In a test the question only says "factorise completely". You decide the method. Work through the checks in order; each one makes the next easier. **Step** | **Look for** | **Method** 1 | a factor common to every term (if the first term is negative, take out -1 as well) | take out the HCF 2 | two terms that are squares, with a minus sign between them | a2-b2=(a+b)(a-b) 3 | three terms with square end terms and middle term +-2ab | (a+- b)2 4 | any other three-term quadratic | multiplication frame or factor pairs 5 | four terms with a common factor in each pair | grouping in pairs 6 | four terms where no grouping into pairs works, but three terms make (a+- b)2 or -(a+- b)2 and the fourth is a square | write the three as one square, then use a2-b2 7 | each bracket in the answer | factorise again if possible A repeated bracket counts as one letter. In 72-2(x-y)2, let A=x-y: 72-2A2=2(36-A2)=2(6+A)(6-A), so the answer is 2(6+x-y)(6-x+y). If you prefer, expand the brackets first and then factorise. Method - Take out the HCF of all the terms. - Count the terms in what remains and choose the method from the table. - Look at every bracket again and factorise it further if you can. - Check by expanding the answer in full; the middle term is the one most often wrong. Example Factorise completely 6x2+3x-45, 2x2+7x-15 and 4x2-4xy+y2-9. 6x2+3x-45: common factor 3 (step 1), then a three-term quadratic that is not a perfect square (step 4). 6x2+3x-45 = 3(2x2+x-15) = 3(2x-5)(x+3) 2x2+7x-15: no common factor, three terms, 2x2 is not a square (step 4). 2x2+7x-15 = 2x2+10x-3x-15 = 2x(x+5)-3(x+5) = (2x-3)(x+5) Check the middle term: (2x-3)(x+5)=2x2+10x-3x-15=2x2+7x-15. 4x2-4xy+y2-9: four terms, but no way of pairing them gives a common bracket. The first three are (2x-y)2 and 9=32 (step 6). 4x2-4xy+y2-9 = (2x-y)2-32 = (2x-y+3)(2x-y-3) The three terms need not come first. In 9x2-y2+4y-4, the last three are -(y2-4y+4)=-(y-2)2, so the expression is (3x)2-(y-2)2=(3x-y+2)(3x+y-2). A common error is to stop too early: 3(2x2+x-15) is not complete, and neither is 3(4x2-9), where the bracket is still a difference of two squares.

Choosing a factorisation method

Secondary 2

In a test the question only says "factorise completely". You decide the method. Work through the checks in order; each one makes the next easier.

Step	Look for	Method
$1$	a factor common to every term (if the first term is negative, take out $-1$ as well)	take out the HCF
$2$	two terms that are squares, with a minus sign between them	$a^2-b^2=(a+b)(a-b)$
$3$	three terms with square end terms and middle term $\pm2ab$	$(a\pm b)^2$
$4$	any other three-term quadratic	multiplication frame or factor pairs
$5$	four terms with a common factor in each pair	grouping in pairs
$6$	four terms where no grouping into pairs works, but three terms make $(a\pm b)^2$ or $-(a\pm b)^2$ and the fourth is a square	write the three as one square, then use $a^2-b^2$
$7$	each bracket in the answer	factorise again if possible

A repeated bracket counts as one letter. In $72-2(x-y)^2$, let $A=x-y$: $72-2A^2=2(36-A^2)=2(6+A)(6-A)$, so the answer is $2(6+x-y)(6-x+y)$. If you prefer, expand the brackets first and then factorise.

Method

Take out the HCF of all the terms.
Count the terms in what remains and choose the method from the table.
Look at every bracket again and factorise it further if you can.
Check by expanding the answer in full; the middle term is the one most often wrong.

Example

Factorise completely $6x^2+3x-45$, $2x^2+7x-15$ and $4x^2-4xy+y^2-9$.

$6x^2+3x-45$: common factor $3$ (step 1), then a three-term quadratic that is not a perfect square (step 4).

$\begin{aligned} 6x^2+3x-45 &= 3(2x^2+x-15) \\ &= 3(2x-5)(x+3) \end{aligned}$

$2x^2+7x-15$: no common factor, three terms, $2x^2$ is not a square (step 4).

$\begin{aligned} 2x^2+7x-15 &= 2x^2+10x-3x-15 \\ &= 2x(x+5)-3(x+5) \\ &= (2x-3)(x+5) \end{aligned}$

Check the middle term: $(2x-3)(x+5)=2x^2+10x-3x-15=2x^2+7x-15$.

$4x^2-4xy+y^2-9$: four terms, but no way of pairing them gives a common bracket. The first three are $(2x-y)^2$ and $9=3^2$ (step 6).

$\begin{aligned} 4x^2-4xy+y^2-9 &= (2x-y)^2-3^2 \\ &= (2x-y+3)(2x-y-3) \end{aligned}$

The three terms need not come first. In $9x^2-y^2+4y-4$, the last three are $-(y^2-4y+4)=-(y-2)^2$, so the expression is $(3x)^2-(y-2)^2=(3x-y+2)(3x+y-2)$.

A common error is to stop too early: $3(2x^2+x-15)$ is not complete, and neither is $3(4x^2-9)$, where the bracket is still a difference of two squares.

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Step	Look for	Method
\(1\)	a factor common to every term (if the first term is negative, take out \(-1\) as well)	take out the HCF
\(2\)	two terms that are squares, with a minus sign between them	\(a^2-b^2=(a+b)(a-b)\)
\(3\)	three terms with square end terms and middle term \(\pm2ab\)	\((a\pm b)^2\)
\(4\)	any other three-term quadratic	multiplication frame or factor pairs
\(5\)	four terms with a common factor in each pair	grouping in pairs
\(6\)	four terms where no grouping into pairs works, but three terms make \((a\pm b)^2\) or \(-(a\pm b)^2\) and the fourth is a square	write the three as one square, then use \(a^2-b^2\)
\(7\)	each bracket in the answer	factorise again if possible