Wikipedia:Reference desk/Archives/Mathematics/2014 January 20

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

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

but this is of no help whatsoever in factoring $\scriptstyle {p^{4}(1+4q+10q^{2}+20q^{3})+q^{4}(1+4p+10p^{2}+20p^{3})}$ (without the -1 term). In fact there is no multinomial in $\scriptstyle {p}$ and $\scriptstyle {q}$ 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)