2022 AMC 10A Problem 16

Let $h, l,$ and $w$ be the dimensions of the old box. Then the volume of the new box is $$(h + 2)(l + 2)(w + 2).$$ Expanding, we get $$hlw + 2(hl + hw + lw)$$$$+ 4(l + h + w) + 8.$$ We can use Vieta’s formulas to find the terms in this expression. We get that $$hlw = -\dfrac{D}{A} = \dfrac{3}{5},$$ $$hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10},$$ and $$l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}.$$ Plugging these values into the expression, we get $$\dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30.$$ Thus, D is the correct answer.