Question

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

Answer 1

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`