17.03 Factorising trinomials (Extended)
We call expressions of the form $x^2+mx+n$x2+mx+n, where $x$x is a pronumeral and $m$m and $n$n are numbers, monic quadratic trinomials. In order to factorise these, we want to use the rule $AC+AD+BC+BD=\left(A+B\right)\left(C+D\right)$AC+AD+BC+BD=(A+B)(C+D), but there are three terms instead of four.
The first term is $x^2$x2. Since $x$x is a pronumeral, we can't really split it up, so to fit the distributive law we want $A=x$A=x and $C=x$C=x. If we also let $B=p$B=p and $D=q$D=q, then we get $\left(x+p\right)\left(x+q\right)=x^2+px+qx+pq$(x+p)(x+q)=x2+px+qx+pq by expansion. We can then factorise $x$x from the two middle terms to get $x^2+\left(p+q\right)x+pq$x2+(p+q)x+pq.
Comparing this to the monic quadratic expression we have $x^2+mx+n=x^2+\left(p+q\right)x+pq$x2+mx+n=x2+(p+q)x+pq. Equating the coefficients of $x$x tells us $m=p+q$m=p+q and $n=pq$n=pq. This means that there are two numbers, $p$p and $q$q which add to give $m$m and multiply to give $n$n. If we can find these two numbers we can factorise the monic quadratic expression.
Worked example
Fully factorise $x^2+7x-18$x2+7x−18.
Think: In this question, $m=7$m=7 and $n=-18$n=−18. In order to factorise the expression we want to find numbers $p$p and $q$q which add to $7$7 and multiply to $-18$−18.
Do: There are infinite pairs of numbers which add to $7$7, but there are a finite pairs of numbers which multiply to $-18$−18. Since $-18$−18 is negative, we know that one factor is positive and one is negative. First, let's list all of the positive factor pairs of $18$18:
- $1$1 and $18$18
- $2$2 and $9$9
- $3$3 and $6$6
Now we can try making one of the factors negative (so that the product is $-18$−18) and then adding the two numbers to see if the sum is $7$7.
| $p$p | $q$q | $p+q$p+q |
|---|---|---|
| $1$1 | $-18$−18 | $-17$−17 |
| $-1$−1 | $18$18 | $17$17 |
| $2$2 | $-9$−9 | $-7$−7 |
| $-2$−2 | $9$9 | $7$7 |
| $3$3 | $-6$−6 | $-3$−3 |
| $-3$−3 | $6$6 | $3$3 |
From this, we can see that $-2+9=7$−2+9=7 and $-2\times9=-18$−2×9=−18, which means that $p=-2$p=−2 and $q=9$q=9. From this we can factorise the original expression:
$x^2+7x-18=\left(x-2\right)\left(x+9\right)$x2+7x−18=(x−2)(x+9)
Reflect: All we needed to do to factorise the expression was find two numbers which add to $7$7 and multiply to $-18$−18. It doesn't matter which order the numbers go in, but it does matter what the signs are. Consider that if the constant term was positive, then $p$p and $q$q would either both be positive or both be negative.
An expression of the form $x^2+mx+n$x2+mx+n is a monic quadratic trinomial.
To factorise expressions like this, we find a pair of numbers $p$p and $q$q such that $p+q=m$p+q=m and $pq=n$pq=n.
Then the factorisation is $x^2+mx+n=\left(x+p\right)\left(x+q\right)$x2+mx+n=(x+p)(x+q).
Practice questions
Question 1
Factorise: $x^2+11x+24$x2+11x+24
Question 2
Factorise: $x^2-4x+4$x2−4x+4
Question 3
Factorise: $x^2-3x-70$x2−3x−70
Question 4
Factorise: $35-12x+x^2$35−12x+x2