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January 20

Algebraic factorisation

This expression arose in some probability calculations I was doing on a square grid:

p^4 (1 + 4q + 10q^2 + 20q^3) + q^4 (1 + 4p + 10p^2 + 20p^3)

For p+q=1 it should have value 1 (checked with some trial values), suggesting that some power of (p+q) is a factor. Is it possible to factorise it in these terms?→31.54.113.130 (talk) 22:11, 20 January 2014 (UTC).

p+q is not a factor. Setting q=-p does not give zero. Sławomir Biały (talk) 23:30, 20 January 2014 (UTC)

The fact that $p^4(1+4q+10q^2+20q^3)+q^4(1+4p+10p^2+20p^3)=1$ when $p+q=1$ means that $p+q-1$ is a factor of $p^4(1+4q+10q^2+20q^3)+q^4(1+4p+10p^2+20p^3)-1$. The factorization is

$p^4(1+4q+10q^2+20q^3)+q^4(1+4p+10p^2+20p^3)-1=(q+p-1)(20q^3{p}^3+10q^3{p}^2+4q^{3}p+q^3+10q^2{p}^3+6q^2{p}^2+3q^{2}p+q^2+4q{p}^3+3q{p}^2+2qp+q+p^3+p^2+p+1)$

but this is of no help whatsoever in factoring $p^4(1+4q+10q^2+20q^3)+q^4(1+4p+10p^2+20p^3)$ (without the -1 term). In fact there is no multinomial in $p$ and $q$ with with integer or rational coefficients which is is a factor of your original expression (other than 1 and the expression itself). --catslash (talk) 00:25, 21 January 2014 (UTC)

Thanks, I wasn't thinking straight. I now realise that if (p+q) is taken out as a factor of some of the terms it can then be set to 1 and vanish, whereupon the expression shortens and can be simplified in a like manner until eventually it does become a power of (p+q) and thus seen to have value 1. This is less tedious than substituting for one of p and q as 1 minus the other and then expanding all of the powers.→31.54.113.130 (talk) 19:53, 21 January 2014 (UTC)