<s>[INST] 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}. $
 [/INST] 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.
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