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

Mar 23, 2016

If $f \left(x\right)$ is a polynomial with integer coefficients and if $\frac{p}{q}$ is a zero of $f \left(x\right)$ i.e. $f \left(\frac{p}{q}\right) = 0$,

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

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

Now dividing $f \left(x\right)$ by $\left(x - 1\right)$, we get $2 {x}^{2} - 2 x + 3$.

As discriminant for $2 {x}^{2} - 2 x + 3$ is ${\left(- 2\right)}^{2} - 4 \times 2 \times 3 = - 20$, we cannot factorize it further.@

Hence, $1$ is the only zero of $f \left(x\right) = 2 {x}^{3} - 4 {x}^{2} + 5 x - 3$.

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 $a {x}^{2} + b x + c$, discriminant is given by ${b}^{2} - 4 a c$.