Given that $-1 - 4\sqrt{2}$ is a root of the equation \[x^3 + ax^2 + bx + 31 = 0\]and that $a$ and $b$ are rational numbers, compute $a.$
Because the coefficients of the polynomial are rational, the radical conjugate $-1+4\sqrt2$ must also be a root of the polynomial. By Vieta's formulas, the product of the roots of this polynomial is $-31,$ and the product of these two roots is $(-1-4\sqrt2)(-1+4\sqrt2) = -31,$ so the remaining root must be $\frac{-31}{-31} = 1.$ Then by Vieta's formulas again, we have \[a = -[1 + (-1-4\sqrt2) + (-1+4\sqrt2)] = \boxed{1}.\]
1
Level 3
Intermediate Algebra
Problem: Given that $-1 - 4\sqrt{2}$ is a root of the equation \[x^3 + ax^2 + bx + 31 = 0\]and that $a$ and $b$ are rational numbers, compute $a.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \( a \), we start by recognizing that if \( -1 - 4\sqrt{2} \) is a root of the polynomial, then its conjugate \( -1 + 4\sqrt{2} \) must also be a root because the coefficients \( a \) and \( b \) are rational. Let the third root be \( r \). 

Using Vieta's formulas, we know that the sum of the roots of the polynomial \( x^3 + ax^2 + bx + 31 = 0 \) is given by:

\[
(-1 - 4\sqrt{2}) + (-1 + 4\sqrt{2}) + r = -a
\]

Simplifying the left side, the terms involving \( \sqrt{2} \) cancel out:

\[
-1 - 4\sqrt{2} - 1 + 4\sqrt{2} + r = -a
\]

This reduces to:

\[
-2 + r = -a
\]

Thus, we can express \( r \) in terms of \( a \):

\[
r = -a + 2
\]

Next, we apply Vieta's formulas again to find the sum of the products of the roots taken two at a time, which is equal to \( b \):

\[
(-1 - 4\sqrt{2})(-1 + 4\sqrt{2}) + (-1 - 4\sqrt{2})r + (-1 + 4\sqrt{2})r = b
\]

Calculating the first term:

\[
(-1 - 4\sqrt{2})(-1 + 4\sqrt{2}) = 1 - (4\sqrt{2})^2 = 1 - 32 = -31
\]

Now substituting \( r = -a + 2 \) into the remaining terms:

\[
(-1 - 4\sqrt{2})(-a + 2) + (-1 + 4\sqrt{2})(-a + 2)
\]

Calculating each product:

1. For \( (-1 - 4\sqrt{2})(-a + 2) \):

\[
a - 2 + 4a\sqrt{2} - 8\sqrt{2}
\]

2. For \( (-1 + 4\sqrt{2})(-a + 2) \):

\[
a - 2 - 4a\sqrt{2} + 8\sqrt{2}
\]

Adding these two results together:

\[
(a - 2 + 4a\sqrt{2} - 8\sqrt{2}) + (a - 2 - 4a\sqrt{2} + 8\sqrt{2}) = 2a - 4 + 0\sqrt{2} = 2a - 4
\]

Thus, we have:

\[
-31 + (2a - 4) = b
\]

This simplifies to:

\[
2a - 35 = b
\]

Now we have two equations: 

1. \( r = -a + 2 \)
2. \( b = 2a - 35 \)

To find \( a \), we need to ensure that the polynomial has rational coefficients. The product of the roots, given by Vieta's formulas, is:

\[
(-1 - 4\sqrt{2})(-1 + 4\sqrt{2})r = -31
\]

Substituting the product of the first two roots:

\[
-31r = -31
\]

This implies:

\[
r = 1
\]

Substituting \( r = 1 \) into the equation for \( r \):

\[
1 = -a + 2
\]

Solving for \( a \):

\[
a = 1
\]

Thus, the final answer is \( 1 \).<|END_OF_LATENT|> The final answer is $1$.