# Thread: real zeros of a polynomial function

1. ## real zeros of a polynomial function

i understand how four was found. i dont understand how 8/3 was found.

2. $\displaystyle a_0= -\frac{1}{3}$, $\displaystyle a_1= \frac{1}{3}$, $\displaystyle a_2= -\frac{4}{3}$, $\displaystyle a_3= \frac{5}{3}$, and $\displaystyle a_4= -\frac{1}{3}$ again.

Taking absolute values, $\displaystyle |a_0|= \frac{1}{3}$, $\displaystyle |a_1|= \frac{1}{3}$, $\displaystyle |a_2|= \frac{4}{3}$, $\displaystyle |a_3|= \frac{5}{3}$, and $\displaystyle |a_4|= \frac{1}{3}$ so we really only have three different values, $\displaystyle \frac{1}{3}$, $\displaystyle \frac{4}{3}$, and $\displaystyle \frac{5}{3}$.

The largest (max) of those is $\displaystyle \frac{5}{3}$ so $\displaystyle 1+ max(|a_0, a_1, a_2, a_3, a_4)$ is $\displaystyle 1+ \frac{5}{3}= \frac{3}{3}+ \frac{5}{3}= \frac{8}{3}$.

The difference between the two is that in the first one, you add all the fractions, to get 4, then take the larger of 1 and 4. In the second, you take the largest of the fractions, then add 1 to that.

3. Originally Posted by nando29

i understand how four was found. i dont understand how 8/3 was found.

The maximum of $\displaystyle \{\left|-\frac{1}{3}\right|,\left|\frac{5}{3}\right|, etc.\}$ is $\displaystyle \frac{5}{3}$ , and this added to 1 gives $\displaystyle \frac{8}{3}$

Tonio

4. thank you very much!