# Rational Zeros Theorem

• Dec 6th 2009, 11:15 AM
Shonard
Rational Zeros Theorem
I have been given the equation:

$f(x) = 81x^7 + 72x^6 - 389x^5 - 360x^4 - 80x^3$

and asked to find a) and b) using the Rational Zeros Theorem

a) Factor f(x) into linear factors.

b) List the zeros of f(x) and their multiplicities

I understand that to use the rational zeroes theorem requires that there is a constant term so I factor out x^3, but at that point it seems there are no rational zeros. Any thoughts?
• Dec 6th 2009, 11:59 AM
VonNemo19
Quote:

Originally Posted by Shonard
I have been given the equation:

$f(x) = 81x^7 + 72x^6 - 389x^5 - 360x^4 - 80x^3$

and asked to find a) and b) using the Rational Zeros Theorem

a) Factor f(x) into linear factors.

b) List the zeros of f(x) and their multiplicities

I understand that to use the rational zeroes theorem requires that there is a constant term so I factor out x^3, but at that point it seems there are no rational zeros. Any thoughts?

The possible zeros are of the form $\frac{p}{q}$ where $p$ represents all the numbers that divide -80, and $q$ represents all the integers that divide 81.

This particular expression is a nightmare. Try $\frac{-4}{9}$
• Dec 6th 2009, 02:03 PM
Shonard
Awesome that worked.

I now have:

$(x + 4/9)(81x^3 + 36x^2 - 405x - 180)$

unfortunately this sequence is about as easy to work with as the first, is there any way to narrow down the possibility's?
• Dec 6th 2009, 03:53 PM
Plato