## Algebra and Trigonometry 10th Edition

$x=1,\pm \sqrt{3}$
$f(x)=x^3-x^2-3x+3$ Since we are solving for all the real values of $x$ when $f(x) = 0$, set the function $f(x)=x^3-x^2-3x+3$ equal to zero and we get the following equation: $x^3-x^2-3x+3=0$ Notice that this needs to be solved using factoring by grouping. Start by grouping the left two terms and the right two terms. $(x^3-x^2)+(-3x+3)=0$ Then, factor as much out of the left two terms as possible. In this example, factor out $x^2$: $x^2(x-1)+(-3x+3)=0$ Next, factor the $(x-1)$ term out of the right two terms: $x^2(x-1)+(x-1)(-3)=0$ Now, notice that you can factor an $(x-1)$ term out of the entire left side of the equation: $(x-1)(x^2-3)=0$ Next, you can see that the above equation holds true when $x-1=0$ or $x^2-3=0$. Solve for $x$ in each equation individually as shown below: $x-1=0$ Add 1 to both sides. $x-1+1=0+1$ $x=1$ Now for the other equation: $x^2-3=0$ Add 3 to both sides of the equation: $x^2-3+3=0+3$ $x^2=3$ Now solve for x by taking the square root of both sides. $\sqrt{x^2}=\sqrt{3}$ Remember that taking the square root of both sides means we need to include the $\pm$ on the right side of the equation. $x=\pm \sqrt{3}$ So, the final solution is $x=1,\pm \sqrt{3}$