Question

Question

Answer 1

Answer:

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

Step-by-step 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:

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