To find the roots of the polynomial equation x³ + 2x² + 16x + 32 = 0, we can start by looking for rational roots using the Rational Root Theorem. The theorem suggests that any rational root, in the form of p/q (where p is a factor of the constant term and q is a factor of the leading coefficient), could be possible solutions.
The constant term here is 32 and the leading coefficient is 1. The factors of 32 are ±1, ±2, ±4, ±8, ±16, and ±32. We can test these values in the polynomial to see if any yield zero.
Let’s evaluate the polynomial for these factors:
- x = -2:
(-2)³ + 2(-2)² + 16(-2) + 32 = -8 + 8 – 32 + 32 = 0.
So, x = -2 is a root.
Having found one root, we can perform polynomial long division to divide the polynomial by (x + 2).
After dividing, we get:
- (x³ + 2x² + 16x + 32) ÷ (x + 2) = x² + 16
Now, we need to find the roots of the quadratic equation x² + 16 = 0. Rearranging gives us:
- x² = -16
- x = ±√(-16) = ±4i
So, the complete set of roots for the equation x³ + 2x² + 16x + 32 = 0 are:
- -2 (real root)
- 4i (complex root)
- -4i (complex root)
In conclusion, the roots of the polynomial include one real root and two complex roots.