Let $r_1,$ $r_2,$ and $r_3$ be the roots of
\[x^3 - 3x^2 + 8 = 0.\]Find the monic polynomial, in $x,$ whose roots are $2r_1,$ $2r_2,$ and $2r_3.$
Let $y = 2x.$  Then $x = \frac{y}{2},$ so
\[\frac{y^3}{8} - \frac{3y^2}{4} + 8 = 0.\]Multiplying by 8, we get $y^3 - 6y^2 + 64 = 0.$  The corresponding polynomial in $x$ is then $\boxed{x^3 - 6x^2 + 64}.$
x^3-6x^2+64
Level 4
Intermediate Algebra
Problem: Let $r_1,$ $r_2,$ and $r_3$ be the roots of
\[x^3 - 3x^2 + 8 = 0.\]Find the monic polynomial, in $x,$ whose roots are $2r_1,$ $2r_2,$ and $2r_3.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the monic polynomial whose roots are $2r_1,$ $2r_2,$ and $2r_3$, we start with the original polynomial $x^3 - 3x^2 + 8 = 0$. The roots of this polynomial, $r_1, r_2, r_3$, can be used to derive the new polynomial.

First, we apply Vieta's formulas to the original polynomial. The sum of the roots $r_1 + r_2 + r_3$ is equal to the coefficient of $x^2$ with the opposite sign, which gives us $r_1 + r_2 + r_3 = 3$. The sum of the products of the roots taken two at a time, $r_1r_2 + r_2r_3 + r_3r_1$, is equal to the coefficient of $x$ (which is 0 in this case), so $r_1r_2 + r_2r_3 + r_3r_1 = 0$. The product of the roots $r_1r_2r_3$ is equal to the constant term with the opposite sign, yielding $r_1r_2r_3 = -8$.

Next, we need to find the new roots, which are $2r_1, 2r_2, 2r_3$. The sum of the new roots is $2(r_1 + r_2 + r_3) = 2 \cdot 3 = 6$. The sum of the products of the new roots taken two at a time is $2^2(r_1r_2 + r_2r_3 + r_3r_1) = 4 \cdot 0 = 0$. The product of the new roots is $2^3(r_1r_2r_3) = 8 \cdot (-8) = -64$.

Using these results, we can construct the new polynomial. The polynomial with roots $2r_1, 2r_2, 2r_3$ can be expressed in the form:

\[ x^3 - ( \text{sum of roots} ) x^2 + ( \text{sum of products of roots taken two at a time} ) x - ( \text{product of roots} ). \]

Substituting the values we found:

1. The sum of the roots is $6$.
2. The sum of the products of the roots taken two at a time is $0$.
3. The product of the roots is $-64$.

Thus, the polynomial becomes:

\[ x^3 - 6x^2 + 0 \cdot x + 64 = x^3 - 6x^2 + 64. \]

This polynomial is monic, as the leading coefficient is 1. Therefore, the final answer is $x^3 - 6x^2 + 64$.<|END_OF_LATENT|> The final answer is $x^3-6x^2+64$.