Two of the roots of the equation \[ax^3+bx^2+cx+d=0\]are $3$ and $-2.$ Given that $a \neq 0,$ compute $\frac{b+c}{a}.$
Setting $x=3$ and $x=-2,$ we get the two equations \[\begin{aligned} 27a+9b+3c+d &= 0, \\ -8a+4b-2c+d &= 0. \end{aligned}\]Subtracting these two equations eliminates $d$ and gives \[35a + 5b + 5c = 0.\]Thus, $b+c=-7a,$ so $\frac{b+c}{a} = \boxed{-7}.$
-7
Level 3
Intermediate Algebra
Problem: Two of the roots of the equation \[ax^3+bx^2+cx+d=0\]are $3$ and $-2.$ Given that $a \neq 0,$ compute $\frac{b+c}{a}.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve for \(\frac{b+c}{a}\), we start by recognizing that the polynomial can be expressed in terms of its roots. Since two roots are given as \(3\) and \(-2\), we can denote the third root as \(r\). The polynomial can be factored as:

\[
a(x - 3)(x + 2)(x - r) = 0
\]

Expanding this product will help us identify the coefficients \(b\) and \(c\). First, we expand \((x - 3)(x + 2)\):

\[
(x - 3)(x + 2) = x^2 - 3x + 2x - 6 = x^2 - x - 6
\]

Next, we multiply this result by \((x - r)\):

\[
(x^2 - x - 6)(x - r) = x^3 - rx^2 - x^2 + rx - 6x + 6r
\]

Combining like terms gives:

\[
x^3 + (-r - 1)x^2 + (r - 6)x + 6r
\]

Thus, we can identify the coefficients:

- \(b = a(-r - 1)\)
- \(c = a(r - 6)\)

Now, we need to compute \(b + c\):

\[
b + c = a(-r - 1) + a(r - 6) = a(-r - 1 + r - 6) = a(-7)
\]

This simplifies to:

\[
b + c = -7a
\]

To find \(\frac{b+c}{a}\), we divide both sides by \(a\):

\[
\frac{b+c}{a} = \frac{-7a}{a} = -7
\]

Thus, the final answer is \(-7\).<|END_OF_LATENT|> The final answer is $-7$.