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:

  • $\br{-1} \times \br{3^4 \times 7^2}$
  • $\br{-3} \times \br{3^3 \times 7^2}$
  • $\br{-7} \times \br{3^4 \times 7}$
  • $\br{-3^2} \times \br{3^2 \times 7^2}$
  • $\br{-3\times 7} \times \br{3^3 \times 7}$
  • $\br{-3^3} \times \br{3 \times 7^2}$
  • $\br{-7^2} \times \br{3^4}$
  • $\br{-3^2 \times 7} \times \br{3^2 \times 7}$

(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!