Question

If a polynomial function f(x) has roots –8, 1, and 6i, what must also be a root of f(x)? A. –6

Answer 1

Answer:

B. –6i

Step-by-step explanation:

Complex roots come in pairs.  The pairs are complex conjugates.  If you have a+bi, then you will have a-bi

Since we have 6i,  the other root must be -6i

Answer 2

Answer:

-6i

Step-by-step explanation:

Complex roots always come in pairs, and those pairs are made up of a positive and a negative version. If 6i is a root, then its negative value, -6i, is also a root.

If you want to know the reasoning, it's along these lines: to even get a complex/imaginary root, we take the square root of a negative value. When you take the square root of any value, your answer is always "plus or minus" whatever the value is. The same thing holds for complex roots. In this case, the polynomial function likely factored to f(x) = (x+8)(x-1)(x^2+36). To solve that equation, you set every factor equal to zero and solve for the x's.

x + 8 = 0

x = -8

x - 1 = 0

x = 1

x^2 + 36 = 0

x^2 = -36 ... take the square root of both sides to get x alone

x = √-36 ... square root of an imaginary number produces the usual square root and an "i"

x = ±6i