Solve the equation $x^{4}-2x^{3}+4x^{2}+6x-21=0$ Solve the equation $$x^{4}-2x^{3}+4x^{2}+6x-21=0$$ given that two of its roots are equal in magnitude but opposite in sign.
I don't know how to solve it. The roots are given as $\pm\sqrt{3},1\pm i\sqrt{6}. $
 A: You are given the information that some polynomial $x^2-c$ divides $x^4-2x^3+4x^2+6x-21$. Now polynomial division gives:
$x^4-2x^3+4x^2+6x-21 = (x^2-c) \cdot (x^2-2x+c+4) + 2 (3-c) x+ (c^2+4c-21).$
But the remainder is supposed to be zero, so we get:
$2 (3-c) x+ (c^2+4c-21) = 0$
and in particular $c=3$. Thus, 
$x^4-2x^3+4x^2+6x-21 = (x^2-3) \cdot (x^2-2x+7)$
and the roots are easy to find.
A: Hint
Let $a,-a$ be the two roots. Then 
$$x^{4}-2x^{3}+4x^{2}+6x-21=(x^2-a^2)(x^2+bx+c)$$
Opening the brackets you get
$$x^{4}-2x^{3}+4x^{2}+6x-21=x^4+bx^3+(c-a^2)x^2-a^2bx-a^2c$$
Equaling the coefficients you get a system which is very easy to solve.
Note that the coefficinets of $x^3$ and of $x$ already give you both $a$ and $b$.
A: Hint:
We know that $$(x-a)(x+a) = (x^2-a^2)$$ is a factor. And
 $$x^2-2x^3+4x^2+6x-21 = (x^2-a^2)(x^2+bx+c)$$ From here we can equate coefficients,  $$21 = a^2c$$
$$4 = -a^2+c$$
$$-a^2b = 6$$
I'll leave the rest for you.
A: As $x$ and $-x$ are two roots,
$$x^{4}-2x^{3}+4x^{2}+6x-21=0$$and $$x^{4}+2x^{3}+4x^{2}-6x-21=0$$both hold, hence
by addition $$x^{4}+4x^{2}-21=0.$$
This yields $x^2=3$ or $x^2=-7$. Taking the first case, we have $x=\pm\sqrt3$ and $x^2-3$ is a factor.
By synthetic division, the other factor is $$x^2-2x+7.$$
A: Hint We can write the two special roots as $a, -a$, and so the polynomial contains has a factor of the form $x^2 - a^2$. So, we can write the polynomial as
$$x^4 - 2 x^3 + 4 x^2 + 6 x - 21 = (x^2 - a^2) (x^2 + B x + C)$$
for some $a, B, C$. (We know the leading coefficient of the second term has to be $1$ just by expanding and matching.) Expanding the r.h.s. and comparing, e.g., the coefficients of the $x^3$ terms gives $B = -2$, and comparing the other like coefficients lets us determine $a$ and $C$, too. Nowo, the solutions are $a$, $-a$, and the roots of $x^2 + B x + C$, which (having already determined $B, C$) we can find as usual with the quadratic equation.
