## Algebra 2 (1st Edition)

$2\pi x^3-13\pi x^2+8\pi x+48\pi$
Plugging in the values we get: $V=\pi(x-4)(x-4)(2x+3)=\pi(x^2-4x-4x+16)(2x+3)=\pi(x^2-8x+16)(2x+3)=\pi(2x^3-16x^2+32x+3x^2-24x+48)=\pi(2x^3-13x^2+8x+48)=2\pi x^3-13\pi x^2+8\pi x+48\pi$