2022 AMC 10A Problem 16

Attempt Problem 16 of the 2022 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10A solutions, or check the answer key.

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16.

The roots of the polynomial 10x339x2+29x610x^3 - 39x^2 + 29x - 6 are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by 22 units. What is the volume of the new box?

245\dfrac{24}{5}

425\dfrac{42}{5}

815\dfrac{81}{5}

3030

4848

Answer: D
Concepts:Vieta’s Formulaspolynomialvolume
Difficulty rating: 1420
Solution:

Let h,l,h, l, and ww be the dimensions of the old box. Then the volume of the new box is (h+2)(l+2)(w+2). (h + 2)(l + 2)(w + 2). Expanding, we get hlw+2(hl+hw+lw) hlw + 2(hl + hw + lw) +4(l+h+w)+8.+ 4(l + h + w) + 8. We can use Vieta's formulas to find the terms in this expression. We get that hlw=DA=35, hlw = -\dfrac{D}{A} = \dfrac{3}{5}, hl+hw+lw=CA=2910, hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10}, and l+h+w=BA=3910. l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}.

Plugging these values into the expression, we get 35+22910+43910+8=30. \dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30.

Thus, D is the correct answer.

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