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zvonat [6]
3 years ago
11

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

Mathematics
1 answer:
Mila [183]3 years ago
6 0

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

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