Find a polynomial equation with real coefficients that has the given roots. 4 and -9i (imaginary)

Answer 1

Imaginary roots occur in conjugate pairs.

So, -9i is another root.

Therefore:

y = a(x - 4)(x - 9i)(x - (-9i))

y = a(x - 4)(x - 9i)(x + 9i)

Pick a = 1

y = (x - 4)(x - 9i)(x + 9i)

Recall: Difference of two squares.

a^2 - b^2 = (a - b)( a + b)

Therefore:

y = (x - 4)*[ x^2 - (9i)^2 ]

y = (x - 4)*[ x^2 - 81(-1) ]

y = (x - 4)(x^2 + 81)

Expand:

y = x^3 + 81x - 4x^2 - 324

y = x^3 - 4x^2 + 81x - 324

Answer 2

(x-4)(x^2+81) = x^3 -4x^2 +81x -324

if 9i is a root, so is -9i, they come in conjugate pairs

if you want real coefficients.

and (x-9i)(x+9i) = x^2+81

multiply that by x-4 and you get a 3rd degree polynomial with real coefficients.