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}. $

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.
