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):
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:
Factorise $xy-5y-2x+10$xy−5y−2x+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 $xy-5y=y\left(x-5\right)$xy−5y=y(x−5) 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
$xy-5y-2x+10$xy−5y−2x+10 | $=$= | $y\left(x-5\right)-2\left(x-5\right)$y(x−5)−2(x−5) | |
$=$= | $\left(x-5\right)\left(y-2\right)$(x−5)(y−2) | using the basic factorisation where $x-5$x−5 is the HCF |
Factorise $3x^2-3x-90$3x2−3x−90 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.
$3x^2-3x-90=3\left(x^2-x-30\right)$3x2−3x−90=3(x2−x−30)
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.
Therefore:
$3\left(x^2-x-30\right)=3\left(x+5\right)\left(x-6\right)$3(x2−x−30)=3(x+5)(x−6)
Factorise $10x+50$10x+50.
Factorise $k^2-81$k2−81.
Factorise $x^2+12x+36$x2+12x+36.