Question

Which of the number $1,-1$ and $-3$ are zeroes of the polynomial $2x^4+9x^3+11x^2+4x-6$?

• A. 1
• B. -1
• C. -3
• D. None of these

Answer 1

Answer: (C)

-3

Let $f\left(x\right)=2x^4+9x^3+11x^2+4x-6$.

$f\left(1\right)=2(1{)}^4+9(1{)}^3+11(1{)}^2+4(1)-6$
$=2+9+11+4-6=20\ne 0$.
$\therefore$ $1$ is not a zero of the polynomial $f\left(x\right)$.

Again $f(-1)=2(-1{)}^4+9(-1{)}^3+11(-1{)}^2+4(-1)-6$
$=2-9+11-4-6=-6\ne 0$0
$\therefore$ $-1$ is not a zero of the polynomial $f\left(x\right)$.

Also $f(-3)=2(-3{)}^4+9(-3{)}^3+11(-3{)}^2+4(-3)-6$
$=162-243+99-12-6=0$0
$\therefore$ $-3$ is a zero of the polynomial $f\left(x\right)$.

Thus $1$ and $-1$ are not zeros of $f\left(x\right)$, whereas $-3$ is zero of $f\left(x\right)$.
Therefore, option $C$ is correct.