# How do you use the Intermediate Value Theorem and synthetic division to determine whether or not the following polynomial P(x) = x^3 - 3x^2 + 2x - 5 have a real zero between the numbers 2 and 3?

Jun 29, 2018

Use the Intermediate Value Theorem to find that it does have a zero in $\left[2 , 3\right]$...

#### Explanation:

Given:

$P \left(x\right) = {x}^{3} - 3 {x}^{2} + 2 x - 5$

we find:

$P \left(2\right) = {\left(\textcolor{b l u e}{2}\right)}^{3} - 3 {\left(\textcolor{b l u e}{2}\right)}^{2} + 2 \left(\textcolor{b l u e}{2}\right) - 5 = 8 - 12 + 4 - 5 = - 5$

$P \left(3\right) = {\left(\textcolor{b l u e}{3}\right)}^{3} - 3 {\left(\textcolor{b l u e}{3}\right)}^{2} + 2 \left(\textcolor{b l u e}{3}\right) - 5 = 27 - 27 + 6 - 5 = 1$

So:

$P \left(2\right) = - 5 < 0 < 1 = P \left(3\right)$

The intermediate value theorem tells us that if $f \left(x\right)$ is continuous on $\left[a , b\right]$ then $f \left(x\right)$ takes every value between $f \left(a\right)$ and $f \left(b\right)$ somewhere in the interval $\left[a , b\right]$.

Hence we can deduce that there is some $x \in \left[2 , 3\right]$ such that $P \left(x\right) = 0$.

We do not need to use synthetic division.