Compute $m$ so that the polynomial $R(x)$ has a root at $x=-3$ Apologies upfront, this is quite an elementary question, but I'm stuck here.
I'm asked to figure out the value of $m$ so that this polynomial: $$R(x)=x^2-mx+3$$ has a root at $x=-3$.
Following the definition of the root of a polynomial, $x=-3$ is a root if and only if $R(-3)=0$, which yields $\boxed{m=-12}$.
But in the other hand, following the Factor Theorem, if $x=-3$ is a root, then the polynomial $S(x)=(x+3)$ needs to be a factor of $R(x)$. Therefore, the remainder of $R(x)/S(x)$ needs to be $0$.
Applying Ruffini's rule with $x=-4$ I get the condition: $$3+3(3+m)=0$$ which is satisfied only in case $\boxed{m=-4}$
So I'm getting 2 different values for $m$, which is driving me crazy. I know this is a very basic question but I'm still wondering where I went wrong...
Thanks

It looks like you incorrectly performed your original calculation using the definition of a root of a polynomial. The calculation should look like this:
$$(-3)^2 - m(-3) + 3=0$$
$$9 + 3m + 3 = 0$$
$$12 + 3m = 0$$
$$12 = -3m$$
$$m=-4$$
