factorize polynomials

Question

(x+1)(x+3)(x+5)(x+7)+16

Answer 1

This question needed a bit of thinking outside the box.

If the existing factors are grouped thus: ((x+1)(x+7))((x+3)(x+5)) we have the product of two quadratics: x^2+8x+7 and x^2+8x+15. The only difference is the constant term.

Let y=x^2+8x, then the expression can be written:

(y+7)(y+15)+16=y^2+22y+105+16=y^2+22y+121=(y+11)^2=(x^2+8x+11)^2.

The quadratic can be factorised with irrational zeroes: x=-4+sqrt(5) giving us:

((x+4+sqrt(5))(x+4-sqrt(5)))^2

as the final factorisation.