# real zeros of a polynomial function

• July 29th 2010, 12:50 AM
nando29
real zeros of a polynomial function
Attachment 18373

Attachment 18372

i understand how four was found. i dont understand how 8/3 was found.
• July 29th 2010, 02:37 AM
HallsofIvy
$a_0= -\frac{1}{3}$, $a_1= \frac{1}{3}$, $a_2= -\frac{4}{3}$, $a_3= \frac{5}{3}$, and $a_4= -\frac{1}{3}$ again.

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

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

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.
• July 29th 2010, 02:40 AM
tonio
Quote:

Originally Posted by nando29
Attachment 18373

Attachment 18372

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

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

Tonio
• July 29th 2010, 10:34 AM
nando29
thank you very much!