)$ has been decomposed over $\mathbb{C}$. The non-real poles come in complex-conjugate pairs $\lambda$, $\bar{\lambda}$ (since $B \in \mathbb{R}[X]$), and the corresponding partial-fraction coefficients are also complex conjugate: if the coefficient of $1 / (X - \lambda)^k$ is $a$, then the coefficient of $1 / (X - \bar{\lambda})^k$ is $\bar{a}$. Group the two conjugate terms: $$ \frac{a}{(X - \lambda)^k} + \frac{\bar{a}}{(X - \bar{\lambda})^k}. $$ Bring to a common denominator $((X - \lambda)(X - \bar{\lambda}))^k = (X^2 + p X + q)^k$ where $p = -(\lambda + \bar{\lambda}) = -2 \mathrm{Re}(\lambda) \in \mathbb{R}$ and $q = \lambda \bar{\lambda} = |\lambda|^2 \in \mathbb{R}$. The resulting numerator $a (X - \bar{\lambda})^k + \bar{a} (X - \lambda)^k$ is real (sum of a complex number and its conjugate) and of degree $\leq k$.
Caveat for $k \geq 2$. For $k = 1$ the numerator has degree $\leq 1$ and the grouped expression is already a 2nd-species element. For $k \geq 2$ the numerator may have degree $> 1$; the expression must then be rewritten uniquely as a sum of 2nd-species elements with denominators $(X^2 + p X + q)^v$ for $v = 1, \ldots, k$: $$ \sum_{v = 1}^{k} \frac{\alpha_v X + \beta_v}{(X^2 + p X + q)^v}. $$ In this chapter we use the grouping method mainly in the simple-pole case.

Method — Use parity

Let $R \in \mathbb{R}(X)$ and consider the change of variable $X \mapsto -X$. The substitution permutes the poles of $R$: a pole $\lambda$ of order $m$ becomes a pole $-\lambda$ of order $m$. By unicity of the decomposition, parity links the coefficients at paired poles.

1st-species coefficient pairing. For a real pole $\lambda \neq 0$ of order $m$, let $a_{\lambda, k}$ be the coefficient of $1 / (X - \lambda)^k$ for $k \in \{1, \ldots, m\}$ (the pole at $-\lambda$ has the same order $m$). Then $$ \textcolor{colorprop}{\begin{aligned} R \text{ even} &\Longrightarrow a_{-\lambda, k} = (-1)^k \, a_{\lambda, k} \\ R \text{ odd} &\Longrightarrow a_{-\lambda, k} = (-1)^{k + 1} \, a_{\lambda, k}. \end{aligned}} $$ In particular for simple poles ($k = 1$): $R$ even gives $a_{-\lambda} = -a_\lambda$, $R$ odd gives $a_{-\lambda} = a_\lambda$.
2nd-species coefficient pairing for even-symmetric quadratics. For an irreducible quadratic factor of the form $(X^2 + q)^v$ (i.e. $p = 0$, hence even in $X$), the corresponding 2nd-species element $\dfrac{\alpha_{j, v} X + \beta_{j, v}}{(X^2 + q)^v}$ inherits the parity of $R$: if $R$ is even, $\alpha_{j, v} = 0$ ; if $R$ is odd, $\beta_{j, v} = 0$.

This typically halves the number of unknown coefficients. (Quadratic factors with $p \neq 0$ are not even in $X$ and do not benefit from this shortcut directly.)

Example — Recognition only

The fraction $R = \dfrac{1}{X^2 + 1}$ is already a 2nd-species simple element of $\mathbb{R}(X)$ (with $\alpha = 0$, $\beta = 1$, $p = 0$, $q = 1$, $k = 1$). Its real partial-fraction decomposition is itself.

Example

Decompose $R = \dfrac{1}{X (X^2 + 1)}$ as a sum of real simple elements (use the direct method).

Answer

The denominator factors over $\mathbb{R}$ as $X (X^2 + 1)$, with one simple real pole at $0$ and one irreducible quadratic factor $X^2 + 1$ (discriminant $-4 < 0$). Both worked examples in this section feature an irreducible quadratic with $p = 0$ (a deliberate didactic simplification per the program's « toute technicité dans les exemples est exclue »); the general $p \neq 0$ case is exercised later in the primitive example with denominator $x^2 + x + 1$, and more systematically in the exo file. The expected real form is $$ R = \frac{a}{X} + \frac{\alpha X + \beta}{X^2 + 1}. $$ Clear fractions. Multiply by $X (X^2 + 1)$: $$ 1 = a (X^2 + 1) + (\alpha X + \beta) X = (a + \alpha) X^2 + \beta X + a. $$ Identify coefficients of like powers of $X$.

Coefficient of $X^0$: $a = 1$.
Coefficient of $X^1$: $\beta = 0$.
Coefficient of $X^2$: $a + \alpha = 0$, so $\alpha = -1$.

Conclusion. $$ \frac{1}{X (X^2 + 1)} = \frac{1}{X} - \frac{X}{X^2 + 1}. $$

Example

Decompose $R = \dfrac{X}{(X^2 + 1)(X^2 + 4)}$ as a sum of real simple elements, using parity to reduce the unknowns.

Answer

The denominator has two irreducible quadratic factors $X^2 + 1$ and $X^2 + 4$, both with negative discriminant. The expected real form is $$ R = \frac{\alpha_1 X + \beta_1}{X^2 + 1} + \frac{\alpha_2 X + \beta_2}{X^2 + 4}. $$ Use parity. The fraction $R = X / ((X^2 + 1)(X^2 + 4))$ satisfies $R(-X) = -R(X)$ --- it is odd. The denominator is even (only even powers of $X$); for the decomposition to be odd, the numerator at each 2nd-species term must be odd, i.e. only the linear part $\alpha_i X$ survives: $$ \beta_1 = \beta_2 = 0. $$ This drops the unknowns from four to two: $\alpha_1$ and $\alpha_2$.
Clear fractions. Multiply by $(X^2 + 1)(X^2 + 4)$: $$ X = \alpha_1 X (X^2 + 4) + \alpha_2 X (X^2 + 1) = X \big[ \alpha_1 (X^2 + 4) + \alpha_2 (X^2 + 1) \big]. $$ Dividing by $X$ (valid as a polynomial identity): $$ 1 = \alpha_1 (X^2 + 4) + \alpha_2 (X^2 + 1) = (\alpha_1 + \alpha_2) X^2 + (4 \alpha_1 + \alpha_2). $$ Identify coefficients.

Coefficient of $X^2$: $\alpha_1 + \alpha_2 = 0$, so $\alpha_2 = -\alpha_1$.
Coefficient of $X^0$: $4 \alpha_1 + \alpha_2 = 1$, so $4 \alpha_1 - \alpha_1 = 3 \alpha_1 = 1$, hence $\alpha_1 = 1/3$ and $\alpha_2 = -1/3$.

Conclusion. $$ \frac{X}{(X^2 + 1)(X^2 + 4)} = \frac{1/3 \, X}{X^2 + 1} - \frac{1/3 \, X}{X^2 + 4} = \frac{1}{3} \left( \frac{X}{X^2 + 1} - \frac{X}{X^2 + 4} \right). $$

The partial-fraction decomposition over $\mathbb{R}$ makes primitives of rational functions constructively accessible: each 1st-species element integrates to a $\ln$ or to a rational power term, and each 2nd-species element of first power decomposes (via the « $(2 X + p)$ split ») into a $\ln$ of the irreducible quadratic plus an $\arctan$ term. Higher powers $(\alpha X + \beta) / (X^2 + p X + q)^k$ with $k \geq 2$ also admit primitives via reductions or recurrence formulas, but this technical case is not developed in the cours examples. Primitives are stated on intervals avoiding poles --- a fraction with a real pole is not defined at the pole, hence has different primitives on the two sides. The same decomposition also gives explicit $k$-th derivatives for fractions with simple poles.

Proposition — Primitive of $1 / (x - a)^n$

Let $a \in \mathbb{R}$, $n \in \mathbb{N}^*$, and $I \subset \mathbb{R}$ an interval avoiding $a$. On $I$: $$ \textcolor{colorprop}{\begin{aligned} \int \frac{dx}{x - a} &= \ln |x - a| + C && \text{(case } n = 1 \text{)} \\ \int \frac{dx}{(x - a)^n} &= \frac{-1}{(n - 1)(x - a)^{n - 1}} + C && \text{(case } n \geq 2 \text{)}. \end{aligned}} $$

Proof

Case $n = 1$. On $I$, the function $x \mapsto x - a$ has constant sign, so $\ln |x - a|$ is well-defined. Differentiate: $\dfrac{d}{dx} \ln |x - a| = \dfrac{1}{x - a}$.
Case $n \geq 2$. Differentiate the candidate primitive: $$ \frac{d}{dx} \left[ \frac{-1}{(n - 1)(x - a)^{n - 1}} \right] = \frac{-1}{n - 1} \cdot (-(n - 1)) (x - a)^{-n} = \frac{1}{(x - a)^n}. $$

Proposition — Primitive of a 2nd-species element

Let $\alpha, \beta, p, q \in \mathbb{R}$ with $p^2 - 4 q < 0$. Set $\omega = \dfrac{\sqrt{4 q - p^2}}{2} > 0$. Then on $\mathbb{R}$, $$ \textcolor{colorprop}{\int \frac{\alpha x + \beta}{x^2 + p x + q} \, dx = \frac{\alpha}{2} \ln(x^2 + p x + q) + \frac{1}{\omega} \left( \beta - \frac{\alpha p}{2} \right) \arctan\!\left( \frac{x + p/2}{\omega} \right) + C.} $$

Proof

Numerator splitting. The derivative of $x^2 + p x + q$ is $2 x + p$. Write the numerator as $$ \alpha x + \beta = \frac{\alpha}{2} (2 x + p) + \left( \beta - \frac{\alpha p}{2} \right). $$ This is the « $(2 X + p)$ split »: the first term is a multiple of the derivative of the denominator.
Integrate the first piece. It integrates as $(\alpha / 2) \ln(x^2 + p x + q)$ (the argument is positive on $\mathbb{R}$ since the discriminant is negative, so no absolute value).
Integrate the second piece. Complete the square in the denominator: $$ x^2 + p x + q = \left( x + \frac{p}{2} \right)^2 + \left( q - \frac{p^2}{4} \right) = \left( x + \frac{p}{2} \right)^2 + \omega^2, $$ with $\omega^2 = q - p^2 / 4 = (4 q - p^2) / 4 > 0$. The substitution $u = (x + p/2) / \omega$ gives $$ \int \frac{dx}{x^2 + p x + q} = \int \frac{dx}{(x + p/2)^2 + \omega^2} = \frac{1}{\omega} \int \frac{du}{u^2 + 1} = \frac{1}{\omega} \arctan(u) + C = \frac{1}{\omega} \arctan\!\left( \frac{x + p/2}{\omega} \right) + C. $$ Sum the two pieces. Multiply this last result by the constant $\beta - \alpha p / 2$ and add to the $\ln$ piece.

Method — Primitive of a rational function

Given $R = A / B \in \mathbb{R}(X)$:

Compute the partie entière $E$ of $R$ (Euclidean division of $A$ by $B$). The polynomial $E$ integrates trivially.
Compute the partial-fraction decomposition of the proper part $R - E$ over $\mathbb{R}$.
Integrate term by term: each 1st-species element via Proposition on $1 / (x - a)^n$; each 2nd-species element of first power (i.e. $(\alpha x + \beta) / (x^2 + p x + q)$) via the previous proposition. Higher powers of irreducible quadratics, $(\alpha x + \beta) / (x^2 + p x + q)^k$ with $k \geq 2$, are handled by case-by-case reductions or recurrence formulas; they are not developed in this cours (« toute technicité dans les exemples est exclue »), and are deferred to the exo file.

State the interval on which the primitive is given --- avoid the real poles.

Example

Compute a primitive of $f(x) = \dfrac{1}{x^2 - 1}$ on each of the three intervals $\,]\!-\!\infty, -1[$, $\,]\!-\!1, 1[$, $\,]1, +\infty[$.

Answer

Decompose. From the earlier example, $$ \frac{1}{x^2 - 1} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1}. $$ Integrate term by term. On any interval avoiding $\pm 1$: $$ \int \frac{dx}{x^2 - 1} = \frac{1}{2} \ln|x - 1| - \frac{1}{2} \ln|x + 1| + C = \frac{1}{2} \ln\!\left| \frac{x - 1}{x + 1} \right| + C. $$ The expression is the same on all three intervals; the additive constant $C$ may be chosen independently on each interval.
Sign of the argument.

On $]\!-\!\infty, -1[$: $x - 1 < 0$, $x + 1 < 0$, so $(x - 1)/(x + 1) > 0$. The absolute value can be dropped: $\frac{1}{2} \ln \frac{x - 1}{x + 1}$.
On $]\!-\!1, 1[$: $x - 1 < 0$, $x + 1 > 0$, so $(x - 1)/(x + 1) < 0$. We write $\frac{1}{2} \ln \frac{1 - x}{1 + x}$.
On $]1, +\infty[$: both factors positive, so $\frac{1}{2} \ln \frac{x - 1}{x + 1}$.

Example

Compute a primitive of $f(x) = \dfrac{x + 1}{x^2 + x + 1}$ on $\mathbb{R}$.

Answer

The denominator $x^2 + x + 1$ has discriminant $1 - 4 = -3 < 0$, so it is irreducible over $\mathbb{R}$ --- no real pole, primitive defined on all of $\mathbb{R}$.
Apply the $(2 x + p)$ split. Here $p = 1$, $q = 1$, so $2 x + p = 2 x + 1$, and $$ x + 1 = \frac{1}{2} (2 x + 1) + \frac{1}{2}. $$ Integrate the $\ln$ piece. $$ \int \frac{(2 x + 1)/2}{x^2 + x + 1} \, dx = \frac{1}{2} \ln(x^2 + x + 1) + C_1. $$ Integrate the $\arctan$ piece. Complete the square: $x^2 + x + 1 = (x + 1/2)^2 + 3/4$, so $\omega^2 = 3/4$, $\omega = \sqrt{3}/2$. Then $$ \int \frac{1/2}{x^2 + x + 1} \, dx = \frac{1}{2} \cdot \frac{1}{\omega} \arctan\!\left( \frac{x + 1/2}{\omega} \right) + C_2 = \frac{1}{\sqrt{3}} \arctan\!\left( \frac{2 x + 1}{\sqrt{3}} \right) + C_2. $$ Sum. A primitive on $\mathbb{R}$ is $$ F(x) = \frac{1}{2} \ln(x^2 + x + 1) + \frac{1}{\sqrt{3}} \arctan\!\left( \frac{2 x + 1}{\sqrt{3}} \right) + C. $$

Method — $k$-th derivative of a rational fraction with simple poles

Suppose $R \in \mathbb{R}(X)$ has only simple real poles, and decompose $$ R = E + \sum_i \frac{a_i}{X - \lambda_i}. $$ Then for any $k \geq 1$, on each interval avoiding the poles, $$ R^{(k)}(x) = E^{(k)}(x) + \sum_i \frac{(-1)^k k! \, a_i}{(x - \lambda_i)^{k + 1}}. $$ The first term vanishes once $k > \deg E$.

Example

Compute the $k$-th derivative of $f(x) = \dfrac{1}{x^2 - 1}$ on each interval avoiding $\pm 1$.

Answer

Decompose. $$ \frac{1}{x^2 - 1} = \frac{1/2}{x - 1} - \frac{1/2}{x + 1}. $$ Differentiate $k$ times each term. For any $\lambda$ and on an interval avoiding $\lambda$: $$ \frac{d^k}{dx^k} \left[ \frac{1}{x - \lambda} \right] = \frac{(-1)^k k!}{(x - \lambda)^{k + 1}}. $$ Combine. $$ \frac{d^k}{dx^k} \left[ \frac{1}{x^2 - 1} \right] = \frac{1}{2} \cdot \frac{(-1)^k k!}{(x - 1)^{k + 1}} - \frac{1}{2} \cdot \frac{(-1)^k k!}{(x + 1)^{k + 1}} = \frac{(-1)^k k!}{2} \left( \frac{1}{(x - 1)^{k + 1}} - \frac{1}{(x + 1)^{k + 1}} \right). $$ The closed form is exact and applies on each of the three intervals $]\!-\!\infty, -1[$, $]\!-\!1, 1[$, $]1, +\infty[$.

Skills to practice

Decomposing with simple poles
Decomposing with a double pole
Using the $P' / P$ formula
Decomposing with a quadratic factor
Bridging $\mathbb{C}$ to $\mathbb{R}$ by grouping conjugate poles
Using parity
Computing primitives via decomposition
Computing $k$-th derivatives