Algebra 2 (1st Edition)

No solution; $-4$ would cause a denominator to evaluate to 0.
For cross multiplication, we know: $$\frac{a}{d} =\frac{b}{c} \rightarrow ac=bd$$ Thus, we find: $$4\left(x-4\right)\left(x+4\right)=\left(x^2+2x-8\right)\cdot \:4\\ 4x^2-64=4x^2+8x-32\\ -8x=32 \\ x=-4$$ Checking our solution by plugging the given value back into the equation, we assure that the solution IS extraneous.