## Algebra and Trigonometry 10th Edition

Zeros of $f(x)$: $x=\pm \frac{1}{2},6$
$f(x)=4x^3-24x^2-x+6$ 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 $x=\pm \frac{1}{2},6$