## College Algebra (10th Edition)

$\color{blue}{x = \left\{-2, 0\right\}}$
Let $u = x+3$ Replacing $x+3$ with $u$ gves: $(x+3)^2-4(x+3)+3=0 \\u^2-4u+3=0$ Factor the trinomial to obtain: $(u-3)(u-1)=0$ Use the Zero-Product Property (which states that if $xy=0$, then either $x=0$ or $y=0$ or both are zero) by equating each factor to zero to obtain: $u-3=0 \text{ or } u-1=0$ Solve each equation to obtain: $u=3$ or $u=1$ Replace $u$ with $x+3$ to obtain: \begin{array}{ccc} &u=3 &\text{or} &u=1 \\&x+3=3 &\text{or} &x+3=1 \\&x=3-3 &\text{or} &x=1-3 \\&x=0 &\text{or} &x=-2 \end{array} Therefore, the solutions to the given equation are: $\color{blue}{x = \left\{-2, 0\right\}}$