# Intermediate value theorem/x^3-4x+2/Bisection method/Example

We want to determine approximately a zero for the polynomial

${\displaystyle {}f(x)=x^{3}-4x+2\,}$

with the help of the method given by the Intermediate value theorem. We have ${\displaystyle {}f(1)=-1}$ and ${\displaystyle {}f(2)=2}$, hence by fact, there must be a zero inside the interval ${\displaystyle {}[1,2]}$. We compute the value of the function at the arithmetic mean of the interval, which is ${\displaystyle {}{\frac {3}{2}}}$, and get

${\displaystyle {}f{\left({\frac {3}{2}}\right)}={\frac {27}{8}}-4\cdot {\frac {3}{2}}+2={\frac {27-48+16}{8}}={\frac {-5}{8}}<0\,.}$

Hence, we have to continue with the right half of the interval ${\displaystyle {}[{\frac {3}{2}},2]}$. The arithmetic mean thereof is ${\displaystyle {}{\frac {7}{4}}}$. The value of the function at this point is

${\displaystyle {}f{\left({\frac {7}{4}}\right)}={\left({\frac {7}{4}}\right)}^{3}-4\cdot {\frac {7}{4}}+2={\frac {343}{64}}-5={\frac {343-320}{64}}={\frac {23}{64}}>0\,.}$

So now we have to continue with the left part of the interval, namely ${\displaystyle {}[{\frac {3}{2}},{\frac {7}{4}}]}$. Its arithmetic mean is ${\displaystyle {}{\frac {13}{8}}}$. The value of the function at this point is

${\displaystyle {}f{\left({\frac {13}{8}}\right)}={\left({\frac {13}{8}}\right)}^{3}-4\cdot {\frac {13}{8}}+2={\frac {2197}{512}}-{\frac {13}{2}}+2={\frac {2197-3328+1024}{512}}={\frac {-107}{512}}<0\,.}$

Therefore, we know that there is a zero between ${\displaystyle {}{\frac {13}{8}}}$ and ${\displaystyle {}{\frac {7}{4}}={\frac {14}{8}}}$.