Factorisation

Factorising reverses expansion

When brackets are expanded, multiplication changes a product into a sum or difference:

$3(x-3)=3x-9.$

Factorising travels in the opposite direction. It rewrites the expression as a product:

$3x-9=3(x-3).$

The expressions on the two sides are equivalent; only their forms differ. A reliable check on any factorisation is therefore to expand the proposed answer.

Taking out a common factor

Begin by finding a factor shared by every term. Divide each term by that factor to determine what belongs inside the bracket:

$12x^2y-8xy^2=4xy(3x-2y).$

Here $4xy$ is common to both terms. Dividing gives $12x^2y\div4xy=3x$ and $-8xy^2\div4xy=-2y$.

To factorise fully, take out the highest common factor. For example,

$12x^2+8x=4x(3x+2),$

rather than stopping at $2x(6x+4)$, because the terms inside the latter bracket still share a factor of $2$.

Sometimes taking out a negative factor makes the bracket easier to read. For example,

$-6x^2+9x=-3x(2x-3).$

Expanding $-3x(2x-3)$ restores both original signs.

Factorising $x^2+bx+c$

A monic quadratic can arise by expanding two brackets:

$(x+p)(x+q)=x^2+(p+q)x+pq.$

So, to factorise $x^2+bx+c$, find two numbers $p$ and $q$ such that

$p+q=b \qquad\text{and}\qquad pq=c.$

For example, to factorise $x^2-x-6$, seek two numbers with product $-6$ and sum $-1$. They are $-3$ and $2$, so

$x^2-x-6=(x-3)(x+2).$

Expanding checks the result: the middle terms are $2x-3x=-x$, while the constants multiply to $-6$.

The signs help to narrow the search:

• if $c$ is positive, the two numbers have the same sign;
• if $c$ is negative, they have opposite signs;
• their sum must still have the sign and value of $b$.

For instance,

$x^2-5x+6=(x-2)(x-3),$

because $(-2)+(-3)=-5$ and $(-2)(-3)=6$.

A difference of two squares

The case $b=0$ includes expressions such as $x^2-16$. The identity

$u^2-v^2=(u-v)(u+v)$

shows that

$x^2-16=(x-4)(x+4).$

This works because the middle terms cancel when the brackets are expanded. The same structure can include irrational values:

$x^2-3=(x-\sqrt3)(x+\sqrt3).$

Do not confuse a difference of squares with a repeated bracket: $(x-4)^2$ expands to $x^2-8x+16$, not $x^2-16$.

Factorising $ax^2+bx+c$ when $a\ne0$ and $a\ne1$

When the coefficient of $x^2$ is not $1$, the brackets need leading terms whose product is $ax^2$. A systematic approach is to split the middle term.

For $ax^2+bx+c$:

1. calculate $ac$;
2. find two numbers whose product is $ac$ and whose sum is $b$;
3. use them to split $bx$ into two terms;
4. factorise the resulting pairs by grouping.

Consider

$2x^2+3x-2.$

Here $ac=2\times(-2)=-4$. The numbers $4$ and $-1$ multiply to $-4$ and add to $3$. Split the middle term:

$2x^2+4x-x-2.$

Factorise the first pair and the second pair:

$2x(x+2)-1(x+2).$

The repeated bracket is now a common factor:

$2x^2+3x-2=(2x-1)(x+2).$

Expanding gives $2x^2+4x-x-2=2x^2+3x-2$, confirming the factorisation.

As another example,

$6x^2+x-2$

has $ac=-12$. The numbers $4$ and $-3$ add to $1$, so

$6x^2+x-2=6x^2+4x-3x-2$

$=2x(3x+2)-1(3x+2)$

$=(2x-1)(3x+2).$

The grouping works only when the same bracket appears in both parts. If it does not, recheck the pair of numbers, the signs and any common factors taken from each group.

Choosing an efficient route

Before treating an expression as a quadratic, first remove any common factor. For example,

$5x^2-20=5(x^2-4)=5(x-2)(x+2).$

This is fully factorised: first the common factor $5$ was removed, then the remaining difference of two squares was factorised. The overall habit is therefore: common factor first, identify the remaining structure, then expand to check.

Core idea

Factorising is the reverse of expanding: rewrite a sum or difference as a product of factors.

Common factors

Find the HCF of every term, place it outside a bracket, and divide each term by it:

$12x^2y-8xy^2=4xy(3x-2y).$

Check that no factor remains common to every term inside the bracket.

Monic quadratics

For

$x^2+bx+c=(x+p)(x+q),$

choose $p$ and $q$ so that

$p+q=b,\qquad pq=c.$

Example:

$x^2-x-6=(x-3)(x+2).$

Difference of two squares:

$u^2-v^2=(u-v)(u+v).$

Non-monic quadratics

For $ax^2+bx+c$ with $a\ne0$ and $a\ne1$:

1. find two numbers with product $ac$ and sum $b$;
2. split the middle term using those numbers;
3. factorise by grouping;
4. take out the repeated bracket.

$2x^2+3x-2=2x^2+4x-x-2=(2x-1)(x+2).$

Final check

Expand the answer. It must reproduce the original coefficients, signs and constant term exactly.