Factorisation techniques
Before solving quadratic equations let's review common factorisation techniques. Looking for factorisation first when solving quadratics is a quick and effective strategy before leaping to the quadratic formula. Here's a list:
- Highest Common Factor (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.
- Difference of two squares: $A^2-B^2=\left(A+B\right)\left(A-B\right)$A2−B2=(A+B)(A−B), so look for the difference of two terms which are both perfect squares.
- Grouping in pairs: Look for four terms where you can split them up into two pairs and factorise separately, then finally factorise using HCF factorisation afterwards.
- Perfect squares: $A^2+2AB+B^2=\left(A+B\right)^2$A2+2AB+B2=(A+B)2, $A^2-2AB+B^2=\left(A-B\right)^2$A2−2AB+B2=(A−B)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.
- Sum and product for 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).
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:
Worked examples
example 1
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 HCF factorisation where $x-5$x−5 is the highest common factor |
example 2
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)
Practice questions
Question 1
Factorise $k^2-81$k2−81.
Question 2
Factorise $x^2+12x+36$x2+12x+36.
Factorisation for non-monic quadratics
So far, most of the quadratics we've dealt with are monic, meaning their $x^2$x2 term only has a coefficient of $1$1. If the coefficient is not $1$1, then we've usually found we can factorise out that coefficient from the whole quadratic.
eg. $2x^2-4x+6=2\left(x^2-2x+3\right)$2x2−4x+6=2(x2−2x+3).
But how do we factorise quadratics that can't be simplified in this way? First let's have a look at how a non-monic quadratic is composed:
Now we are more familiar with these tricky quadratics let's have a look at the three different methods below.
Cross method
We've already encountered the cross method once before with monic quadratics, and it's easy to see how this extends into non-monic territory.
For example, let's have a look at $5x^2+11x-12$5x2+11x−12. We must draw a cross with a possible pair of factors of $5x^2$5x2 on one side and another possible factor pair of $-12$−12 on the other side.
Let's start with the factor pairs of $5x$5x & $x$x on the left, and $-6$−6 & $2$2 on the other:
$5x\times2+x\times\left(-6\right)=4x$5x×2+x×(−6)=4x, which is incorrect, so let's try again with another two pairs:
$5x\times3+x\times\left(-4\right)=11x$5x×3+x×(−4)=11x which is the right answer. By reading across in the two circles, the quadratic must then factorise to $\left(5x-4\right)\left(x+3\right)$(5x−4)(x+3).
PSF method
The PSF (Product, Sum, Factor) method uses a similar idea we had with monic quadratics where we think about sums and products, but slightly differently.
For a quadratic in the form $ax^2+bx+c$ax2+bx+c:
1. Find two numbers, $m$m & $n$n, that have a SUM of $b$b and a PRODUCT of $ac$ac.
2. Rewrite the quadratic as $ax^2+mx+nx+c$ax2+mx+nx+c.
3. Use grouping in pairs to factorise the four-termed expression.
Worked example
example 3
Using the same example as above, factorise $5x^2+11x-12$5x2+11x−12 using the PSF method.
Think about what the sum and product of $m$m & $n$n should be
Do
We want the sum of of $m$m & $n$n to be $11$11, and the product to be $5\times\left(-12\right)=-60$5×(−12)=−60.
The two numbers work out to be $-4$−4 & $15$15, so:
| $5x^2+11x-12$5x2+11x−12 | $=$= | $5x^2-4x+15x-12$5x2−4x+15x−12 |
| $=$= | $x\left(5x-4\right)+3\left(5x-4\right)$x(5x−4)+3(5x−4) | |
| $=$= | $\left(5x-4\right)\left(x+3\right)$(5x−4)(x+3) |
This is the same answer that we got before!
PSF variation
The above two methods are the most often used. However, a slightly different method can also be used to factorise directly if you can remember the formula.
$ax^2+bx+c=\frac{\left(ax+m\right)\left(ax+n\right)}{a}$ax2+bx+c=(ax+m)(ax+n)a, where $m+n=b$m+n=b & $mn=ac$mn=ac
Worked example
example 4
Factorise $5x^2-36x+7$5x2−36x+7 completely
Think about whether it is easier to consider the product or the sum of $m$m & $n$n first
Do
| $m+n$m+n | $=$= | $b$b |
| $=$= | $-36$−36 | |
| $mn$mn | $=$= | $ac$ac |
| $=$= | $5\times7$5×7 | |
| $=$= | $35$35 |
It's much easier to look at the product first as there are less possible pairs that multiply to give $35$35 than those that add to give $-36$−36. We can easily see that $m$m & $n$n $=$= $-1$−1 & $-35$−35. Then:
| $5x^2-36x+7$5x2−36x+7 | $=$= | $\frac{\left(5x-1\right)\left(5x-35\right)}{5}$(5x−1)(5x−35)5 |
| $=$= | $\frac{\left(5x-1\right)\left(x-7\right)\times5}{5}$(5x−1)(x−7)×55 | |
| $=$= | $\left(5x-1\right)\left(x-7\right)$(5x−1)(x−7) |
Practice questions
Question 3
Factorise the following trinomial:
$6x^2+13x+6$6x2+13x+6
Question 4
Factorise $-12x^2-7x+12$−12x2−7x+12.