Find the constant $b$ such that
$$\left(5x^2-3x+\frac{7}{3}\right)(ax^2+bx+c) = 15x^4 - 14x^3 + 20x^2 - \frac{25}{3}x + \frac{14}{3}$$
We can use the coefficient of the $x^3$ term to find $b$. On the right we have $-14x^3$, and on the left, the only $x^3$ terms we will get when we expand are $-3x(ax^2)$ and $5x^2(bx)$.

So we must have
$$-3ax^3 + 5bx^3 = -14x^3$$which means
$$5b - 3a = -14$$To find $a$, we use the same reasoning and look at the $x^4$ terms. On the right we have $15x^4$, and on the left, the only $x^4$ term we will get when we expand is $5x^2(ax^2)$. Then we know that
$$5ax^4 = 15x^4$$which means that $a=3$.

Then $5b -3(3) = -14$ and $b = \boxed{-1}$.
-1
Level 2
Intermediate Algebra
Problem: Find the constant $b$ such that
$$\left(5x^2-3x+\frac{7}{3}\right)(ax^2+bx+c) = 15x^4 - 14x^3 + 20x^2 - \frac{25}{3}x + \frac{14}{3}$$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the constant \( b \), we need to expand the left-hand side of the equation. We start with the polynomial multiplication of \( (5x^2 - 3x + \frac{7}{3})(ax^2 + bx + c) \). 

First, we distribute \( 5x^2 \) across \( ax^2 + bx + c \):
- \( 5x^2 \cdot ax^2 = 5a x^4 \)
- \( 5x^2 \cdot bx = 5b x^3 \)
- \( 5x^2 \cdot c = 5c x^2 \)

Next, we distribute \( -3x \):
- \( -3x \cdot ax^2 = -3a x^3 \)
- \( -3x \cdot bx = -3b x^2 \)
- \( -3x \cdot c = -3c x \)

Then, we distribute \( \frac{7}{3} \):
- \( \frac{7}{3} \cdot ax^2 = \frac{7a}{3} x^2 \)
- \( \frac{7}{3} \cdot bx = \frac{7b}{3} x \)
- \( \frac{7}{3} \cdot c = \frac{7c}{3} \)

Now, we combine all these results:
- The coefficient of \( x^4 \) is \( 5a \).
- The coefficient of \( x^3 \) is \( 5b - 3a \).
- The coefficient of \( x^2 \) is \( 5c - 3b + \frac{7a}{3} \).
- The coefficient of \( x \) is \( -3c + \frac{7b}{3} \).
- The constant term is \( \frac{7c}{3} \).

We set these coefficients equal to the corresponding coefficients from the right-hand side \( 15x^4 - 14x^3 + 20x^2 - \frac{25}{3}x + \frac{14}{3} \):
1. \( 5a = 15 \) implies \( a = 3 \).
2. \( 5b - 3a = -14 \) becomes \( 5b - 9 = -14 \), leading to \( 5b = -5 \) and thus \( b = -1 \).
3. \( 5c - 3b + \frac{7a}{3} = 20 \) can be solved with \( c \) once \( b \) is known.
4. \( -3c + \frac{7b}{3} = -\frac{25}{3} \) will also help find \( c \).
5. Finally, \( \frac{7c}{3} = \frac{14}{3} \) gives \( c = 2 \).

The calculations confirm that the value of \( b \) is indeed \( -1 \).<|END_OF_LATENT|> The final answer is $-1$.