Question

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

Answer 1

The polynomial is:

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

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 = -(12)/(12)`

`a = -1`

Thus

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

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