Equation $3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$ Solve the equation 
$$3x^4 + 2x^3 + 9x^2 + 4x + 6 = 0$$
Having a complex root of modulus $1$.
To get the solution, I tried to take a complex root $\sqrt{\frac{1}{2}} + i \sqrt{\frac{1}{2}}$ but couldn't get the solution right. Please help me.

Note that you have an equation with integer coefficients (real coefficients would do), so that any complex root $a$ of modulus $1$ implies a conjugate root $\bar {a }$ with $a\bar a=1$. You will therefore have a real factor of the form $x^2-bx+1$ with $b=a+\bar a$. Making obvious deductions from the leading and constant terms, there is a real factorisation of the original quartic of the form $$3x^4 + 2x^3 + 9x^2 + 4x + 6 = (x^2-bx+1)(3x^2+cx+6)$$
Equating coefficients of $x^3$ and $x$ gives $c-3b=2$ and $c-6b=4$, so that $c=0, b=-\frac 23$.
That gives two quadratic equations to be solved. [Note the factor $3$ can be redistributed to give a factorisation with integer coefficients]
