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

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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 $- 1$ $-1$ and $- 2$ $-2$ are hence the equation can take the form

$(x + 1) (x + 2) (x^{2} + c x + d)$ $(x+1)(x+2)(x^2+cx+d)$

Equate the two and after some calculations you will find that

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

Hence the solutions are

$x_{1} = - 1$ $x_1=-1$ ,
$x_{2} = - 2$ $x_2=-2$ ,
$x_{3} = \sqrt{3}$ $x_3=sqrt3$
$x_{4} = - \sqrt{3}$ $x_4=-sqrt3$

How do you solve x4+3x3−x2−9x−6=0x^4 + 3x^3 - x^2 - 9x - 6 = 0?