Ask Uncle Colin: A factorising trick

Dear Uncle Colin,

How would you factorise $63x^2+32x-63$? I tried the method where you multiply $a$ and $c$ (it gives you -3969) - but I’m not sure how to find factors of that that sum to 32!

Factors Are Troublesomely Oversized, Urgh

Hi, FATOU, and thanks for your message!

When the numbers in a quadratic like this get large (in mental arithmetic terms, at least), I try not to think about the number itself, but about its prime factorisation.

In this case, we know -3969 is $-63\times 63$, making it $-3^2\times 7\times 3^2\times 7$ or $-3^4\times 7^2$.

How does that help?

One approach is to go systematically through the possible factor splits like this:

• $(-1)\times (3^4\times 7^2)$
• $(-3)\times (3^3\times 7^2)$
• $(-7)\times (3^4\times 7)$
• $(-3^2)\times (3^2\times 7^2)$
• $(-3\times 7)\times (3^3\times 7)$
• $(-3^3)\times (3\times 7^2)$
• $(-7^2)\times \left(3^4\right)$
• $(-3^2\times 7)\times (3^2\times 7)$

(There are $(4+1)\times (2+1)=15$ factors of $3^4\times 7^2$ - we’ve found eight pairs, one of which is a ‘double’, so we have them all).

We could then add up each pair to see what we wind up with.

But there’s a trick!

We know that 32 is not a multiple of 3, and not a multiple of 7. Any factor pair with a 3 in each factor, or a 7 in each factor, cannot possibly sum to 32! That reduces our workload considerably.

The only possibilities are the first - and $-1+3969$ is definitely not 32 - and the penultimate: $-49+81$ does indeed make 32.

(Had the middle number been a multiple of 3, or of 7, or of both, we could have used a similar idea: in those cases, the 3s, the 7s, or both must have been split across the factors.)

Finishing it off

We can write the quadratic as $63x^2+81x-49x-63$, which is $9x(7x+9)-7(7x+9)$, or $(9x-7)(7x+9)$.

Hope that helps!