## Algebra and Trigonometry 10th Edition

$x=\pm 3,4$
$f(x)=x^3-4x^2-9x+36$ 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 $x=\pm 3,4$