Factorising Expressions (Edexcel International AS Maths: Pure 1): Revision Note

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: $x^2+8x+16=\left(x+4\right)\left(x+4\right)={\left(x+4\right)}^2$

    • ‘Hidden’ quadratics: $3^{2x}-12\times 3^x+27={\left(3^x\right)}^2-12\left(3^x\right)+27=(3^x-3)(3^x-9)$

    • 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. $6x^3+{3x}^2-9x=3x\left(2x^2+x-3\right)=3x\left(2x+3\right)\left(x-1\right)$

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 $a{x}^2+bx+c$ 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: $2x\left(3x-1\right)+3\left(3x-1\right)$

  • 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 $a{x}^2+bx+c\equiv \left(px+r\right)\left(qx+s\right)$

  • then expanding and simplifying gives

    •  $a{x}^2+bx+c\equiv pq{x}^2+psx+qrx+rs\equiv pq{x}^2+\left(ps+qr\right)x+rs$

• 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