International Baccalaureate · MYP 4

Factorisation

⌚ ~90 min ▢ 9 blocks ✓ 44 exercises ➣ Prerequisites : Algebraic Expansion, Expansion and Factorisation, Radicals and Surds

I Factorising with the Identities

I.1 The Difference of Two Squares

Factorisation is the reverse process of expansion. While expanding converts a product into a sum, factorisation converts a sum into a product.
Mastering factorisation is a crucial skill for simplifying complex expressions and solving higher-degree equations. Unless stated otherwise, we work with real numbers.

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Proposition — Difference of Squares

$\textcolor{colordef}{a}^2-\textcolor{colorprop}{b}^2 = (\textcolor{colordef}{a}-\textcolor{colorprop}{b})(\textcolor{colordef}{a}+\textcolor{colorprop}{b})$

Example

Factorise $x^2-9$.

Answer

$$\begin{aligned} x^2-9&= \textcolor{colordef}{x}^2-\textcolor{colorprop}{3}^2\\ &= (\textcolor{colordef}{x}-\textcolor{colorprop}{3})(\textcolor{colordef}{x}+\textcolor{colorprop}{3}). \end{aligned}$$

Note

Any non-negative real number $c$ can be written as a square: $c=(\sqrt{c})^2$. This allows us to rewrite expressions such as $x^2 - c$ as a difference of squares and apply the formula above.

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Example

Factorise $x^2-3$.

Answer

We can write $3$ as $(\sqrt{3})^2$ to create a difference of squares. $$\begin{aligned} x^2-3 &= \textcolor{colordef}{x}^2 - \left(\textcolor{colorprop}{\sqrt{3}}\right)^2 \\ &= \left(\textcolor{colordef}{x}-\textcolor{colorprop}{\sqrt{3}}\right)\left(\textcolor{colordef}{x}+\textcolor{colorprop}{\sqrt{3}}\right). \end{aligned}$$

Note

An expression $x^2+c$ with $c>0$ can never be written $(x+p)(x+q)$ with real numbers $p$ and $q$.

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Example

Factorise if possible: $x^2+1$.

Answer

Since $x^2+1=x^2+1^2$ is a sum of squares, it cannot be factorised into linear factors with real coefficients.

Skills to practice

Factorising a Difference of Two Squares: Level 1
Factorising a Difference of Two Squares: Level 2

I.2 Perfect Square Trinomials

Proposition — Perfect Square Trinomials

$$\begin{aligned} \textcolor{colordef}{a}^{2}+2 \textcolor{colordef}{a} \textcolor{colorprop}{b}+\textcolor{colorprop}{b}^{2} &= (\textcolor{colordef}{a}+\textcolor{colorprop}{b})^{2}\\ \textcolor{colordef}{a}^{2}-2 \textcolor{colordef}{a} \textcolor{colorprop}{b}+\textcolor{colorprop}{b}^{2} &= (\textcolor{colordef}{a}-\textcolor{colorprop}{b})^{2}\\ \end{aligned}$$

Example

Factorise $x^2+2x+1$.

Answer

We check if this fits the form $a^2+2ab+b^2$.

The first term is $x^2$, so let $a=x$.
The last term is $1=1^2$, so let $b=1$.
We check if the middle term is $2ab$: $2(x)(1) = 2x$. It matches.

Therefore, the expression is a perfect square: $$\begin{aligned} x^2+2x+1 &= \textcolor{colordef}{x}^{2}+2(\textcolor{colordef}{x})(\textcolor{colorprop}{1})+\textcolor{colorprop}{1}^{2}\\ &= (\textcolor{colordef}{x}+\textcolor{colorprop}{1})^{2}. \end{aligned}$$

Skills to practice

Factorising Perfect Square Trinomials

II Factorising Trinomials

II.1 Factorising Trinomials by Sum and Product

A perfect square trinomial is a special case: most trinomials, such as $x^2 + 5x + 6$, do not have that shape — yet many of them can still be factorised into two brackets. The key is to run the expansion of $(x+p)(x+q)$ backwards.

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Proposition — Sum and Product

If two numbers $p$ and $q$ satisfy $$p + q = b \qquad \text{and} \qquad p \times q = c,$$ then$$x^2 + bx + c = (x+p)(x+q).$$

Proof

Expand the right side: $$\begin{aligned}[t] (x+p)(x+q) &= x^2 + qx + px + pq && {\scriptstyle\text{(expanding the two brackets)}} \\ &= x^2 + (p+q)x + pq && {\scriptstyle\text{(collect the } x\text{-terms)}} \\ &= x^2 + bx + c. && {\scriptstyle\left(p+q = b,\ pq = c\right)} \end{aligned}$$

Example

Factorise $x^2 + 5x + 6$.

Answer

We look for two numbers with sum $5$ and product $6$: the pair $2$ and $3$ works. $$\begin{aligned}[t] x^2 + 5x + 6 &= (x+2)(x+3). && {\scriptstyle\left(2 + 3 = 5,\ 2 \times 3 = 6\right)} \end{aligned}$$ Check by expanding: $(x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6$.

Example

Factorise $x^2 - 2x - 15$.

Answer

We look for two numbers with sum $-2$ and product $-15$: the product is negative, so the two numbers have opposite signs. The pair $-5$ and $3$ works: $$\begin{aligned}[t] x^2 - 2x - 15 &= (x-5)(x+3). && {\scriptstyle\left(-5 + 3 = -2,\ (-5) \times 3 = -15\right)} \end{aligned}$$ The signs are read off the target: a negative product forces opposite signs; the sum then decides which of the two is the larger in absolute value.

Proposition — Common Factor Laws

For any numbers $a$, $b$ and $c$: $\textcolor{colorprop}{a}b+\textcolor{colorprop}{a}c=\textcolor{colorprop}{a}(b+c)\quad$ and $\quad\textcolor{colorprop}{a}b-\textcolor{colorprop}{a}c=\textcolor{colorprop}{a}(b-c)$

Method — Factorising with a Common Bracket

A common factor is not always a number or a letter: it can be a whole bracket appearing in every term.

Spot the bracket that appears in each term.
Write this bracket in front, and collect in a second bracket what multiplies it in each term.
Simplify the second bracket.

When the second product is subtracted, the minus sign applies to everything that enters the second bracket: in $(x+1)(2x+3) - (x+1)(x-4)$, the second bracket is $(2x+3) - (x-4)$.
A term made of the common bracket alone is multiplied by $1$: in $(x+3)(2x-1) + (x+3)$, the second term is $(x+3) \times 1$, so the second bracket is $(2x-1) + 1$.

Method — Factorising by Grouping

When an expression has four terms and no factor common to all of them:

Group the terms in two pairs.
Factorise each pair using its own common factor.
If the same bracket appears in both pairs, take it out as a common factor.

Example

Factorise $ab + 3a + 2b + 6$.

Answer

$$\begin{aligned}[t] ab + 3a + 2b + 6 &= a(b + 3) + 2(b + 3) && {\scriptstyle\text{(factorise each pair)}} \\ &= (b+3)(a+2). && {\scriptstyle\text{(common bracket } b+3)} \end{aligned}$$

Method — Factorising $ax^2+bx+c$ by Splitting the Middle Term

When the coefficient of $x^2$ is not $1$:

Compute the product $a \times c$.
Find two numbers with product $ac$ and sum $b$.
Split the middle term $bx$ into these two pieces.
Factorise by grouping the four terms in pairs.

Example

Factorise $2x^2 + 7x + 3$.

Answer

Here $ac = 2 \times 3 = 6$; two numbers with product $6$ and sum $7$: the pair $6$ and $1$. $$\begin{aligned}[t] 2x^2 + 7x + 3 &= 2x^2 + 6x + x + 3 && {\scriptstyle\text{(split } 7x = 6x + x)} \\ &= 2x(x + 3) + (x + 3) && {\scriptstyle\text{(factorise each pair)}} \\ &= (x+3)(2x+1). && {\scriptstyle\text{(common bracket } x+3)} \end{aligned}$$

Skills to practice

Factorising Trinomials
Splitting the Middle Term

II.2 Factorising Completely

Method — Factorising Completely

An expression is factorised completely when none of its factors can be factorised further. The working order:

Common factor first — always look for it before anything else.
Then examine each bracket: difference of squares? perfect square? sum-and-product trinomial?
Repeat until no factor can be broken down further.

Example

Factorise $2x^3 - 8x$ completely.

Answer

$$\begin{aligned}[t] 2x^3 - 8x &= 2x\left(x^2 - 4\right) && {\scriptstyle\text{(common factor } 2x \text{ first)}} \\ &= 2x(x-2)(x+2). && {\scriptstyle\text{(difference of squares)}} \end{aligned}$$ Stopping at $2x\left(x^2-4\right)$ would not be complete: the bracket still hides a difference of squares.

Example

Factorise $3x^2 + 6x + 3$ completely.

Answer

$$\begin{aligned}[t] 3x^2 + 6x + 3 &= 3\left(x^2 + 2x + 1\right) && {\scriptstyle\text{(common factor } 3 \text{ first)}} \\ &= 3(x+1)^2. && {\scriptstyle\text{(perfect square)}} \end{aligned}$$ Taking the common factor out first is what makes the perfect square visible.

Skills to practice

Factorising by Grouping and Completely

III Going further

Skills to practice

Reasoning and Problem-Solving
Investigating and Solving Problems
Solving Exam-Style Problems
Quiz