2022 AMC 10A Problem 16

Let the roots be $r, s, t$. Dividing the polynomial by its leading coefficient, $x^3 - \frac{39}{10}x^2 + \frac{29}{10}x - \frac{6}{10}$, Vieta's formulas give
$$r + s + t = \frac{39}{10}, \qquad rs + st + tr = \frac{29}{10}, \qquad rst = \frac{6}{10}.$$

The new box has edges $r+2, s+2, t+2$, so its volume is
$$(r+2)(s+2)(t+2) = rst + 2(rs + st + tr) + 4(r + s + t) + 8.$$
Substituting:
$$\frac{6}{10} + 2 \cdot \frac{29}{10} + 4 \cdot \frac{39}{10} + 8 = \frac{6 + 58 + 156}{10} + 8 = 22 + 8 = 30.$$

The answer is $\boxed{\textbf{(D)}\ 30}$.

Any symmetric expression in the roots, such as a product of shifted roots, can be written in terms of the elementary symmetric sums, and Vieta hands those over directly from the coefficients. There is never a need to find the roots themselves.

Since $P(x) = 10(x - r)(x - s)(x - t)$, plugging in $x = -2$ gives $P(-2) = 10(-2-r)(-2-s)(-2-t) = -10(r+2)(s+2)(t+2)$. Compute $P(-2) = -80 - 156 - 58 - 6 = -300$, so the volume is $\frac{300}{10} = 30$. (The roots happen to be $\frac12, \frac25, 3$, and $\frac52 \cdot \frac{12}{5} \cdot 5 = 30$ checks.)

• Dropping the leading coefficient 10 when reading Vieta's formulas, or getting the sign of rst wrong (it is +6/10, since the constant term is -6).
• Factoring the cubic by hand to find the roots, which works but costs several minutes compared with a direct Vieta or $P(-2)$ evaluation.