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 $-3^4 \times 7^2$.
One approach is to go systematically through the possible factor splits like this:
(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.
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.)
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!