## Algebra and Trigonometry 10th Edition

$x=6, -3$
$f(x)=x^2+2x+1$ and $g(x)=5x+19$ Since we are solving for all the values of $x$ when $f(x) = g(x)$, set the function $f(x)=x^2+2x+1$ equal to $g(x)=5x+19$ and we get the following equation: $x^2+2x+1=5x+19$ Subtract $5x+19$ from both sides of the equation so that all the terms are on the left side: $x^2+2x+1-(5x+19)=5x+19-(5x+19)$ $x^2+2x+1-5x-19=0$ $x^2-3x-18=0$ Then, factor the left side of the equation. To do this, we need to find two numbers that multiply to get -18 and add to get -3. Those numbers are -6 and 3. $(x-6)(x+3)=0$ Next, you can see that the above equation holds true when $x-6=0$ or $x+3=0$. 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: $x+3=0$ Subtract 3 from both sides of the equation: $x+3-3=0-3$ $x=-3$ So, the solution is $x=6, -3$