The polynomial $ax^4 + bx^3 + 32x^2 - 16x + 6$ has a factor of $3x^2 - 2x + 1.$  Find the ordered pair $(a,b).$
We can write
\[ax^4 + bx^3 + 32x^2 - 16x + 6 = (3x^2 - 2x + 1)(cx^2 + dx + 6).\]Expanding, we get
\[ax^2 + bx^3 + 32x^2 - 16x + 6 = 3cx^4 + (-2c + 3d)x^3 + (c - 2d + 18) x^2 + (d - 12) x + 6.\]Comparing coefficients, we get
\begin{align*}
a &= 3c, \\
b &= -2c + 3d, \\
32 &= c - 2d + 18, \\
-16 &= d - 12.
\end{align*}Solving, we find $a = 18,$ $b =-24,$ $c = 6,$ and $d = -4,$ so $(a,b) = \boxed{(18,-24)}.$
(18,-24)
Level 4
Intermediate Algebra
Problem: The polynomial $ax^4 + bx^3 + 32x^2 - 16x + 6$ has a factor of $3x^2 - 2x + 1.$  Find the ordered pair $(a,b).$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the coefficients \(a\) and \(b\), we start by recognizing that since \(3x^2 - 2x + 1\) is a factor of the polynomial, we can express the polynomial as the product of this factor and another quadratic polynomial \(px^2 + qx + r\). 

We can write:
\[
ax^4 + bx^3 + 32x^2 - 16x + 6 = (3x^2 - 2x + 1)(px^2 + qx + r)
\]

Next, we expand the right-hand side:
\[
(3x^2 - 2x + 1)(px^2 + qx + r) = 3px^4 + (3q - 2p)x^3 + (3r - 2q + p)x^2 + (-2r + q)x + r
\]

Now, we equate the coefficients of the corresponding powers of \(x\) from both sides of the equation. 

1. For \(x^4\): 
   \[
   3p = a
   \]

2. For \(x^3\): 
   \[
   3q - 2p = b
   \]

3. For \(x^2\): 
   \[
   3r - 2q + p = 32
   \]

4. For \(x^1\): 
   \[
   -2r + q = -16
   \]

5. For the constant term: 
   \[
   r = 6
   \]

Now, we substitute \(r = 6\) into the equations. 

From the constant term:
\[
r = 6
\]

Substituting \(r\) into the equation for \(x^1\):
\[
-2(6) + q = -16 \implies -12 + q = -16 \implies q = -4
\]

Next, substitute \(q\) into the equation for \(x^2\):
\[
3(6) - 2(-4) + p = 32 \implies 18 + 8 + p = 32 \implies p = 32 - 26 = 6
\]

Now we have \(p = 6\), \(q = -4\), and \(r = 6\). 

Substituting \(p\) into the equation for \(x^4\):
\[
3p = a \implies 3(6) = a \implies a = 18
\]

Substituting \(p\) and \(q\) into the equation for \(x^3\):
\[
3q - 2p = b \implies 3(-4) - 2(6) = b \implies -12 - 12 = b \implies b = -24
\]

Thus, the ordered pair \((a, b)\) is \((18, -24)\).<|END_OF_LATENT|> The final answer is $(18,-24)$.