#### Answer

Zeros of $f(x)$: $x=\pm \frac{1}{2},6$

#### Work Step by Step

$f(x)=4x^3-24x^2-x+6$
The zeros of a function are the x-values for when y-values equal zero. In other words, zeros are when $y=f(x)=0$. Since we are solving for the zeros of $f(x)$, set the function $f(x)=4x^3-24x^2-x+6$ equal to zero. We get the following equation:
$4x^3-24x^2-x+6=0$
Notice that this needs to be solved using factoring by grouping. Start by grouping the left two terms and the right two terms.
$(4x^3-24x^2)+(-x+6)=0$
Then, factor as much out of the left two terms as possible. In this example, factor out $4x^2$:
$4x^2(x-6)+(-x+6)=0$
Next, factor the $(x-6)$ term out of the right two terms:
$4x^2(x-6)+(x-6)(-1)=0$
Now, notice that you can factor an $(x-6)$ term out of the entire left side of the equation:
$(x-6)(4x^2-1)=0$
Next, you can see that the above equation holds true when $x-6=0$ or $4x^2-1=0$. Solve for $x$ in each equation individually as shown below:
$x-6=0$
Add 6 to both sides.
$x-6+6=0+6$
$x=6$
Now for the other equation:
$4x^2-1=0$
Add 1 to both sides of the equation:
$4x^2-1+1=0+1$
Divide both sides of the equation by 4:
$\frac{4x^2}{4}=\frac{1}{4}$
$x^2=\frac{1}{4}$
Now solve for x by taking the square root of both sides.
$\sqrt{x^2}=\sqrt{\frac{1}{4}}$
Remember that taking the square root of both sides means we need to include the $\pm$ on the right side of the equation.
$x=\pm \sqrt{\frac{1}{4}}$
$x=\pm \frac{1}{2}$
So, the zeros are $x=\pm \frac{1}{2},6$