Find the third-degree polynomial function that has zeros −2 and −15i, and a value of 1,170 when x=3.

The third degree polynomial function = x³ + 27x² + 200x + 300

Step-by-step explanation:

The third-degree polynomial function has zeros −2 and −15.

From the above, we have been given two factors of the polynomial function. Let's derive the factors from the two zeros of the polynomial given.

The two given zeros of the polynomial can be written as:

x= -2

x+2 = 0

(x+2) is a factor of the polynomial

x= -15

x+15 = 0

(x+15) is a factor of the polynomial

So we have two factors of the polynomial (x+2) and (x+15). But since it is a third degree polynomial, we have to find the third factor.

Let (x-b) be the third factor and f(x) represent the third degree polynomial

f(x) = (x-b) (x+2) (x+15)

Expanding (x+2) (x+15) = x² + 2x + 15x + 30

(x+2) (x+15) = x² + 17x + 30

f(x) = (x-b) (x² + 17x + 30)

From the given information, a value of 1,170 is obtained when x=3

f(3) = 1170

Insert 3 for x in f(x)

f(3) = (3-b) (3² + 17(3) + 30)

1170 = (3-b) (9 + 51 + 30)

1170 = (3-b) (90)

1170/90 = 3-b

3-b = 13

b = 3-13 = -10

Insert value of b in f(x)

f(x) = [x-(-10)] (x² + 17x + 30)

f(x) = (x+10) (x² + 17x + 30)

f(x) = x³ + 17x² + 30x + 10x² + 170x + 200x + 300

f(x) = x³ + 27x² + 200x + 300

The third degree polynomial function = x³ + 27x² + 200x + 300

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step-by-step explanation: