Question

Verify that the numbers given along side of the polynomial are their zeros. $x^4+2x^3-7x^2-8x+12;-3,-2,1,2$

• A. Yes, $-3,-2,1,2$ are the zeros of given polynomial
• B. No, $-3,-2,1,2$ are the zeros of given polynomial
• C. Ambiguous
• D. Data insufficient

Answer 1

The correct option is

B

Yes, $-3,-2,1,2$ are the zeros of given polynomial

Here the polynomial $p\left(x\right)$ is
$x^4+2x^3-7x^2-8x+12$
Value of the polynomial $x^4+2x^3-7x^2-8x+12$
when $x=-3$
$P(-3)=(-3{)}^4+2(-3{)}^3-7(-3{)}^2-8(-3)+12$
$=81-54-63+24+12=0$
So, $-3$ is a zero of $p\left(x\right)$.

On putting $x=-2$ in the given polynomial, we have
$P(-2)=(-2{)}^4+2(-2{)}^3-7(-2{)}^2-8(-2)+12$
$=16-16+28+16+12=0$
So, $-2$ is a zero of $p\left(x\right)$.

On putting $x=1$ in the given polynomial, we have
$P\left(1\right)=(1{)}^4+2(1{)}^3-7(1{)}^2-8(1)+12$
$=1+2-7-8+12=0$
So, $2$ is a zero of $p\left(x\right)$.

On putting $x=2$ in the given polynomial, we have
$P\left(2\right)=(2{)}^4+2(2{)}^3-7(2{)}^2-8(2)+12$
$=16+16-28-16+12=0$

So, $2$ is a zero of $p\left(x\right)$.
Hence, $-3,-2,1,2$ are the zeros of given polynomial.