Question

p(x)=Third-degree, with zeros of −3, −1, and 2, and passes through the point (1,12).

Answer 1

Answer:

The polynomial is:

$p(x) = -x^3 - 2x^2 + 5x + 6$

Step-by-step explanation:

Zeros of a function:

Given a polynomial f(x), this polynomial has roots $x_{1}, x_{2}, x_{n}$ such that it can be written as: $a(x - x_{1})*(x - x_{2})*...*(x-x_n)$, in which a is the leading coefficient.

Zeros of −3, −1, and 2

This means that $x_1 = -3, x_2 = -1, x_3 = 2$. Thus

$p(x) = a(x - x_{1})*(x - x_{2})*(x-x_3)$

$p(x) = a(x - (-3))*(x - (-1))*(x-2)$

$p(x) = a(x+3)(x+1)(x-2)$

$p(x) = a(x^2+4x+3)(x-2)$

$p(x) = a(x^3+2x^2-5x-6)$

Passes through the point (1,12).

This means that when $x = 1, p(x) = 12$. We use this to find a.

$12 = a(1 + 2 - 5 - 6)$

$-12a = 12$

$a = -\frac{12}{12}$

$a = -1$

Thus

$p(x) = -(x^3+2x^2-5x-6)$

$p(x) = -x^3 - 2x^2 + 5x + 6$