Factorising quadratics when a isn't 1

I got taught then when factorising a quadratic when a isn't 1 to just basically do trial and error and I wanted to know if there was a more efficient method than just trying a bunch of combinations.

What do you mean by trial and error?

Assuming roots of a quadratic are rational (meaning your leading coefficient is tyically not 1) then you can factorise it.

Let's say the roots are $x = -\dfrac{p}{q}$ and $x = - \dfrac{r}{s}$ and these correspond to linear factors $(qx+p)$ and $(sx+r)$ being zero.

Therefore, the two roots satisfy the quadratic equation $(qx+p)(sx+r)=0$, i.e. $qs x^2 + (qr+ps) x + pr = 0$

The leading coefficient is $qs$.

Here $qr,ps$ are two magic numbers which add to give the middle coefficient, but they also multiply to give $qrps$ which is exactly the same as the leading coefficient $qs$ multiplying the constant term $pr$.

All this is to say that there some logic involved here in factorising. When trying to factorise $ax^2 + bx + c = 0$ you should be splitting your $b$ into two magic numbers which add to $b$ yet multiply to $ac$.

As they multiply to $ac$, you just need to look at the factor pairs of this product and see what adds to $b$.

Example: Factorise $2x^2 + 5x + 2$.

Look for two numbers which add to $5$ and multiply to $2\times 2 = 4$.

These are $1,4$. Then rewrite your quadratic as $(2x^2 + x) + (4x+2)$ [we have split $5x=x+4x$] and factorise fully each bracket to get

$x(2x+1) + 2(2x+1)$

Finally note that $(2x+1)$ is a bracket occuring twice in this expression therefore it can be factored out overall to give

$(2x+1)(x+2)$

This is the factorisation. No trial and error needed. Just find the two magic numbers by looking at factor pairs of $ac$ which add to $b$.

Thanks for the explanation!
We got taught to find factors of a and put them into the brackets first and then fill in the rest of the bracket with factors of c and do trial and error until the x values add up to b.

OK, I see.

That works too, but this the above method you might prefer it. Significantly fewer options to check when looking at factors of ac in one go rather than choosing a factor pair of a then looking at factors of c.

Try both methods on this one example:

$12x^2 + 7x + 12$

to get $(3x+4)(4x-3)$.