## College Algebra 7th Edition

$x=\left\{-4, 2\right\}$
Subtract $8e^{2x}$ on both sides of the equation to obtain: $x^2e^{2x} + 2xe^{2x}-8e^{2x} = 0$ Factor the trinomial to obtain: $e^{2x}(x+4)(x-2)=0$ RECALL: $a\cdot b \cdot c = 0 \longrightarrow a = 0 \text{ or } b=0 \text{ or } c=0$ Use the rule above by equating each factor to zero then solving each equation to obtain: $\begin{array}{ccccc} &e^{2x}=0 &\text{ or } &x+4=0 &\text{ or } &x-2=0 \\& & &x=-4 &\text{ or } &x=2 \end{array}$ Note that $e^{2x}=0$ has no real number solution as there is no real number that will make $e^{2x}=0$. Thus, the solutions are $x=\left\{-4, 2\right\}$.