Factorisation of algebraic expressions

Factorisation is a consequence of the distributive law

$a(b+c)\equiv ab+ac$a(b+c)≡ab+ac.

Instead of expanding an expression with brackets to make a sum, we find factors common to the separate terms so that a sum of terms becomes a product.

Sometimes, we recognise the form of a sum expression as one that has a known factorisation. At other times some manipulation and experimentation may be needed.

common factors

If the terms in an expression have one or more factors in common, those factors can be taken outside a bracket.

Example 1

Every term of the sum $3at+9a^2s$3at+9a2s has the factors $3$3 and $a$a in common. So, these can be taken outside a bracket to give the equivalent expression $3a\left(t+3as\right)$3a(t+3as).

Notice that the terms inside the bracket no longer have common factors. We took out as many factors as possible, in equal quantities from each term.

Example 2

The common factors may themselves be compound terms. Consider the expression $pq(x+2y)+2rs(x+2y)$pq(x+2y)+2rs(x+2y). The bracket $x+2y$x+2y is common to both terms in the sum. It can be taken outside another bracket to give $(x+2y)\left(pq+2rs\right)$(x+2y)(pq+2rs).

difference of squares

Any expression in the form $A^2-B^2$A2−B2 can be factorised to the form $\left(A+B\right)\left(A-B\right)$(A+B)(A−B). This may be surprising because the terms in the expression do not appear to have common factors. However, we can carry out the following manipulation which does make use of the common factor idea.

$A^2-B^2$A2−B2	$=$=	$A^2+AB-AB-B^2$A2+AB−AB−B2
	$=$=	$A(A+B)-B(A+B)$A(A+B)−B(A+B)
	$=$=	$(A+B)(A-B)$(A+B)(A−B)

Example 3

Factorise the trigonometric expression $\cos^2x-\sin^2x$cos2x−sin2x.

Recognising the difference of two squares, we write $(\cos x+\sin x)(\cos x-\sin x)$(cosx+sinx)(cosx−sinx).

Example 4

Factorise $4y^2-3$4y2−3 using the difference of two squares.

The first of the two terms is obviously a square but the second has to be thought of as $\left(\sqrt{3}\right)^2$(√3)2. Then, the expression can be written $\left(2y+\sqrt{3}\right)\left(2y-\sqrt{3}\right)$(2y+√3)(2y−√3).

 

perfect squares

If we expand $(a+b)^2$(a+b)2, we get $a^2+2ab+b^2$a2+2ab+b2. So, any expression in this expanded form can be factorised as a squared bracket.

More generally, the expansion of $(ax+by)^2$(ax+by)2 is $a^2x^2+2axby+b^2y^2$a2x2+2axby+b2y2. So, again, an expression in this form has an easy factorisation.

Example 5

Factorise $16x^2-8xy+y^2$16x2−8xy+y2.

This is the same as $(4x)^2+2(4x)(-y)+(-y)^2$(4x)2+2(4x)(−y)+(−y)2 which we recognise as $\left(4x-y\right)^2$(4x−y)2.

 

Grouping in Pairs

If we expand the product $\left(w+x\right)\left(y+z\right)$(w+x)(y+z) we get $wy+wz+xy+xz$wy+wz+xy+xz.

Similarly, we can factorise an expression with four terms like this by grouping in pairs, e.g. $\left(wy+wz\right)+\left(xy+xz\right)$(wy+wz)+(xy+xz). Taking the HCF from each pair gives $w\left(y+z\right)+x\left(y+z\right)$w(y+z)+x(y+z). Factorising the HCF of these pairs gives $\left(w+x\right)\left(y+z\right)$(w+x)(y+z).

Example 6

Factorise $x^3-2x^2-x+2$x3−2x2−x+2.

Notice that the first two terms have a common factor of $x^2$x2 and the second two have a common factor of $-1$−1. So $x^3-2x^2-x+2=x^2\left(x-2\right)+x-2$x3−2x2−x+2=x2(x−2)+x−2. Factorising $\left(x-2\right)$(x−2) gives $\left(x^2+1\right)\left(x-2\right)$(x2+1)(x−2).

 

Quadratics

Some but not all quadratic expressions in the real numbers can be factorised into two linear factors. Sometimes this can be done by inspection. In other cases, the completing the square process is needed.

A quadratic expression that cannot be factorised is said to be irreducible.

Example 7

The factorised form of the expression $2x^2-7x-15$2x2−7x−15 can be found with a little experimentation to be $(2x+3)(x-5)$(2x+3)(x−5). The thinking that would lead to this result would include the realisation that the terms in $x$x at the beginning of the brackets have to have the product $2x^2$2x2 and the constant terms have to multiply to $-15$−15. With these considerations, there are only a few possibilities that need to be tried.

Example 8

Factorise $2x^2-6x-14$2x2−6x−14.

Although the coefficients in this example are not very different from those in Example $6$6, the factorisation is more difficult to find. We take out the common factor $2$2 as a first step, then 'complete the square' and finally use the difference of two squares factorisation.

$2x^2-6x-14$2x2−6x−14	$=$=	$2\left(x^2-3x-7\right)$2(x2−3x−7)
	$=$=	$2\left(x^2-3x+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2-7\right)$2(x2−3x+(32​)2−(32​)2−7)
	$=$=	$2\left(\left(x-\frac{3}{2}\right)^2-\left(\frac{9}{4}+7\right)\right)$2((x−32​)2−(94​+7))
	$=$=	$2\left(\left(x-\frac{3}{2}\right)^2-\frac{37}{4}\right)$2((x−32​)2−374​)
	$=$=	$2\left(x-\frac{3}{2}+\frac{\sqrt{37}}{2}\right)\left(x-\frac{3}{2}-\frac{\sqrt{37}}{2}\right)$2(x−32​+√372​)(x−32​−√372​)
	$=$=	$2\left(x-\frac{3-\sqrt{37}}{2}\right)\left(x-\frac{3+\sqrt{37}}{2}\right)$2(x−3−√372​)(x−3+√372​)

 

cubics

Sums and differences of cubes can be written as the product of a linear term and a quadratic term.

$A^3+B^3\equiv(A+B)\left(A^2-AB+B^2\right)$A3+B3≡(A+B)(A2−AB+B2)

$A^3-B^3\equiv(A-B)\left(A^2+AB+B^2\right)$A3−B3≡(A−B)(A2+AB+B2)

These identities can be checked by expanding the expressions on the right.

Example 9

Factorise the sum of cubes $8y^3+125z^3$8y3+125z3.

Thinking of this as $(2y)^3+(5z)^3$(2y)3+(5z)3, we use the appropriate identity to write $(2y+5z)\left((2y)^2-10yz+(5z)^2\right)$(2y+5z)((2y)2−10yz+(5z)2). That is, 

$(2y+5z)\left(4y^2-10yz+25z^2\right)$(2y+5z)(4y2−10yz+25z2).

 

It can also be useful to recognise the binomial expansion $(a+b)^3\equiv a^3+3a^2b+3ab^2+b^3$(a+b)3≡a3+3a2b+3ab2+b3.

 

Worked Questions 

Question 1

Question 2

Question 3