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 $P$?
I know that if $x-5$ is a factor of $P$, then $P(5)=0$, by the factor theorem. So I can plug in $x=5$ into $P(x)$ and set it equal to zero, and then solve for $c$. I get $P(5)=5^3+2(5^2)+c(5)+10=0$. Simplifying, I get $125+50+5c+10=0$. Combining like terms, I get $185+5c=0$. Subtracting $185$ from both sides, I get $5c=-185$. Dividing both sides by $5$, I get $c=-37$. 