17.04 Factorising non-monic trinomials (Extended)
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 sometimes found we can factorise out that coefficient from the whole quadratic first.
For example: $2x^2-4x-6=2(x^2-2x-3)=2(x-3)(x+1)$2x2−4x−6=2(x2−2x−3)=2(x−3)(x+1).
But how do we factorise quadratics such as $2x^2-x-3$2x2−x−3 that can't be factorised with a common factor first? 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 may have already encountered the cross method before with monic quadratics, and it's easy to see how this extends into non-monic territory.
Worked example
example 1
Factorise $5x^2+11x-12$5x2+11x−12.
Think: 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.
Do: 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 different.
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 2
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 3
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 1
Factorise the trinomial:
$7x^2-75x+50$7x2−75x+50
Question 2
Factorise the following trinomial:
$6x^2+13x+6$6x2+13x+6
Question 3
Factorise $-12x^2-7x+12$−12x2−7x+12.