To find the roots of the polynomial equation x³ + 2x² + 16x + 32 = 0, we can use various methods such as factoring, synthetic division, or the Rational Root Theorem. Let’s start by trying to find any rational roots using the Rational Root Theorem, which suggests that any rational solution p/q (where p is a factor of the constant term and q is a factor of the leading coefficient) could be a candidate.
The constant term here is 32, and the leading coefficient is 1. The factors of 32 are ±1, ±2, ±4, ±8, ±16, and ±32.
Let’s test these values to see if any of them are roots:
- x = -2:
Substituting -2 into the equation:
(-2)³ + 2(-2)² + 16(-2) + 32 = -8 + 8 – 32 + 32 = 0
-2 is a root!
- Now, we will factor the polynomial:
Since -2 is a root, we can perform synthetic division to factor x + 2 out of x³ + 2x² + 16x + 32.
-2 | 1 2 16 32 | -2 0 -32 ------------------- | 1 0 16 0
This results in the quotient being x² + 16. So, we can rewrite the original equation as:
(x + 2)(x² + 16) = 0
Now we need to find the roots of x² + 16 = 0.
By moving 16 to the other side:
x² = -16
Taking the square root of both sides gives us:
x = ±4i
In conclusion, the roots of the polynomial equation x³ + 2x² + 16x + 32 = 0 are:
- x = -2 (real root)
- x = 4i (imaginary root)
- x = -4i (imaginary root)
Thus, the roots are -2, 4i, and -4i.