Question

Write a polynomial function of minimum degree with real coefficients whose zeros include those listed. Write the polynomial in standard form.

8, -14, and 3 + 9i

Answer 1

If we have a polynomial with real coefficients but some complex zeros (aka roots), the complex roots must come in conjugate pairs. Otherwise, we'll have coefficients that include odd powers of \(i\), and that fails the "real coefficients" constraint.

The conjugate of a complex number \(a+bi\) is simply \(a-bi\) and vice versa.

That means our roots are 8, -14, 3+9i, and 3-9i. We construct the polynomial by multiplying together the equations that specify those roots:
x-8=0
x+14=0
x-3-9i = 0
x-3+9i = 0

respectively those become
(x-8)
(x+14)
(x-3-9i)
(x-3+9i)

so a polynomial of minimum degree with the specified roots is
\[(x-8)(x+14)(x-3-9i)(x-3+9i)\]

When expanding that, I suggest multiplying the two complex terms first, which will get rid of the \(i\)'s.

Answer 2

Note that we could also put an arbitrary constant in front of that expression, because multiplying the polynomial by a constant will not change the roots, but will change the vertical scale of the curve.

Answer 3

You would use that constant if you had a problem where you were to find a polynomial with given roots that passed through a specified point, for example.