Deciding the Factorisation Method (Cambridge International IGCSE Mathematics)

Quadratics Factorising Methods

How do I know if an expression factorises?

• The easiest way to check if ax² + bx + c factorises is to check if you can find a pair of integers which:

  • Multiply to give ac
  • Sum to give b
  • If you can find integers to satisfy this, the expression must factorise

• There are some alternate methods to check:

  • Method 1: Use a calculator to solve the quadratic expression equal to 0

    • Only some calculators have this functionality
    • If the solutions are integers or fractions (without square roots), then the quadratic expression will factorise

  • Method 2: Find the value under the square root in the quadratic formula

    • b² – 4ac
    • If this number is a square number, then the quadratic expression will factorise

Which factorisation method should I use for a quadratic expression?

• Does it have 2 terms only?

  • Yes, like $x^{2}-7x$

    • Factorise out the highest common factor, x
    • $x ( x - 7 )$

  • Yes, like $x^{2}-9$

    • Use the "difference of two squares" to factorise
    • $( x + 3 ) ( x - 3 )$

Does it have 3 terms?

• Yes, starting with x² like $x^{2}-3x-10$

  • Use "factorising simple quadratics" by finding two numbers that add to -3 and multiply to -10
  • $( x + 2 ) ( x - 5 )$

• Yes, starting with ax² like $3x^{2}+15x+18$

  • Check to see if the 3 in front of x² is a common factor for all three terms(which it is in this case), then factorise it out of all three terms
  • $3 ( x^{2} + 5 x + 6 )$
  • The quadratic expression inside the brackets is now x² +... , which factorises more easily
  • $3 ( x + 2 ) ( x + 3 )$

• Yes, starting with ax² like $3x^{2}-5x-2$

  • The 3 in front of x² is not a common factor for all three term
  • Use "factorising harder quadratics", for example factorising by grouping or factorising using a grid
  • $( 3 x + 1 ) ( x - 2 )$

What other expressions should I be able to factorise?

• You may have a cubed term like $x^{3}-3x^{2}-10x$

  • Check to see if x is a common factor for all three terms(which it is in this case), so factorise it out of all three terms
  • $x ( x^{2} - 3 x - 10 )$
  • The remaining quadratic can then be factorised
  • $x ( x + 2 ) ( x - 5 )$

• It can also be useful to spot a quadratic in the form $x^{2}+2ax+a^{2}$

  • This factorises to $( x + a )^{2}$
  • E.g. $x^{2} + 6 x + 9 = ( x + 3 )^{2}$

Exam Hint A common mistake in the exam is to divide expressions by numbers, e.g. $2x^{2}+4x+2$ becomes $x^{2}+2x+1$ (which is incorrect).

This can only be done with equations.

e.g. $2x^{2}+4x+2=0$ becomes $x^{2}+2x+1=0$ (dividing "both sides" by 2).

Worked Example Factorise $-8x^{2}+100x-48$.

Answer:

Spot the common factor of -4 and factorise it out

$- 8 x^{2} + 100 x - 48 = - 4 ( 2 x^{2} - 25 x + 12 )$

Check to see if the quadratic in the bracket will factorise using $b^{2}-4ac$

$\begin{gathered} ( - 25 )^{2} - ( 4 \times 2 \times 12 ) \\ = 625 - 96 \\ = 529 \end{gathered}$

529 is a square number (23²) so the expression will factorise

Factorise $2x^{2}-25x+12$

We require a pair of numbers which multiply to ac, and sum to b

$a\times c=2\times12=24$

The only numbers which multiply to 24 and sum to -25 are

-24 and -1

Split the $-25x$ term into $-24x-x$

$\begin{gathered} 2 x^{2} - 24 x - x + 12 \end{gathered}$

Group and factorise the first two terms, using $2x$ as the common factor Group and factorise the last two terms using $-1$ as the common factor

$\begin{gathered} 2 x ( x - 12 ) - 1 ( x - 12 ) \end{gathered}$

These factorised terms now have a common factor of $( x - 12 )$, so this can be factorised out

$( 2 x - 1 ) ( x - 12 )$

Recall that -4 was factorised out at the start

$- 8 x^{2} + 100 x - 48 = - 4 ( 2 x^{2} - 25 x + 12 ) = - 4 ( 2 x - 1 ) ( x - 12 )$

$-4(2x-1)(x-12)$