NZ Level 6 (NZC) Level 1 (NCEA)
Further factorisations
Lesson

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.
6. Non-monic quadratics: Like monic quadratics, but the coefficient of $x^2$x2 is not one, e.g. $Px^2+Qx+R$Px2+Qx+R. Here we need four numbers, $A$A, $B$B, $C$C and $D$D where $A\times B=P$A×B=P, $C\times D=R$C×D=R and $A\times D+B\times C=Q$A×D+B×C=Q. This gives us the factorisation $\left(Ax+C\right)\left(Bx+D\right)$(Ax+C)(Bx+D).
7. Sums and differences of cubes: An expression of the form $A^3+B^3$A3+B3 has the factorisation $\left(A+B\right)\left(A^2-AB+B^2\right)$(A+B)(A2AB+B2). Similarly, an expression of the form $A^3-B^3$A3B3 has the factorisation $\left(A-B\right)\left(A^2+AB+B^2\right)$(AB)(A2+AB+B2).

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:

#### Examples

##### Question 1

Factorise $\left(y+11y^2\right)^2-y^2$(y+11y2)2y2 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)
##### Question 2

Factorise and simplify $\left(3m-n\right)\left(4n+7m\right)-2n+6m$(3mn)(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(n3m). If we take out $2$2 then we get $2\left(-n+3m\right)=2\left(3m-n\right)$2(n+3m)=2(3mn).

Since we have $\left(3m-n\right)$(3mn) 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)
##### Question 3

Factorise and simplify completely: $j^3-27j^2k+50jk^2$j327j2k+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 $\left(j^2-26jk+50k^2\right)$(j226jk+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)

##### Question 4

Factorise the following expression:

$x^2-6x+9-y^2$x26x+9y2.

##### Question 5

Factorise the expression $20x^2-25x-30$20x225x30.

##### Question 6

Factor the polynomial $\left(k+3\right)^3+8$(k+3)3+8 by using the substitution $u=k+3$u=k+3.

### Outcomes

#### NA6-5

Form and solve linear equations and inequations, quadratic and simple exponential equations, and simultaneous equations with two unknowns

#### NA6-6

Generalise the properties of operations with rational numbers, including the properties of exponents

#### 91027

Apply algebraic procedures in solving problems