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

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


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


 * but this is of no help whatsoever in factoring $$\scriptstyle{p^{4} (1 + 4 q + 10 q^{2} + 20 q^{3}) + q^{4} (1 + 4 p + 10 p^{2} + 20 p^{3})}$$ (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)