Find the constant $c$ such that the remainder when $2x+7$ divides $2x^3+cx^2-11x+39$ is $4$.
I know that the remainder theorem tells me that the remainder when a polynomial $f(x)$ is divided by $x-a$ is equal to $f(a)$. So, in this case, I want to find the value of $x$ that makes $2x+7=0$, and then plug that into the original polynomial and set it equal to $4$. To find the value of $x$ that makes $2x+7=0$, I can subtract $7$ from both sides and divide by $2$. I get $x=-\frac{7}{2}$. Now, I plug this into the original polynomial and set it equal to $4$. I get $2(-\frac{7}{2})^3+c(-\frac{7}{2})^2-11(-\frac{7}{2})+39=4$. I can simplify this equation by multiplying out the powers and fractions. I get $-\frac{343}{4}+\frac{49}{4}c+\frac{77}{2}+39=4$. I can further simplify this by multiplying both sides by $4$ to clear the fractions. I get $-343+49c+154+156=16$. I can combine the like terms on the left side. I get $49c-33=16$. I can add $33$ to both sides. I get $49c=49$. I can divide both sides by $49$. I get $c=1$. 