Question

Find a fourth degree polynomial function having only real coefficients with 3, 1, and (-1+3i) as zeros and f(2)=-36?

Answer 1

Start with the complex factor: x = -1+3i
(x+1) = 3i
Square:
(x+1)^2 = (3i)^2
x^2+2x+1 = -9
x^2+2x+10 = 0

The other two factors are (x-3) and (x-1)
multiply them all and expand: (x^2+2x+10) * (x-3) * (x-1)

Answer 2

The fourth degree polynomial is: f(x) = a * (x^2+2x+10) * (x-3) * (x-1)
Put x = 2 to evaluate f(2). Equate it to -36 and solve for a.

Answer 3

I did just as you had stated, and ended up with x^4-2x^3+5x^2-34x+30
When I plugged in f(2) I ended up with -18. Can you tell me where I went wrong?

Answer 4

f(x) = a * (x^2+2x+10) * (x-3) * (x-1) where 'a' is a constant to be determined.
Find f(2):
f(2) = a* (4+4+10)(-1)(1)
= a * (-18)

But we are given f(2) = -36
Therefore, -18a = -36
a = 2.

So f(x) = 2(x^4-2x^3+5x^2-34x+30) or
f(x) = 2x^4 - 4x^3 + 10x^2 - 68x + 60