#### Answer

$x=\pm 3,4$

#### Work Step by Step

$f(x)=x^3-4x^2-9x+36$
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 $f(x)=x^3-4x^2-9x+36$ equal to zero. We get the following equation:
$x^3-4x^2-9x+36=0$
Notice that this needs to be solved using factoring by grouping. Start by group the left two terms and the right two terms.
$(x^3-4x^2)+(-9x+36)=0$
Then, factor as much out of the left two terms as possible. In this example, factor out $x^2$:
$x^2(x-4)+(-9x+36)=0$
Next, factor the $(x-4)$ term out of the right two terms:
$x^2(x-4)+(x-4)(-9)=0$
Now, notice that you can factor an $(x-4)$ term out of the entire left side of the equation:
$(x-4)(x^2-9)=0$
Next, you can see that the above equation holds true when $x-4=0$ or $x^2-9=0$. Solve for $x$ in each equation individually as shown below:
$x-4=0$
Add 4 to both sides. $x-4+4=0+4$
$x=4$
Now for the other equation:
$x^2-9=0$
Add 9 to both sides of the equation:
$x^2-9+9=0+9$
$x^2=9$
Now solve for x by taking the square root of both sides.
$\sqrt{x^2}=\sqrt{9}$
Remember that taking the square root of both sides means we need to include the $\pm$ on the right side of the equation.
$x=\pm \sqrt{9}$ $x=\pm 3$
So, the zeros are $x=\pm 3,4$