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Stage 4 - Stage 5

Mixed factorisations


Now that we know how to factorise using various methods, let's try to apply them to different examples and figure out which would work for each one! Here's a list of all we've learnt so far (click on the links to read more about them):

  1. HCF factorisation: $AB+AC+\dots=A\left(B+C+\dots\right)$AB+AC+=A(B+C+), can be with any number of terms, not just two. Just a case of finding the HCF. 
  2. Difference of two squares: $A^2-B^2=\left(A+B\right)\left(A-B\right)$A2B2=(A+B)(AB), so look for the difference of two terms which are both perfect squares. 
  3. Grouping in pairs: Look for four terms where you can split them up into two pairs and factorise separately, then finally factorise using basic factorisation afterwards.
  4. Perfect squares: $A^2+2AB+B^2=\left(A+B\right)^2$A2+2AB+B2=(A+B)2, $A^2-2AB+B^2=\left(A-B\right)^2$A22AB+B2=(AB)2, so look for three terms where the first and third terms are perfect squares, and the middle term is twice the product of their square roots. 
  5. Monic Quadratics: Look for three terms of the form $x^2+Px+Q$x2+Px+Q, where $P$P and $Q$Q are any numbers (and $x$x could be another variable!). Try and see if you can solve using the perfect square method first, otherwise find two numbers $A$A and $B$B that have a sum of $P$P and a product of $Q$Q, and the answer would be $\left(x+A\right)\left(x+B\right)$(x+A)(x+B). OR you could use the cross method as well. 

The key when facing questions involving these techniques is to figure out which to use and when. Remember to always check if your answer can be further factorised to finish answering the question! Let's have a look at some examples:



Question 1

Factorise $xy-5y-2x+10$xy5y2x+10 completely

Think about whether to take out positive or negative factors

Do: These four terms have no common factors so let's try grouping in pairs.

$y$y goes into the first two pairs and $2$2 goes into the last two.

So $xy-5y=y\left(x-5\right)$xy5y=y(x5) but $-2x+10=2\left(-x+5\right)$2x+10=2(x+5)

Therefore taking $2$2 out does not help us factorise further so let's try $-2$2 instead

$xy-5y-2x+10$xy5y2x+10  $=$= $y\left(x-5\right)-2\left(x-5\right)$y(x5)2(x5)  
  $=$= $\left(x-5\right)\left(y-2\right)$(x5)(y2) using the basic factorisation where $x-5$x5 is the HCF


Question 2

Factorise $3x^2-3x-90$3x23x90 completely

Think about which 3 term method to use here, and whether you'll need to use another method first

Do: This is a quadratic but non-monic.

We can not use the perfect squares technique here, but the three terms do have an HCF of $3$3, so let's factor that out first.


The expression in the brackets is a monic quadratic that is also not a perfect square.

Factor pairs of $-30$30 are $1$1 & $-30$30, $-1$1 & $30$30, $2$2 & $-15$15, $-2$2 & $15$15, $3$3 & $-10$10, $-3$3 & $10$10, $5$5 & $-6$6, and $-5$5 & $6$6

The only pair with a sum of $-1$1 is $5$5 & $-6$6.




Question 3

Factorise $10x+50$10x+50.


Question 4

Factorise $k^2-81$k281.


Question 5

Factorise $x^2+12x+36$x2+12x+36.

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