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.