Question

find all zeros of the function $f(x) = 8x^3 - 6x^2 - 17x - 6$. enter the zeros separated by commas.

Explanation:

Step1: Use Rational Root Theorem

The Rational Root Theorem states that possible rational roots are $\pm\frac{p}{q}$, where $p$ is a factor of the constant term and $q$ is a factor of the leading coefficient. For $f(x)=8x^3 - 6x^2 - 17x - 6$, $p = \pm1,\pm2,\pm3,\pm6$ and $q=\pm1,\pm2,\pm4,\pm8$. Testing possible roots:

• Test $x = -\frac{1}{2}$: $f(-\frac{1}{2})=8(-\frac{1}{2})^3 - 6(-\frac{1}{2})^2 - 17(-\frac{1}{2}) - 6=8(-\frac{1}{8}) - 6(\frac{1}{4})+\frac{17}{2}-6=-1-\frac{3}{2}+\frac{17}{2}-6=-1 + 7 - 6 = 0$. So $x = -\frac{1}{2}$ is a root.

Step2: Factor the Polynomial

Since $x = -\frac{1}{2}$ is a root, $(2x + 1)$ is a factor. Use polynomial division or synthetic division to divide $8x^3 - 6x^2 - 17x - 6$ by $(2x + 1)$. Using synthetic division with root $-\frac{1}{2}$:
Coefficients: $8\ -6\ -17\ -6$
Bring down 8. Multiply by $-\frac{1}{2}$: $-4$. Add to -6: $-10$. Multiply by $-\frac{1}{2}$: $5$. Add to -17: $-12$. Multiply by $-\frac{1}{2}$: $6$. Add to -6: $0$.
So the quotient is $8x^2 - 10x - 12$. Factor out a 2: $2(4x^2 - 5x - 6)$. Now factor $4x^2 - 5x - 6$: find two numbers $a,b$ such that $a\times b=4\times(-6)=-24$ and $a + b=-5$. We get $a = -8$, $b = 3$. So $4x^2 - 8x + 3x - 6 = 4x(x - 2)+3(x - 2)=(4x + 3)(x - 2)$. Thus, $f(x)=(2x + 1)\times2\times(4x + 3)(x - 2)=2(2x + 1)(4x + 3)(x - 2)$.

Step3: Find Zeros

Set each factor equal to zero:

• $2x + 1 = 0\Rightarrow x = -\frac{1}{2}$
• $4x + 3 = 0\Rightarrow x = -\frac{3}{4}$
• $x - 2 = 0\Rightarrow x = 2$

Answer:

$-\frac{3}{4}, -\frac{1}{2}, 2$