Factorising Quadratics (Edexcel IGCSE Further Maths)

Spec point — 2A

Factorising Quadratics

How can I factorise simple quadratics?

• If there is no constant term then just factorise out (a multiple of) $x$

  • x²−9x=x (x−9)
  • 5x²+30x=5x (x+6)

• Factorise quadratics of the form $x^{2}+bx+c$ by inspection

  • Find a pair of numbers, $p$ and $q$, that multiply to give $c$ and add to give $b$

    • E.g. for $x^{2}-21x-100$ the numbers would be $4$ and $-25$

  • The quadratic will factorise as (x+p)(x+q)

    • So x²−21x−100=(x+4)(x−25)

How can I factorise harder quadratics?

• A harder quadratic is of the form $ax^{2}+bx+c$ where $a$ is not equal to 1 (or 0)

  • E.g. $12x^{2}-11x-5$

• These can also be factorised by inspection

  • This requires a lot of practice and there are no simple rules to follow

• They can be factorised reliably by grouping

  • Find a pair of numbers that multiply to $ac$ and add to $b$

    • For $12x^{2}-11x-5$, $ac=-60$ and $b=-11$
    • So the two numbers are $4$ and $-15$

  • Rewrite the middle $bx$ term using those two numbers

    • $12x^{2}+4x-15x-5$

  • Group and factorise the first two terms and the last two terms by pulling out common factors

    • 4x (3x+1)−5 (3x+1)

  • Those two terms now have a common factor (in brackets) that can be factorised out

    • (4x−5)(3x+1)

How do I factorise a difference of two squares

• A difference of two squares refers to any expression of the form $a^{2}-b^{2}$

  • I.e. 'something squared subtracted from something else squared'

  • For example,

    • $x^{2}-36$
    • $9^{2}-5^{2}$
    • (x+1)²−(x−4)²
    • $4m^{2}-25n^{2}$ which is equal to (2m)²−(5n)²

• Such expressions will factorise as (a+b)(a−b)

  • This is because (a+b)(a−b)=a²−a b+a b−b²=a²−b²

  • So

    • x²−36=(x+6)(x−6)
    • 9²−5²=(9+5)(9−5) which is equal to $14\times4=56$
    • (x+1)²−(x−4)²=((x+1)+(x−4))((x+1)−(x−4)) which is equal to (2x−3)(5)=10x−15
    • 4m²−25n²=(2m+5n)(2m−5n)

Exam Hint

• As a check, expand your answer and make sure you get the same expression as the one you were trying to factorise.
• You should be able to recognise a difference of squares in both factorised and unfactorised form

Worked Example (a) Factorise $x^{2}-4x-21$.

•       We will factorise by inspection*
  

•      We need two numbers that multiply to *$-21$ *and add to* $-4$
  

•     *$+3$ *and* $-7$ *satisfy this*
  

•      Write down the brackets*
  

$(x+3)(x-7)$

(b) Factorise $6x^{2}-7x-3$.

•    We will factorise by splitting the middle term and grouping*
  

•    We need two numbers that multiply to  *`6×(−3)=−18` *and  add to* $-7$* *
  

$+2$ and $-9$ satisfy this

•    Split the middle term*
  

$6x^{2}+2x-9x-3$

•    Factorise *$2x$ *out of the first two terms, and* $-3$ *out of the last two terms*
  

2x (3x+1)−3 (3x+1)

•    These have a common factor of *`(3x+1)` *which can be factored out*
  

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

(c) Factorise $9x^{2}-16$.

Recognise that this is a difference of two squares, because * *9x²−16=(3x)²−(4)²

•     Use the relation  *`a²−b²=(a+b)(a−b)`
  

$(3x+4)(3x-4)$