Factorisation techniques
Factorisation means to write an expression as a product of factors. For example when we write $12=2\times6$12=2×6, the factors are $2$2 and $6$6. When we write $x^2+2x=x\left(x+2\right)$x2+2x=x(x+2), the factors are $x$x and $x+2$x+2. The factors of algebraic expressions are often written in brackets.
Here's a list of the most common techniques you can use to factorise:
- 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 ($a=1$a=1): 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!
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: Always look for a common factor first. In this case $3$3 is the common factor. Then think about which 3 term method to use next.
Do: Taking out a common factor of 3 gives
$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. We look for numbers that multiply to make $-30$−30 and add to make $-1$−1.
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).
A monic quadratic has $a=1$a=1. This means there is no number written in front of the $x^2$x2 term. Eg: $x^2+2x+1$x2+2x+1.
A non-monic quadratic has $a\ne1$a≠1. This means there is a number, that is not a common factor, in front of the $x^2$x2term. Eg: $5x^2+3x+7$5x2+3x+7.
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
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 that multiply to make $5x^2$5x2 on one side and another possible factor pair that multiply to make $-12$−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:
We cross multiply the terms and add the results and this must make the middle term which is $11x$11x. $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, we find the two factors. The quadratic must then factorise to $\left(5x-4\right)\left(x+3\right)$(5x−4)(x+3).
PSF method (or 'splitting the middle term')
The PSF (Product, Sum, Factor) method uses a similar idea we had with monic quadratics where we think about sums and products, but with a slight difference. This method is also called 'splitting the middle term'.
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 our two numbers $m$m & $n$n should be.
Do: We want the sum 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 | Split the $11x$11x into $-4x$−4x + $15x$15x |
| $=$= | $x\left(5x-4\right)+3\left(5x-4\right)$x(5x−4)+3(5x−4) | Use grouping: the first two terms have a common factor of $x$x and the second have a common factor of $3$3 |
|
| $=$= | $\left(5x-4\right)\left(x+3\right)$(5x−4)(x+3) | $5x-4$5x−4 is now a common factor. It goes in one bracket and $x+3$x+3 in the other. |
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: We need two numbers that multiply to make $5\times7=35$5×7=35 and add to make $-36$−36.
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 | Put $5x$5xin each bracket with $m$m in one bracket and $n$n in the other. Divide by $5$5 as you have an 'extra' $5$5 in the numerator at the moment. |
| $=$= | $\frac{\left(5x-1\right)\left(x-7\right)\times5}{5}$(5x−1)(x−7)×55 | Factorise $5$5 as a common factor from the second bracket. | |
| $=$= | $\left(5x-1\right)\left(x-7\right)$(5x−1)(x−7) | Cancel the $5$5 in numerator and denominator to determine the factorised expression. |
Factorising with the CAS calculator
Your CAS calculator is able to factorise expressions for you too. However you will be expected to be able to factorise expressions in the non-calculator paper in your examinations and tests, so you will still need to be able to use the strategies above to factorise without the calculator.
Experiment with your calculator and find out how to factorise using it. On the Casio classpad you choose Action►Transformation►factor►factor from the menu in the main screen. Make sure you have cleared all variables first and the calculator will factorise for you. Try the practice questions below using manual methods then check your answers using the calculator.
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.