#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

Answer:

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

Answer:

#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#