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

**Solution:**

**Every complex number has another complex number associated with it, known as the complex conjugate. **

**A complex conjugate of a complex number is another complex number that has the same real part as the original complex number and the imaginary part has the same magnitude but opposite sign. **

**The product of a complex number and its complex conjugate is a real number.**

Complex roots are in pairs that are both positive and negative.

If 6i is a root then - 6i is also a root

We know that

f (x) = (x + 8) (x - 1) (x^{2} + 36)

Now equate each factor to zero

x + 8 = 0

x = - 8

x - 1 = 0

x = 1

x^{2} + 36 = 0

x^{2} = - 36

x = √-36

So we get

x = ±6i

Therefore, the root of f (x) is ±6i.

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

**Summary:**

If a polynomial function f(x) has roots -8, 1, and 6i, - 6i is also a root of f(x).

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