Q. If $(x + 1)$ is a factor of $x^{4} - (p - 3)x^{3} - (3p - 5)x^{2} + (2p - 7)x + 6$, then the value of $p$ is
A) 4
B) 2
C) 1
D) None of these
Explanation
(a)
If $(x + 1)$ is a factor of
$x^{4} - (p - 3)x^{3} - (3p - 5)x^{2} + (2p - 7)x + 6$, then by putting $x = - 1$, we get
$1 + \ (p - 3) - (3p - 5) - (2p - 7) + 6 = 0$
$\Rightarrow \ - 4p = - 16\ \Rightarrow p = 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.