Question

Find the value of p for which x+1 is a factor of x^4+(p-3)x^3-(3p-5)x^2+2p-9)x+6

Answer 1

$let \: f(x) = {x}^{4} + (p - 3) {x}^{3} - (3p - 5) {x}^{2} + (2p - 9)x + 6 \\ given \: that(x +1)is \: factor \: of \: f(x) \\ than \: \: f(-1) = 0 \\ i.e. {-1}^{4} + (p - 3) {-1}^{3} - (3p - 5) {-1}^{2} + (2p - 9)-1 + 6 = 0 \\ 1 - p + 3 - 3p + 5 - 2p + 9+ 6 = 0 \\ 24-6p= 0 \\6p=24\\p=4\\ hence \: for\: p=4 \: above \: statement \: is \: correct$