Find the zeros of the function algebraically.
$f ( x ) = \frac { 1 } { 8 } x ^ { 3 } - x$
A) $x = 0 , \sqrt { 8 }$
B) $x = 0 , - \sqrt { 8 }$
C) $x = 0 , \pm 8$
D) $x = 0 , \pm \sqrt { 8 }$
E) $x = 0,8$

Question

Find the zeros of the function algebraically. 
 $f ( x ) = \frac { 1 } { 8 } x ^ { 3 } - x$  
A) $x = 0 , \sqrt { 8 }$
B) $x = 0 , - \sqrt { 8 }$
C) $x = 0 , \pm 8$
D) $x = 0 , \pm \sqrt { 8 }$
E) $x = 0,8$

Quizplus · Accepted Answer

To find the zeros of the function $f(x) = \frac{1}{8}x^3 - x$, set $f(x) = 0$:$$0 = \frac{1}{8}x^3 - x$$Factor out $x$:$$x\left(\frac{1}{8}x^2 - 1\right) = 0$$This gives us $x = 0$ and $\frac{1}{8}x^2 - 1 = 0$. Solving the quadratic equation $\frac{1}{8}x^2 - 1 = 0$ gives:$$\frac{1}{8}x^2 = 1$$$$x^2 = 8$$$$x = \pm \sqrt{8}$$Therefore, the zeros of the function are $x = 0, \pm \sqrt{8}$.

Find the Zeros of the Function Algebraically f(x)=18x3−xf ( x ) = \frac { 1 } { 8 } x ^ { 3 } - xf(x)=81​x3−x

Find the Zeros of the Function Algebraically $f ( x ) = \frac { 1 } { 8 } x ^ { 3 } - x$