Question

a polynomial p(x)= x^4+3x^3+x^2+1 is divided by x+2 to obtain a remainder r. find the zeros of the polynomial q (x)=x^2+rx-154

Answer 1

The remainder is - 3

X2-3x-154 =0

X2 - 14x+ 11x - 154=0

X (x - 14) +11 ( x-14) =0

(x-14)(x+11)

X= 14, x = - 11

Zeros of the polynomial are 14 and - 11

Answer 2

Answer:

The zeros of the polynomial $q(x)=x^2-3x-154$ are $14$ and $-11$.

Step-by-step explanation:

Recall the Factor remainder theorem,

"If $p(x)$ is a polynomial and $a$ is a real number, then $p(a)$ is the remainder of the polynomial $p(x)$ when divide by $(x-a)$."

Step 1 of 2

Consider the given polynomial as follows:

$p(x)=x^4+3x^3+x^2+1$ . . . . . (1)

According to the question,

The remainder $r$ is obtained when $p(x)$ is divided by $(x+2)$, i.e.,

$p(x)$ is divided by $(x-(-2))$.

Substitute the value $-2$ for $x$ in the equation (1) as follows:

$p(-2)=(-2)^4+3(-2)^3+(-2)^2+1$

          $=16-24+4+1$

$p(-2)=-3$

By Factor remainder theorem,

$-3$ is the remainder of the polynomial $p(x)$ when divided by $(x+2)$.

Thus, $r=-3$.

Step 2 of 2

To find: The zeros of the polynomial $q(x)$.

Consider the polynomial $q(x)$ as follows:

$q(x)=x^2+rx-154$

Since $r=-3$. Then,

$q(x)=x^2-3x-154$

Using middle term splitting method, we have

⇒ $x^2-14x+11x-154$

⇒ $x(x-14)+11(x-14)$

⇒ $(x-14)(x+11)$

⇒ $x=14$ and $x=-11$

Therefore, the zeros of the polynomial $x^2-3x-154$ are $14$ and $-11$.

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