Question

if the solutions of p(x)=0 are -14 and 11, which function could be p?
(A) p(x)=x^2-3x-154

(B) p(x)=x^2-14x+11

(C) p(x)=x^2+14x+11

(D) p(x)=x^2+3x-154

Answer 1

Answer:d

Step-by-step explanation:

Answer 2

The required function is $P(x)=x^2+3x+154$. So, option D is correct.

Important information:

• The solutions of $p(x)=0$ are $-14$ and 11.

We need to find the function p(x).

Function:

If $x=c$ is a solution of $p(x)=0$, then $(x-c)$ is a factor of p(x).

It means $(x+14)$ and $(x-11)$ are two factor of p(x).

$P(x)=(x+14)(x-11)$

$P(x)=x^2-11x+14x+154$

$P(x)=x^2+3x+154$

Therefore, the correct option is D.

Find out more about 'Function' here:

https://brainly.com/question/19446232