Doubt on 'Rest Theorem' in Polynomial Division I'm dealing with an issue dividing polynomials, I have:
Determine the value of $a$ to make: $x^2 + 2x - a$ divisible by $x + 4$
I don't know even where to start, this $a$ confuses me a lot;
Thanks in advance;
But I'm still confused, I'm trying to divide it in the longer way, I've multiplied with an $x$ coefficient and got a rest of $-2x - a$ but I'm unable to find the other part of the coefficient to cancel that $-2x$ and got the $a$.
 A: Gerry's comment will lead you to consider something like this: If $x+4$ really does divide $x^2+2x-a$, then $$x^2+2x-a=(x+4)(x+c)$$ for some $c$ that we don't really know or care about. What happens if you plug in $x=-4$?
A: The easiest way is to use the fact that $-4$ is a root of the polynomial $P(x)$ iff $x+4$ divides $P(x)$.
A: Hint $\ $ If $\rm\:x\!+\!4\ |\ f(x)\:$ then $\rm\:mod\ x\!+\!4\!:\ 0 \equiv f(x)\equiv f(-4)\: $ by $\rm\: x\equiv -4.$  
Alternatively in divisibility (vs. congruence form), apply the Factor Theorem.
Or, by Vieta, if the roots are $\rm\:r,-4\:$ then their sum is minus the linear coefficient $\rm\: r\!-\!4 = -2,\:$ hence $\rm\: r = 2,\:$ and the root product is the constant coefficient $\rm\: -4\:\!r = -a\:\Rightarrow\: a =\: \ldots$
A: $x^2+2x−a$  is the standard form.
\begin{align}
(x+4)(x+c) &= x^2+2x-a\\
x^2+cx+4x+4c &= x^2+2x-a
\end{align}
mean
$$x^2+(c+4)x+4c = x^2+2x-a$$
\begin{align}
c+4 &= 2 & 4c &= -a\\
c &=-2 & 4(-2) &= -a\\
-a &= -8 & a &= 8
\end{align}
so $x^2+2x-8$,  where the value of $a$ is $8$.
A: P(-4)=0
Then:
 $$(-4)^2+2\cdot{(-4)}-a  = 0$$
$$16-8-a  = 0$$
$$8  = a$$
