Find an equation, in standard form, of a degree 4 polynomial that has 1 as its only real root, (1 + i) as one of its complex roots and has y-intercept of (0, -6).

Answer 1

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

P(x) = a·(x-p) (x-q )(x-r) (x-s)

where a is a constant called the leading coefficient and p, q, r, and s are the roots. For a polynomial with real coefficients, the Conjugate Root Theorem says that if 1+i is a root, so is 1-i. So we have three of the four roots: 1+i, 1-i, and 1. Complex roots always come in pairs, so the remaining root must be real. Since 1 is the only real root, it must have a multiplicity of 2, meaning roots are 1+i, 1-i, 1, and 1. So our polynomial so far is:

P(x) = a·(x-(1+i)) (x-(1-i)) (x-1) (x-1) = a (x-1-i) (x-1+i) (x-1)²

To find the value of a, plug in the given point (0, -6):

-6 = a (0-1-i) (0-1+i) (0-1)² = a (-1-i) (-1+i) (1)

Multiply it out to solve for a. Once you have the value of a, you have your polynomial in factored form. Multiply it out and arrange it in order of descending powers to put it into standard form.

P(x) = ax⁴ + bx³ + cx² + dx + e