## Algebra and Trigonometry 10th Edition

$x=-2,8$
$f(x)=x^2-6x-16$ Since we are solving for all the real values of $x$ when $f(x) = 0$, set the function $f(x)=x^2-6x-16$ equal to zero and we get the following equation: $x^2-6x-16=0$ Now factor the left side of the equation by finding two numbers that multiply to get -16 and add to get -6. These two numbers are 2 and -8. So the factored left side looks as follows: $(x+2)(x-8)=0$ Next, you can see that the above equation holds true when $x+2=0$ or $x-8=0$. Solve each of those two equations individually. $x+2=0$ Subtract 2 from both sides. $x+2-2=0-2$ $x=-2$ Now for the other equation: $x-8=0$ Add 8 to both sides. $x-8+8=0+8$ $x=8$ So, the final solution is $x=-2,8$