Find the constant $c$ such that the remainder when $2x+7$ divides $2x^3+cx^2-11x+39$ is $4$.
\[
\begin{array}{c|cc cc}
\multicolumn{2}{r}{x^2} & +\left(\frac{c-7}{2}\right)x & +5 \\
\cline{2-5}
2x+7 & 2x^3 &+cx^2 &- 11x &+ 39  \\
\multicolumn{2}{r}{-2x^3} & -7x^2  \\ 
\cline{2-3}
\multicolumn{2}{r}{0} & (c-7)x^2 & -11x  \\
\multicolumn{2}{r}{} & -(c-7)x^2 & -x(c-7)\left(\frac{7}{2}\right)   \\ 
\cline{3-4}
\multicolumn{2}{r}{} &  0 & -x\left(\frac{7c-27}{2}\right) & + 39  \\
\multicolumn{2}{r}{} &   & -10x & -35  \\
\cline{4-5}
\multicolumn{2}{r}{} &   & -x\left(\frac{7c-27+20}{2}\right) & 4  \\
\end{array}
\]In the last step of the division, we have $39$ left as the constant term in our dividend and we need a remainder of $4$ at the end. Since our divisor has a term of $7$, the only way to do this is if our quotient has $5$ which gives us  $7\cdot5=35$ to subtract from our dividend and get the right remainder.

Then, we need the rest of our remainder to be $0$. This means
$$\frac{7c-27+20}{2} = 0$$which gives us
$$c = \boxed{1}.$$
1
Level 3
Intermediate Algebra
Problem: Find the constant $c$ such that the remainder when $2x+7$ divides $2x^3+cx^2-11x+39$ is $4$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the constant \( c \), we need to understand the process of polynomial long division or the Remainder Theorem. The Remainder Theorem states that the remainder of a polynomial \( f(x) \) when divided by \( x - r \) is \( f(r) \). In this case, we are dividing by \( 2x + 7 \), which can be rewritten as \( 2(x + \frac{7}{2}) \). 

To apply the Remainder Theorem, we first need to find the root of \( 2x + 7 = 0 \). Solving for \( x \), we get \( x = -\frac{7}{2} \). 

Next, we substitute \( x = -\frac{7}{2} \) into the polynomial \( 2x^3 + cx^2 - 11x + 39 \) to find the remainder. We calculate each term:

1. Calculate \( 2(-\frac{7}{2})^3 \):
   \[
   2 \left(-\frac{343}{8}\right) = -\frac{343}{4}
   \]

2. Calculate \( c(-\frac{7}{2})^2 \):
   \[
   c \left(\frac{49}{4}\right) = \frac{49c}{4}
   \]

3. Calculate \( -11(-\frac{7}{2}) \):
   \[
   \frac{77}{2}
   \]

4. The constant term is \( 39 \).

Now, we combine these results:
\[
-\frac{343}{4} + \frac{49c}{4} + \frac{77}{2} + 39
\]
To combine the terms, convert \( \frac{77}{2} \) and \( 39 \) to have a common denominator of 4:
\[
\frac{77}{2} = \frac{154}{4}, \quad 39 = \frac{156}{4}
\]
Thus, the expression becomes:
\[
-\frac{343}{4} + \frac{49c}{4} + \frac{154}{4} + \frac{156}{4}
\]
Combine the constant terms:
\[
-\frac{343 + 154 + 156}{4} + \frac{49c}{4} = \frac{49c - 33}{4}
\]

According to the problem, this remainder must equal 4:
\[
\frac{49c - 33}{4} = 4
\]
To eliminate the fraction, multiply both sides by 4:
\[
49c - 33 = 16
\]
Now, solve for \( c \):
\[
49c = 49 \quad \Rightarrow \quad c = 1
\]

Thus, the constant \( c \) that satisfies the condition is \( 1 \).<|END_OF_LATENT|> The final answer is $1$.