# 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;

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 $a$.

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Have you ever seen something called "the factor theorem"? That would be an excellent place to start. – Gerry Myerson Apr 20 '12 at 0:40
Are you familiar with polynomial long division? – anon Apr 20 '12 at 0:40
Multiply $x+4$ with $x-\frac{a}{4}$ and compare coefficients. – J. M. Apr 20 '12 at 0:42
@anon yes, my difficult is there, take a look in my edit. – aajjbb Apr 20 '12 at 1:28

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 $x=-4$?

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The easiest way is to use the fact that $-4$ is a root of the polynomial $P(x)$ iff $x+4$ divides $P(x)$.

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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 $\rm\: -4\:\!r = -a\:\Rightarrow\: a =\: \ldots$

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$x^2+2x−a$ 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 $8$.

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