Question

What will be the factor of polynomial (x+1)(x+2)(x+3)(x+4)-8 ?

Answer 1

Since x=0x=0 and x=−5x=−5 are roots of the given equation,

(x+1)(x+2)(x+3)(x+4)−24=x(x+5)⋅q(x)(1)(1)(x+1)(x+2)(x+3)(x+4)−24=x(x+5)⋅q(x)where q(x)q(x) is a monic second-degree polynomial. We may notice that, by De l'Hopital's rule,q(0)=limx→0(x+1)(x+2)(x+3)(x+4)−24x(x+5)=24H45=10(2)(2)q(0)=limx→0(x+1)(x+2)(x+3)(x+4)−24x(x+5)=24H45=10and if q(x)=x2+Kx+10q(x)=x2+Kx+10, in order that the coefficient of x3x3 is the same in both sides of (1)(1)1+2+3+4=K+5