Find a polynomial function with real coefficients that has the given zeros. (There are many correct answers.)

Question

Find a polynomial function with real coefficients that has the given zeros. (There are many correct answers.)

-7/2, -6, 1+√3i

Answer 1

Hi Doug,

Keep in mind that complex roots always occur in conjugate pairs if we are looking for real coefficients, so the roots are:

x = -7/2

2x + 7 = 0

x = -6

x + 6 = 0

x = 1 + i√3

x - (1 + i√3) = 0

x = 1 - i√3

x - (1 - i√3) = 0

So ƒ(x) = (2x + 7)(x + 6)(x - (1+i√3))(x - (1-i√3))

Now all you'll have to do is multiply and expand the function.

ƒ(x) = (2x + 7)(x + 6)(x - (1+i√3))(x - (1-i√3))

= (2x² + 19x + 42)(x² - 2x + 4)

= 2x⁴ + 15x⁴ + 12x² - 8x + 168

Therefore, ƒ(x) = 2x⁴ + 15x⁴ + 12x² - 8x + 168