Question

`f(x) = 2x^4-3x^3-5x^2+9x-3` Given that `sqrt(3)` forms two of the roots, find all the roots of the polynomial?

Answer 1

The roots of the polynonial are: `+-sqrt(3), 1, 1/2`

Explanation:

`f(x) = 2x^4-3x^3-5x^2+9x-3`

The roots of the polynonial are those values of `x` for which `f(x)=0`

Since `f(x)` is a polynomial of degree 4 we know it will have 4 roots (real and/or imaginary).

The question tells us the `sqrt(3)` forms two of the roots so we deduce that `(x^2-3)` must be a factor, and that two of the roots must be given by:

`(x^2-3) =0 -> x= +-sqrt(3)`

Long dividing `2x^4-3x^3-5x^2+9x-3` by `(x^2-3)`

`-> 2x^2-3x+1`

This factorises as: `(2x-1)(x-1)`

Hence the two remaining roots are: `1 , 1/2`

Therefore the four roots of `f(x)` are `+-sqrt(3), 1 , 1/2` (all real this case)

Answer 2

`x=1, 1/2, sqrt3, -sqrt3`

Explanation:

There appears to be error in posting the question.
As mentioned the two zeros are not `sqrt3, and sqrt 3`.
I found that the two zeros must be `sqrt3, and -sqrt 3`
.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.--.-.-.
Given equation is
`f(x)=2x^4-3x^3-5x^2+9x-3` ......(1)
Given that two of its zeros are `x=sqrt3, -sqrt3`
`=>(x+sqrt3), (x-sqrt3)` are two factors of equation (1)
`=>(x+sqrt3)(x-sqrt3)=(x^2-3)` is a factor of the polynomial.

If we divide the equation (1) by the above quadratic by long division method we get another quadratic which is a factor of equation (1)
`:. (2x^4-3x^3-5x^2+9x-3)/(x^2-3)`, we get dividend as
`2x^2-3x+1`

To find factors of second quadratic we use split the middle term method
`2x^2-2x-x+1`, paring and taking out the common factors we get
`2x(x-1)-(x-1)`
`=>(x-1)(2x-1)`
Setting each factor `=0`, we obtain remaining two zeros as
`x=1, 1/2`