An expression is factorised completely when no factor can be factorised again. Always look first for a factor common to every term: a number, a letter or a negative sign. Why: taking out the common factor leaves smaller coefficients, so the quadratic that remains is easier. Leaving it inside a bracket gives an answer that is not complete: (2x+4)(x-1) is equal to 2(x+2)(x-1), but 2x+4 still has the factor 2. **Expression** | **Common factor** | **After taking it out** 4x2-4x-24 | 4 | 4(x2-x-6) x2y+3xy-10y | y | y(x2+3x-10) -x2+x+12 | -1 | -(x2-x-12) Method - Find the HCF of the coefficients and any letter that appears in every term. - If the x2-term is negative, take out a negative factor so that the bracket starts with a positive x2-term. - Factorise the quadratic left in the bracket. - Write the common factor in front and check that no bracket has a common factor left. Example Factorise completely -3x2+6x+24. -3x2+6x+24 = -3(x2-2x-8) = -3(x-4)(x+2) Check: -3(x-4)(x+2)=-3(x2-2x-8)=-3x2+6x+24. The answer can also be written 3(4-x)(x+2). A common error is to stop at (-3x+12)(x+2): it has the right value, but -3x+12 still has the factor -3.

Common factors and complete factorisation

Secondary 2

An expression is factorised completely when no factor can be factorised again. Always look first for a factor common to every term: a number, a letter or a negative sign.

Why: taking out the common factor leaves smaller coefficients, so the quadratic that remains is easier. Leaving it inside a bracket gives an answer that is not complete: $(2x+4)(x-1)$ is equal to $2(x+2)(x-1)$, but $2x+4$ still has the factor $2$.

Expression	Common factor	After taking it out
$4x^2-4x-24$	$4$	$4(x^2-x-6)$
$x^2y+3xy-10y$	$y$	$y(x^2+3x-10)$
$-x^2+x+12$	$-1$	$-(x^2-x-12)$

Method

Find the HCF of the coefficients and any letter that appears in every term.
If the $x^2$-term is negative, take out a negative factor so that the bracket starts with a positive $x^2$-term.
Factorise the quadratic left in the bracket.
Write the common factor in front and check that no bracket has a common factor left.

Example

Factorise completely $-3x^2+6x+24$.

$\begin{aligned} -3x^2+6x+24 &= -3(x^2-2x-8) \\ &= -3(x-4)(x+2) \end{aligned}$

Check: $-3(x-4)(x+2)=-3(x^2-2x-8)=-3x^2+6x+24$. The answer can also be written $3(4-x)(x+2)$. A common error is to stop at $(-3x+12)(x+2)$: it has the right value, but $-3x+12$ still has the factor $-3$.

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Expression	Common factor	After taking it out
\(4x^2-4x-24\)	\(4\)	\(4(x^2-x-6)\)
\(x^2y+3xy-10y\)	\(y\)	\(y(x^2+3x-10)\)
\(-x^2+x+12\)	\(-1\)	\(-(x^2-x-12)\)