Question

For items 16-19, write a polynomial function of nth degree that has the given real or complex zeros?

Answer 1

For n = 3, x = 9, and x = 2i, we can say that one of the factors is (x - 9) as well as (x - 2i).

(x - 2i) is derived from x² = -4 which is equal to (x² + 4). Therefore, the two factors are (x - 9) and (x² + 4). To get the polynomial function, let's multiply the two factors using FOIL Method.

`\begin{gathered} f(x)=(x-9(x^2+4_{}) \\ f(x)=(x)(x^2)+(x)(4)-(9)(x^2)-(9)(4) \\ f(x)=x^3+4x-9x^2-36 \\ \text{Arrange the terms} \\ f(x)=x^3-9x^2+4x-36 \end{gathered}`

The polynomial function of the first bullet is f(x) = x³ - 9x² + 4x - 36.

For n = 3, x = -1, and x = 4 + i, we can say that the factors are:

(x + 1) , (x - (4 + i)), and (x - (4 - i))

Note: Always remember those complex zeros like x = 4 + i come in conjugate pairs.

To solve the polynomial function, let's multiply the three factors.

`\begin{gathered} f(x)=(x+1)(x-4-i)(x-4+i) \\ \text{Multiply first the two factors that has imaginary number i.} \\ f(x)=(x+1)(x^2-4x+ix-4x+16-4i-ix+4i-i^2) \\ \text{Arrange the terms} \\ f(x)=(x+1)(x^2-4x-4x+ix-ix-4i+4i+16+1) \\ \text{Combine like terms} \\ f(x)=(x+1)(x^2-8x+17) \\ \text{Multiply binomial to the trinomial} \\ f(x)=(x)(x^2)+(x)(-8x)+(x)(17)+x^2-8x+17 \\ f(x)=x^3-8x^2+17x+x^2-8x+17 \\ f(x)=x^3-7x^2+9x+17 \end{gathered}`

The polynomial function of the second bullet is f(x) = x³ - 7x² + 9x + 17.