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

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Answer 1 · 2016-04-03T19:05:15

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

$(x+1)(x+2)(x^2+cx+d)$

Equate the two and after some calculations you will find that

$x^4 + 3x^3 - x^2 - 9x - 6 = 0=>(x+1) (x+2) (x^2-3) = 0$

Hence the solutions are

$x_1=-1$ ,
$x_2=-2$ ,
$x_3=sqrt3$
$x_4=-sqrt3$

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