How do you find all the zeros of #f(x)=2x^3-4x^2+5x-3#?

1 Answer
Mar 23, 2016

If #f(x)# is a polynomial with integer coefficients and if #p/q# is a zero of #f(x)# i.e. #f(p/q)=0#,

then #p# is a factor of the constant term of #f(x)# and #q# is a factor of the leading coefficient of #f(x)# i.e. of the highest power of #x# in #f(x).

As here #f(x)=2x^3-4x^2+5x-3#, zeros could be among factors of #-3# i.e. #(1,-1,3,-3)#. It is observed that for #x=1#, #f(1)=0#, hence #(x-1)# is a factor of #f(x)=2x^3-4x^2+5x-3#.

Now dividing #f(x)# by #(x-1)#, we get #2x^2-2x+3#.

As discriminant for #2x^2-2x+3# is #(-2)^2-4xx2xx3=-20#, we cannot factorize it further.@

Hence, #1# is the only zero of #f(x)=2x^3-4x^2+5x-3#.

Note that we are assuming the domain to be real numbers, if domain is complex numbers we will have two more zeros, which are not real numbers.

@ for a polynomial #ax^2+bx+c#, discriminant is given by #b^2-4ac#.