## Algebra and Trigonometry 10th Edition

$S=10x^2+4x-8$ For $x=6$: $S=376$
$V=2x^3+x^2-8x-4$ $V=x^2(2x+1)-4(2x+1)$ $V=(x^2-4)(2x+1)=(x^2-2^2)(2x+1)$ $V=(x+2)(x-2)(2x+1)$ So, the edges of the cube are: $(x+2),~(x-2)~and~(2x+1)$ The surface area: $S=2(x+2)(x-2)+2(x+2)(2x+1)+2(x-2)(2x+1)=$ $S=2(x^2-4)+2(2x^2+x+4x+2)+2(2x^2+x-4x-2)=$ $S=10x^2+4x-8$ For $x=6$: $S=10(6)^2+4(6)-8=360+24-8=376$