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 $\left(y+11y^2\right)^2-y^2$(y+11y2)2−y2 completely
Think: We can treat the expression in the brackets as one term.
Do:
What we can see here is the difference of two squares, where our $A=y+11y^2$A=y+11y2 and $B=y$B=y
Therefore:
$\left(y+11y^2\right)^2-y^2$(y+11y2)2−y2 | $=$= | $\left(\left(y+11y^2\right)+y\right)\left(\left(y+11y^2\right)-y\right)$((y+11y2)+y)((y+11y2)−y) |
$=$= | $\left(2y+11y^2\right)\times11y^2$(2y+11y2)×11y2 | |
$=$= | $y\left(2+11y\right)\times11y^2$y(2+11y)×11y2 | |
$=$= | $11y^3\left(2+11y\right)$11y3(2+11y) |
Factorise and simplify $\left(3m-n\right)\left(4n+7m\right)-2n+6m$(3m−n)(4n+7m)−2n+6m
Think: The first term has always been factorised, so we can factorise the rest of the expression first. Also be aware of negatives.
Do:
Let's concentrate on the $-2n+6m$−2n+6m part first. We can either take out $-2$−2 or $2$2 using HCF factorisation.
If we take out $-2$−2 then we get $-2\left(n-3m\right)$−2(n−3m). If we take out $2$2 then we get $2\left(-n+3m\right)=2\left(3m-n\right)$2(−n+3m)=2(3m−n).
Since we have $\left(3m-n\right)$(3m−n) in the first term, we should take $2$2 out.
$\left(3m-n\right)\left(4n+7m\right)-2\left(n-3m\right)$(3m−n)(4n+7m)−2(n−3m) | $=$= | $\left(3m-n\right)\left(4n+7m\right)+2\left(3m-n\right)$(3m−n)(4n+7m)+2(3m−n) |
$=$= | $\left(3m-n\right)\left(4n+7m+2\right)$(3m−n)(4n+7m+2) |
Factorise and simplify completely: $j^3-27j^2k+50jk^2$j3−27j2k+50jk2
Think: What kind of expression do we have after factorising the HCF $j$j?
Do:
$j^3-27j^2k+50jk^2$j3−27j2k+50jk2 | $=$= | $j\left(j^2-27jk+50k^2\right)$j(j2−27jk+50k2) |
Notice that $\left(j^2-26jk+50k^2\right)$(j2−26jk+50k2) is a quadratic trinomial.
We need two numbers that have a product of $50k^2$50k2. We can either have a $k^2$k2 term and a number, or two $k$k terms. Looking at the fact that we need the sum to be $-27k$−27k, that means we're looking for $2$2 negative $k$k terms.
Possible factor pairs of $50k^2$50k2 are$-50k$−50k & $-k$−k, $-25k$−25k & $-2k$−2k, and $-10k$−10k & $-5k$−5k.
$-25k$−25k & $-2k$−2k give us our sum of $-27k$−27k, therefore:
$j\left(j^2-27jk+50k^2\right)$j(j2−27jk+50k2) | $=$= | $j\left(j-25k\right)\left(j-2k\right)$j(j−25k)(j−2k) |
Factorise the following expression:
$x^2-6x+9-y^2$x2−6x+9−y2.
Factorise the expression $20x^2-25x-30$20x2−25x−30.
Factor the polynomial $\left(k+3\right)^3+8$(k+3)3+8 by using the substitution $u=k+3$u=k+3.