Factorising Expressions (Edexcel IAL Maths)

Spec point — 1.10

Factorising expressions

What is meant by factorising expressions?

• Many expressions in mathematics are written as a sum of terms

  • e.g. $x^{2}+6x-16$ is the sum of three the terms $x^{2}$, $6x$ and $-16$

• Many expressions are written as a product of factors

  • e.g. $(x+8)(x-2)$ is the product of the two (linear) factors $x+8$ and $x-2$

• Factorising is the process of rewriting the sum of terms as a product of factors

  • The other way round is expanding

How do I factorise an expression?

• This will depend on the nature of the expression you are dealing with

• In all cases the first thing to consider is if there is a factor (number and/or letter) of all terms in the expression

  • e.g. $2x^{3}+4x^{2}-8x=2x(x^{2}+2x-4)$

• A quadratic expression may be able to be factorised into two linear factors

  • Look out for special cases

    • No constant term: $x^{2}+5x=x(x+5)$
    • Difference of two squares (no x term and constant is square): $x^{2}-36=(x-6)(x+6)$
    • Perfect squares: $x^{2} + 8 x + 16 = ( x + 4 ) ( x + 4 ) = ( x + 4 )^{2}$
    • ‘Hidden’ quadratics:$3^{2 x} - 12 \times 3^{x} + 27 = ( 3^{x} )^{2} - 12 ( 3^{x} ) + 27 = ( 3^{x} - 3 ) ( 3^{x} - 9 )$
    • More than one variable:$x^{2}-y^{2}=(x-y)(x+y)$

• A cubic expression (at this level) will not contain a constant term

  • This means will $x$ be a factor (and there might be a number as a factor too)

  • The remaining expression will be a quadratic

    • this quadratic may also be able to be factorised
    • e.g. $6 x^{3} + 3 x^{2} - 9 x = 3 x ( 2 x^{2} + x - 3 ) = 3 x ( 2 x + 3 ) ( x - 1 )$

How do I factorise harder quadratics?

• There are many shortcuts to factorise quadratic expressions, but they often only apply under certain conditions (such as when a = 1)

  • the method below works for any quadratic expression
  • it is most useful when the coefficient of the $x^{2}$ term is greater than 1 (and not prime)

• Follow the steps:

  • STEP 1 Starting with $ax^{2}+bx+c$ find the product $ac$

    • For example: for $6x^{2}+7x-3$ $ac=6\times-3=-18$

  • STEP 2 Find two numbers m & n whose product is $ac$ and sum is $b$

    • For example: $9\times-2=-18=ac$ & $9+(-2)=7=b$
    • So m = 9 & n = - 2

  • STEP 3 Split the $bx$ term into $mx+nx$

    • For example: $6x^{2}-2x+9x-3$

  • STEP 4 Factorise the first two terms and the last two terms

    • For example: $2 x ( 3 x - 1 ) + 3 ( 3 x - 1 )$

  • STEP 5 Factorise once more for the final answer

    • For example: $(3x-1)(2x+3)$

• If a and/or c are prime, factorising can be done “by inspection”

  • For example: the only way to split (prime) 3 into factors would be 3 and 1

Why does the 'ac' method work?

• Suppose $a x^{2} + b x + c \equiv ( p x + r ) ( q x + s )$

  • then expanding and simplifying gives

    • $a x^{2} + b x + c \equiv p q x^{2} + p s x + q r x + r s \equiv p q x^{2} + ( p s + q r ) x + r s$

• By comparing coefficients

  • $a=pq$
  • $b=ps+qr$
  • $c=rs$

• Let $m=ps$ and $n=qr$ then:

  • $m+n=ps+qr=b$
  • $m\times n=psqr=ac$
  • Therefore these are the two numbers whose product is ac and sum is b

Worked Example

Exam Hint

• Do use your tried and tested shortcuts for factorising quadratics

  • We’ve explained it in full above to help you understand the process rather than to learn ‘tricks’

• You don't need to learn why the 'ac' method works - but we thought you might think that the algebra is cool