Question

if x+1 is a factor of the polynomial (x-1)(2x^2+4x+p) find p​

Answer 1

Answer:

$x + 1 = 0 \\ x = - 1 \\ \\ (x - 1)(2 {x}^{2} + 4x + p) = 0 \\ \\ ( - 1 - 1)2 { (- 1})^{2} + 4( - 1) + p = 0 \\ \\ - 2 (2 - 4 + p) = 0 \\ \\ - 2( - 2 + p) = 0 \\ \\ 4 - 2p = 0 \\ \\ 2p = 4 \\ \\ p = 2$

Answer 2

$\sf \Large Solution!!$

$\sf Let\:x+1=0\quad x=-1$

$\sf f(x)=(x-1)(2x^{2}+4x+p)$

$\sf f(-1)=(-1-1)(2(-1)^{2}+4(-1)+p)$

$\sf \large Given,\:x+1\:is\:a\:factor,\:so$

$\sf 0=(-2)(2-4+p)$

$\sf 0=-4+8-2p$

$\sf 2p=4$

$\sf p=2$