# How do you solve x^4 + 3x^3 - x^2 - 9x - 6 = 0?

Check the divisors of 6 to see if they are roots .You will find that $- 1$ and $- 2$ are hence the equation can take the form

$\left(x + 1\right) \left(x + 2\right) \left({x}^{2} + c x + d\right)$

Equate the two and after some calculations you will find that

${x}^{4} + 3 {x}^{3} - {x}^{2} - 9 x - 6 = 0 \implies \left(x + 1\right) \left(x + 2\right) \left({x}^{2} - 3\right) = 0$

Hence the solutions are

${x}_{1} = - 1$,
${x}_{2} = - 2$,
${x}_{3} = \sqrt{3}$
${x}_{4} = - \sqrt{3}$