2022 AMC 10A Problem 16

1. What we want. Call the three roots $r$, $s$, $t$; they are the edges of the old box. The new edges are $r + 2$, $s + 2$, $t + 2$, so the new volume is
$$(r + 2)(s + 2)(t + 2).$$
2. Write the polynomial using its roots. A polynomial with leading number $10$ and roots $r, s, t$ can be written
$$10x^3 - 39x^2 + 29x - 6 = 10(x - r)(x - s)(x - t).$$
3. Choose a good value of $x$. We want factors like $r + 2$. Putting $x = -2$ makes each factor $-2 - r = -(r + 2)$, and similarly for $s$ and $t$. Three minus signs multiply to one minus sign, so the right side becomes
$$10 \cdot (-1) \cdot (r+2)(s+2)(t+2).$$
4. Compute the left side at $x = -2$.
$$10(-8) - 39(4) + 29(-2) - 6 = -80 - 156 - 58 - 6 = -300.$$
5. Solve. $-10\,(r+2)(s+2)(t+2) = -300$, so the new volume is $\dfrac{300}{10} = 30$.

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

Writing a polynomial as "leading number times ($x$ minus each root)" lets one plug-in give you a product of shifted roots, with no need to find the roots. Choose the value of $x$ that turns each factor into the shape you want, and keep track of the leading number and the minus signs.

Find the roots. Try small values: at $x = 3$, $10(27) - 39(9) + 29(3) - 6 = 270 - 351 + 87 - 6 = 0$, so $3$ is a root. Dividing out $x - 3$ leaves $10x^2 - 9x + 2 = (2x - 1)(5x - 2)$, so the other roots are $\frac12$ and $\frac25$. The new edges are $5$, $\frac52$, $\frac{12}{5}$, and the volume is $5 \cdot \frac52 \cdot \frac{12}{5} = 30$.

Or use Vieta's formulas. Expanding gives $(r+2)(s+2)(t+2) = rst + 2(rs + st + tr) + 4(r + s + t) + 8$. From the coefficients, $r + s + t = \frac{39}{10}$, $rs + st + tr = \frac{29}{10}$, $rst = \frac{6}{10}$, so the volume is $\frac{6 + 58 + 156}{10} + 8 = 22 + 8 = 30$.