What must be the value of the coefficient $c$ in $P(x)=x^3+2x^2+cx+10$, in order for $x-5$ to be a factor of $P$?
From the Factor Theorem, if $x-5$ is a factor of $P$, then $P(5) = 0$. Using this we have
$$5^3+2(5^2)+5c+10 = 0.$$Solving for $c$ gives us $c = \boxed{-37}$.
-37
Level 2
Intermediate Algebra
Problem: What must be the value of the coefficient $c$ in $P(x)=x^3+2x^2+cx+10$, in order for $x-5$ to be a factor of $P$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the value of the coefficient $c$ such that $x-5$ is a factor of the polynomial $P(x)=x^3+2x^2+cx+10$, we can apply the Factor Theorem. The Factor Theorem states that if $x-a$ is a factor of a polynomial $P(x)$, then $P(a)=0$. 

In this case, we set $a=5$. Therefore, we need to evaluate $P(5)$ and set it equal to zero:

1. Substitute $x=5$ into the polynomial:
   $P(5) = 5^3 + 2(5^2) + c(5) + 10$.

2. Calculate each term:
   - $5^3 = 125$,
   - $2(5^2) = 2(25) = 50$,
   - $c(5) = 5c$,
   - The constant term is $10$.

3. Combine these results:
   $P(5) = 125 + 50 + 5c + 10$.

4. Simplify the expression:
   $P(5) = 185 + 5c$.

5. Set the polynomial equal to zero to satisfy the condition of the Factor Theorem:
   $185 + 5c = 0$.

6. Solve for $c$:
   - Subtract 185 from both sides: $5c = -185$.
   - Divide by 5: $c = -37$.

Thus, the value of the coefficient $c$ must be $-37$ for $x-5$ to be a factor of the polynomial $P(x)$.<|END_OF_LATENT|> The final answer is $-37$.