Question

If $(x-1),(x+1)$ and $(x-2)$ are factors of $x^4+(p-3)x^3-(3p-5)x^2+(2p-9)x+6$, then the value of $p$ is :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

Since we are given three factors of $a$ polynomial, the value of function at each of the $3$ values must come out as $0.$
For $x=1,2$, the function vanishes for all values of $p.$
For $x=-1,$ $(-1{)}^4$ $+(p-3)\times (-1{)}^3$ $-(3p-5)\times (-1{)}^2$ $+(2p-9)\times -1$ $+6=0$
$\therefore$ $1-(p-3)-(3p-5)-(2p-9)+6=0$
$\therefore$ $1-p+3-3p+5-2p+9+6=0$
$\therefore$ $-6p+24=0$
$\therefore$ $p=4$