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?

2 Answers
Jul 18, 2016

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)

Jul 18, 2016

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