Question

What is the width of a box when the length is x+2, the height is x+8, and the volume is x^3+9x^2+6x-16?

Answer 1

(x+2)(x+8) = x^2 +10x +16

(x^2 + 10x + 16)(x-1) = x^3+9x^2+6x-16, so the other dimension is x-1

Answer 2

The given box has the following properties

The length is $x+2$

The height is $x+8$

The volume is $x^3+9x^2+6x-16$

As expression for volume can be setup as below , suppose width is w

$x^3+9x^2+6x-16=(x+8)(x+2)w\\ \text{Make factors on the left side we get}\\ \\ x^3+8x^2+x^2+8x-2x-16=(x+8)(x+2)w\\ \\ x^2(x+8)+x(x+8)-2(x+8)=(x+8)(x+2)w\\ \\ (x+8)(x^2+x-2)=(x+8)(x+2)w\\ \\ (x+8)(x^2+2x-x-2)=(x+8)(x+2)w\\ \\ (x+8)(x(x+2)-1(x+2))=(x+8)(x+2)w\\ \\ (x+8)(x+2)(x-1)=(x+8)(x+2)w$

Hence width w=(x-1)