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.
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).
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.
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!
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
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're 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) |
Factorise the trinomial:
$7x^2-75x+50$7x2−75x+50
Factorise the following trinomial:
$6x^2+13x+6$6x2+13x+6
Factorise $-12x^2-7x+12$−12x2−7x+12.