A polynomial function with real coefficients has the given degree, zeros, and solution point. Write the function a) in completely factored form and b) in polynomial form.

Question

Degree: 4 Zeros: -1, 2, i Solution Point: f(1) = 8

Answer 1

For this type of problem, use the factored form of a polynomial. Since the degree is 4, there are 4 factors:

f(x) = a·(x-p)(x-q)(x-r)(x-s)

where a is the leading coefficient (a constant) and p, q, r, and s are the zeros. You are given three of the zeros: -1, 2, and i. To find the fourth root (aka zero), use the complex conjugate root theorem, which states that for a polynomial with real coefficients, if a + bi is a root, then a − bi is also a root. So if i is a root, then -i is also a root. So our fourth root is -i and the equation in factored form is

f(x) = a·(x+1)(x-2)(x-i)(x+i)

To find the value of a, plug in the given point, f(1) = 8, and solve for a:

8 = a·(1+1)(1-2)(1-i)(1+i)

To put the final answer in standard form, just multiply everything out and simplify.