Answer:
[tex]f(x)=2(x-2)(x^2+64)[/tex]
Step-by-step explanation:
A standard polynomial in factored form is given by:
[tex]f(x)=a(x-p)(x-q)...[/tex]
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:
[tex]f(x)=a(x-(2))(x-(-8i))(x-(8i))[/tex]
Simplify:
[tex]f(x)=a(x-2)(x+8i)(x-8i)[/tex]
Expand the second and third factors:
[tex]=(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[/tex]
Hence, our function is now:
[tex]f(x)=a(x-2)(x^2+64)[/tex]
It equals 320 when x = 4. Therefore:
[tex]320=a(4-2)(4^2+64)[/tex]
Solve for a. Evaluate:
[tex]320=(2)(80)a[/tex]
So:
[tex]320=160a\Rightarrow a=2[/tex]
Our third-degree polynomial equation is:
[tex]f(x)=2(x-2)(x^2+64)[/tex]
