Question

Question

Find the third degree polynomial function that has an output of 320 when x = 4, and has zeros 2 and
-8i

Answer 1

f(x)=2(x-2)(x^2+64)

Explanation:

Answer 2

`f(x)=2(x-2)(x^2+64)`

Explanation:

A standard polynomial in factored form is given by:

`f(x)=a(x-p)(x-q)...`

Where p and q are the zeros.

We want to find a third-degree polynomial with zeros x = 2 and x = -8i and equals 320 when x = 4.

First, by the Complex Root Theorem, if x = -8i is a root, then x = 8i must also be a root.

Therefore, we acquire:

`f(x)=a(x-(2))(x-(-8i))(x-(8i))`

Simplify:

`f(x)=a(x-2)(x+8i)(x-8i)`

Expand the second and third factors:

`=(x+8i)x+(x+8i)(-8i)\\\\=(x^2+8ix)+(-8ix-64i^2)\\\\=(x^2)+(8ix-8ix)+(-64i^2)\\\\=x^2-64(-1)\\\\ =x^2+64`

Hence, our function is now:

`f(x)=a(x-2)(x^2+64)`

It equals 320 when x = 4. Therefore:

`320=a(4-2)(4^2+64)`

Solve for a. Evaluate:

`320=(2)(80)a`

So:

`320=160a\Rightarrow a=2`

Our third-degree polynomial equation is:

`f(x)=2(x-2)(x^2+64)`