Question

(x+1)(x+2)(x+3)(x+4)-120

Answer 1

(x+1)(x+2)(x+3)(x+4)-120

=[(x+4)(x+1)][(x+2)(x+3)]-120

=(x^2+5x+4)(x^2+5x+6)-120

=(x^2+5x)^2+10(x^2+5x)+24-120

=(x^2+5x)^2+10(x^2+5x)-96

=(x^2+5x+16)(x^2+5x-6)

=(x^2+5x+16)(x+6)(x-1)