Find a fourth-degree polynomial function, that has -1, 1, and i as zeros such that f(3) equal 160

Question

I need to make an equation with these clues. Help???

Answer 1

When the problem gives you the zeros, use the factored form. For a 4th degree polynomial, there are four factors:

f(x) = a·(x-p)(x-q)(x-r)(x-s)

where a is a constant (to be determined) and p, q, r, s are the zeros. Three zeros are given: -1, 1, and i. The Complex Conjugate Root Theorem tells us that for real coefficients, if i is a root, so is -i. So we have all four zeros (or roots): -1, 1, i, -i. Our equation is:

f(x) = a·(x-(-1)) (x-1) (x-i) (x-(-i))

f(x) = a·(x+1) (x-1) (x-i) (x+i)

To find the value of a, plug in the given point (3, 160):

160 = a·(3+1) (3-1) (3-i) (3+i)

Solve for a.