Question

a square has sides of 6 units. squares of ( x + 2 ) units are cut out of each corner, and then the sides are folded up to create an open box. express the volume of the box as a polynomial function in terms of ( x ).

a. ( v(x) = 4x^3 - 12x + 8 )

b. ( v(x) = 4x^3 + 8x^2 - 4x - 8 )

c. ( v(x) = 4x^3 + 12x - 8 )

d. ( v(x) = 4x^3 + 16x^2 + 20x + 8 )

Explanation:

Step1: Define box dimensions

The original square has side length 6. When a square of side $x+2$ is cut from each corner, the length and width of the box base are $6 - 2(x+2)$. The height of the box is $x+2$.
Calculate base side:
$6 - 2(x+2) = 6 - 2x - 4 = 2 - 2x$

Step2: Apply volume formula

Volume $V(x) = \text{length} \times \text{width} \times \text{height}$. Since length = width, substitute values:
$V(x) = (2-2x)(2-2x)(x+2)$

Step3: Expand first two factors

Multiply $(2-2x)(2-2x)$:
$(2-2x)^2 = 4 - 8x + 4x^2$

Step4: Multiply by height term

Multiply $4x^2 -8x +4$ by $x+2$:

$$\begin{align*} &(4x^2 -8x +4)(x+2)\\ =&4x^2(x+2) -8x(x+2) +4(x+2)\\ =&4x^3 +8x^2 -8x^2 -16x +4x +8\\ =&4x^3 -12x +8 \end{align*}$$

Answer:

A. $V(x) = 4x^3 -12x +8$