Solution of a quartic equation. Suppose that the equation $x^4-2x^3+4x^2+6x-21=0$ is known to have two roots that are equal in magnitude but opposite in sign. Solve the equation. 
This is what I have been thinking. Suppose $\zeta_1$ $\zeta_2$ are roots. Such that $|\zeta_1|=|\zeta_2|$. Then $(x-\zeta_1)(x-\zeta_2)$ divides the polynomial. I don't know where to go from here. Any advice?
 A: If $\pm a$ are roots of $x^4-2x^3+4x^2+6x-21=0$ then
$a^4-2a^3+4a^2+6a-21=0$
$a^4+2a^3+4a^2-6a-21=0$
Adding these two equations gives you a biquadratic equation for $a$, which you can solve. (Make sure you select the roots of the original equation.)
Once you have found $a$, the other roots are $b$ and $2-b$ and must satisfy $-a^2b(2-b)=-21$.
A: So at the most basic level there is a factor $(x+a)(x-a)=x^2-a^2$ and the factorisation is $$(x^2-a^2)(x^2+bx+c)=x^4-2x^3+4x^2+6x-21$$
Equating powers of $x^3$ gives immediately a value for $b$. Then equating powers of $x$ gives $a^2b=-6$ so $a^2$ is known and the constant term gives $a^2c=21$ so that $c$ can be determined.
A: Let $P(x)= x^4−2x^3+4x^2+6x−21=0$ Let $a,-a $ are the two roots given. So, $P(a)=P(-a)=0$
\begin{align}
a^4−2a^3+4a^2+6a−21 &= &a^4+2a^3+4a^2-6a−21\\
\implies 4a^3-12a =0\\
\implies a(a^2-3)=0\\
a=0,a=\pm\sqrt3 
\end{align}
But $a=0$ is not a root of the polynomial. Hence $\pm \sqrt3$ are the two roots of $P(x)$ because at two points the functional values are zero and the functional values are same only at three points. Now by the long division or Synthetic Division we can factor $P(x)$. $P(x)=(x^2-3)(x^2-2x+7)$. Now we have to find the roots of the quadratic equation $(x^2-2x+7).$  The roots of $x^2-2x+7$ is $1\pm \sqrt6 i.$ So, the roots of the polynomial $P(x)$ are $\pm \sqrt3 ,1\pm i \sqrt6.$
