## Algebra and Trigonometry 10th Edition

$x^2+4x-32 = 0$ $x^2+4x+4-36=0$ $(x+2)^2-6^2 = (x+2-6)(x+2+6)=(x-4)(x+8)$ (as $x^2-y^2=(x-y)(x+y)$ Hence either (x-4)=0 that gives x=4 or (x+8)=0 that gives x=-8