Factorising Quadratic Expressions
Turn a quadratic expression into a product by working backwards from expansion.
Factorising reverses expansion
Expansion changes a product into a sum. Factorising changes the sum back into a product. You can always check your answer by expanding the brackets.
Start with a common factor
Before looking for two brackets, check whether every term shares a number, a variable, or both.
Take out the greatest common factor
Factorise $6x^2-18x$
Step 1 The greatest common factor is $6x$.
Step 2 Divide each term by $6x$.
$6x^2-18x=6x(x-3)$
Check: $6x(x-3)=6x^2-18x$.
Monic quadratic trinomials
For $x^2+bx+c$, find two integers with product $c$ and sum $b$.
$x^2+(p+q)x+pq=(x+p)(x+q)$
Both numbers are positive
Factorise $x^2+7x+12$.
We need a pair with product $12$ and sum $7$. The pair is $3$ and $4$.
$x^2+7x+12=(x+3)(x+4)$
Use the signs
Factorise $x^2-x-12$.
A negative product means the two numbers have opposite signs. The pair $3$ and $-4$ has product $-12$ and sum $-1$.
$x^2-x-12=(x+3)(x-4)$
A quick sign guide
If $c$ is positive, the two signs match. Use the sign of $b$ for both. If $c$ is negative, the signs differ; the number with the larger absolute value determines the sign of $b$.
Recognise special quadratics
A perfect-square trinomial has two equal factors. A difference of squares has two terms separated by subtraction.
$x^2+2ax+a^2=(x+a)^2$
$a^2-b^2=(a-b)(a+b)$
Two special forms
$x^2-10x+25=(x-5)^2$ because $25=5^2$ and $-10x=2(x)(-5)$.
$9x^2-16=(3x-4)(3x+4)$ because $9x^2=(3x)^2$ and $16=4^2$.
When the leading coefficient is greater than 1
For $ax^2+bx+c$, multiply $a$ and $c$. Find two integers whose product is $ac$ and whose sum is $b$, split the middle term, then factor by grouping.
Split the middle term
Factorise $6x^2+11x+3$.
Step 1 $a c=6\times3=18$. The pair $9$ and $2$ has product $18$ and sum $11$.
Step 2 Split $11x$.
$6x^2+9x+2x+3$
Step 3 Factor each pair.
$3x(2x+3)+1(2x+3)$
Step 4 Take out the repeated bracket.
$6x^2+11x+3=(3x+1)(2x+3)$
Factorise completely
If there is a common factor, take it out before factorising the remaining quadratic.
Two stages
Factorise $4x^2-20x+24$ completely.
$4x^2-20x+24=4(x^2-5x+6)$
The pair with product $6$ and sum $-5$ is $-2$ and $-3$.
$4(x^2-5x+6)=4(x-2)(x-3)$
Expanding gives the original expression, so the factorisation is complete.
Practice Problems
Factorise each expression completely, then expand your answer to check.
Practice 1
$8x^2+20x$
Show answer
$4x(2x+5)$
Practice 2
$x^2+9x+20$
Show answer
$(x+4)(x+5)$
Practice 3
$x^2+2x-35$
Show answer
$(x+7)(x-5)$
Practice 4
$4x^2-12x+9$
Show answer
$(2x-3)^2$
Practice 5
$6x^2-x-2$
Show answer
$(3x-2)(2x+1)$