Solve `2x^3-9x^2+3x+4=0`

From the rational root theorem, we know that the only possible rational roots are `+-4,+-2,+-1,+-1/2`

Trying 1 we find that x=1 is a solution. We then can use polynomial long division or synthetic division to get

`2x^3-9x^2+3x+4=(x-1)(2x^2-7x-4)=0`

This factors, or we can use the quadratic formula on the quadratic.

Thus `(x-1)(2x+1)(x-4)=0 => x=1,-1/2,4`

`2x^3 -9x^2 + 3x +4 =`  0

`2x^3 -(9x^2 -3x-4) =` 0

`2x^3 -( 8x^2 +x^2 -3x -4 ) =`  0

`2x^3 -8x^2 -(x^2 -3x-4) =`  0

Now we will factor `2x^2`  from the first two terms and the quadratic equation.

`2x^2 (x-4) - (x-4)(x+1) =`  0

Now we will factor (x-4).

`==gt (x-4)(2x^2 -(x+1)) = 0 `

`==gt (x-4) (2x^2 -x -1) =0 `

`==gt (x-4)(2x+1)(x-1) = 0 `

`==gt x1= 4 `

`==gt x2= -1/2 `

`==gt x3= 1`

`` `==gt x = { -1/2, 1, 4}`

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