Find all of the real zeros of the function. Give exact values whenever possible. Identify each zero as rational or irrational.
-$$f ( x ) = x ^ { 3 } + 6 x ^ { 2 } - 20 x - 32$$
A) $- 8$ (rational), $1 + \sqrt { 5 }$ (irrational), and $1 - \sqrt { 5 }$ (irrational)
B) $- 8$ (rational), $2 + \sqrt { 5 }$ (irrational), and $2 - \sqrt { 5 }$ (irrational)
C) $- 8$ (rational), 6 (rational), $- 4$ (rational)
D) $- 8$ (rational), $\sqrt { 5 }$ (irrational), and $- \sqrt { 5 }$ (irrational)

Question

Quizplus · Accepted Answer

To find the real zeros of the given function $f(x) = x^3 + 6x^2 - 20x - 32$, we can use the Rational Root Theorem, synthetic division, or factoring by grouping if possible. Factoring the given polynomial, we find that $x = -8$ is a root since $f(-8) = 0$. After dividing the polynomial by $x + 8$, we get a quadratic equation, which can be solved using the quadratic formula to find the other two roots, $1 + \sqrt{5}$ and $1 - \sqrt{5}$. These roots are indeed the zeros of the function, with $-8$ being rational and the other two being irrational due to the presence of the square root of 5.

Find All of the Real Zeros of the Function f(x)=x3+6x2−20x−32f ( x ) = x ^ { 3 } + 6 x ^ { 2 } - 20 x - 32f(x)=x3+6x2−20x−32

Find All of the Real Zeros of the Function $f ( x ) = x ^ { 3 } + 6 x ^ { 2 } - 20 x - 32$