# I have been having trouble on this problem and would really appreciate some help

#### Lorynn

I've been trying to work through this problem for hours and I am not sure on how to go about solving this problem.
Some help would be extremely appreciated.

" Show the two synthetic division "lines" for the integral values on either side of the upper real zero of f(x)=x3-9x2-x-5 "

I have tried dividing the equation synthetically by using the possible rational roots: +1,-1,+5, and -5 and each of these give me a remainder, I am so lost currently and would really appreciate some help

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Lorynn

#### chiro

MHF Helper
Hey Lorynn.

What do you mean by synthetic division lines? Are you just trying to factor the polynomial?

#### Lorynn

I'm not exactly sure, that is word for word how the problem was worded and I am not exactly sure on what is meant.

#### Plato

MHF Helper
I'm not exactly sure, that is word for word how the problem was worded and I am not exactly sure on what is meant.
I fear that you are at the mercy of a local set of definitions that are not in general use.

#### Archie

The real root is at $$\displaystyle x \approx 9.2$$. I don't know from the question how you are expected to find that though (other than trial and error).

#### chiro

MHF Helper
There is a formula for finding the roots of cubics.

Have you covered this?

#### skeeter

MHF Helper
" Show the two synthetic division "lines" for the integral values on either side of the upper real zero of f(x)=x3-9x2-x-5 "
This may be an example of locating the zero of a polynomial using the Intermediate Value Theorem.

In the old days "BC" (before calculators), one method to localize a zero of a polynomial was to use synthetic division for integral values of x, looking for a sign change in the remainder.

In this example, for integral values of $x\le 9$, $f(x) < 0$, and for integral values of $x \ge 10$, $f(x) > 0$.

Since f(x) is continuous over its domain, the Intermediate Value Theorem says that a zero for f(x) must exist in the interval $9 < x < 10$.

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