### Home > A2C > Chapter 9 > Lesson 9.3.2 > Problem9-148

9-148.

Spud has done it again. He's lost another polynomial function. This one was a cubic, written in standard form. He knows that there were two complex zeros, $−2±5i$ and one real zero, −1. What could his original function have been?

Use the three roots to find the individual polynomials that make up the cubic equation.

$\left(x + 1\right)\left(x + 2 + 5i\right)\left(x + 2 − 5i\right)$

Now multiply the two complex polynomials.

$\left(x + 2 + 5i\right)\left(x + 2 − 5i\right) = x^{2} + 4x + 29$

Now multiply this equation by $x + 1$.

$\left(x + 1\right)\left(x^{2} + 4x + 29\right)$

$x^{3} + 5x^{2} + 33x + 29$