Question

A polynomial function f(x) with real coefficients has the given degree, zeros, and solution point.

Degree: 3
Zeros: -2,2+2√2i
Solution Point: f(−1) = −68


(a) Write the function in completely factored form.


(b) Write the function in polynomial form.


Help Please my teacher doesn't really teach!!!

Answer 1

you have one complex root, but they always come as conjugate pairs
The other must be 2-2√2i
You also know that one factor is (x+2)

the sum of the complex roots
= 2+2√2i + 2-2√2i = 4
product of the complex roots
= (2+2√2i)(2-2√2i) = 4 + 8 = 12

So the quadratic factor is x^2 - 4x + 12

f(x) = a(x+2)(x^2 - 4x + 12)
we also have a point (-1,-68)

-68 = a(1)(17)
a = -4

f(x) = -4(x+2)(x^2 - 4x + 12)

I will leave it up to you to expand it.

Check of my answer:
https://www.wolframalpha.com/input/?i=-4(x%2B2)(x%5E2+-+4x+%2B+12)%3D0

Answer 2

write a polynomial function of minimum degree with real coefficients whose zeros include those listed. write the polynomial in standard form calculator