Question

Given that x+1 and x-3 are factors of the expression x⁴+px³+5x²+5x+q, find the value of p and q. Hence find the other two factors of expression.

Answer 1

Answer:

$p(x)=x^{4}+px^{3}+5x^{2}+5x+q$

Remainder theorem: if x-a is factor then p(a) =0

x+1 = 0

x=-1

$p(-1) = (-1)^{4}+p(-1)^{3}+5(-1)^2+5(-1)+q\\0 = 1-p+5-5+q\\p-q=1$

x-3=0

x=3

$p(3) = (3)^{4}+p(3)^{3}+5(3)^2+5(3)+q\\0 = 81+27p+45+15+q\\27p+q=-141$

solving 2 equations

$p =-5\\q=-6$

$p(x)=x^{4}-5x^{3}+5x^{2}+5x-6$

other 2 factors

(x-1) and (x-2)