Polynomials - The sum of two roots If the sum of two roots of
$$x^4  + 2x^3 - 8x^2 - 18x - 9 = 0$$ is
$0$, find
the roots of the equation
 A: Generally, if there's two roots whose sum is zero, then it means that two factors are $x-a$ and $x+a$, which means that $x^2-a^2$ must be a factor. So clearly
$$
(x^2-a^2)(x^2+bx+c)=x^4+2x^3-8x^2-18x-9=0
$$
Find the values of $a^2$, $b$, and $c$ that satisfy the left equality, and you'll have found factors that you can then solve for all roots. (This works even if there's no rational root)
A: Hints: 
The constant coefficient (constant term) is $-9$. 
$$\pm 1, \pm 3\pm 9 \;\text{divide}\; -9 $$ 
If there are rational roots to the polynomial, they will be among these values. $\large (\star)$ 
Two of the roots  $x_i, x_j$ be must be factors of $-9$: and they must be such that $x_i = -x_j$. Lets call this pair $a, \pm a$. That means $(x−a)$ and $x-(-a) = (x + a)$ will be factors of the polynomial. 
Now, what do we get when we multiply $(x+a)(x-a)$? 

$(\large \star)$ Why do we know this? See the Rational Root Theorem, where you'll see why, and where you'll find additional tremendously useful things to know when you need to factor a polynomial (and you learn how to tell whether or not any rational roots exist!)
A: Hint: $x^4+2x^3+x^2-9(x^2+2x+1)=x^2(x+1)^2-9(x+1)^2=(x^2-9)(x+1)^2$.
A: I'd do it this way. If $x=\pm a$ are roots of the equation, then $f(x)=x^2-a^2$ is a factor of the polynomial. Note that this contains only even powers of $x$.
Now suppose we multiply $f(x)$ by another polynomial $p(x)$ so that $f(x)p(x)=P(x)$. Let e(x) be the polynomial obtained by picking out the even powers of $p(x)$ with $o(x)$ as the odd powers, so that $p(x)=e(x)+o(x)$ and $$f(x)p(x)=f(x)e(x)+f(x)o(x)=E(x)+O(x)=P(x)$$ gives a decomposition of $P(x)$ into the sum of even powers of $x$ plus the sum of odd powers of $x$, with $f(x)$ a factor of both $E(x)$ and $O(x)$.
In brief, we know that, in the case presented $$x^4-8x^2-9$$ and$$2x^3-18x$$
have a common factor of the kind we are seeking. In a more complex example we could use the Euclidean algorithm to pin this down further (there might be other common factors). But here the even and odd parts are easy to factorise as $$(x^2-9)(x^2+1)$$ (treat this as a quadratic in $x^2$) and$$2x(x^2-9)$$
From there we can easily finish the factorisation.
Note: this is an entirely systematic approach, which simplifies the problem without having to guess.
Note: this is essentially the approach suggested by Erick Wong in a comment on another answer, which I spotted after I had written this all out.
A: Let those two roots be $a,-a$ and the other two roots are $b,c$
$$\text{Now, }(x-a)(x+a)(x-b)(x-c)=(x^2-a^2)\{x^2-(b+c)x+bc\}$$
$$=x^4-x^3(b+c)+(bc-a^2)x^2+a^2(b+c)x-a^2bc$$
Comparing the coefficients of $x^3$ of this with that of $x^4  + 2x^3 - 8x^2 - 18x - 9 = 0,$
$b+c=-2$
Comparing the coefficients of $x,a^2(b+c)=-18\implies 2a^2=18\implies a=\pm3$
Comparing the  constants, $a^2bc=9\implies bc=1$
So, $b,c$ are the roots of the equation $t^2-(-2)t+(1)=0\implies t=-1,-1$
