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# Descartes Rule of Signs Help

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By McGraw-Hill Professional
Updated on Oct 4, 2011

## Introduction to Descartes Rule of Signs and the Upper and Lower Bounds Theorem

There are a couple of algebra facts that can help eliminate some of the possible rational zeros. The first we will learn is Descartes’ Rule of Signs. The second is the Upper and Lower Bounds Theorem . Descartes’ Rule of Signs counts the number of positive zeros and negative zeros. For instance, according to the rule f(x) = x 3 + x 2 + 4 x + 6 has no positive zeros at all. This shrinks the list of possible rational zeros from ±1, ±2, ±3, and ±6 to −1, −2, −3, and −6. Another advantage of the sign test is that if we know that there are two positive zeros and we have found one of them, then we know that there is exactly one more.

The Upper and Lower Bounds Theorem gives us an idea of how large (in both the positive and negative directions) the zeros can be. For example, we can use the Upper and Lower Bounds Theorem to show that all of the zeros for f(x) = 5 x 3 + 20x 2 − 9 x − 36 are between −5 and 5. This shrinks the list of possible rational zeros from ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36, , and to ±1, ±2, ±3, ±4, , , and .

### Descartes' Rule of Signs

Descartes’ Rule of Signs counts the number of positive zeros and the number of negative zeros by counting sign changes. The maximum number of positive zeros for a polynomial function is the number of sign changes in f(x) = a n x n + a n −1 x n −1 + ... + a 1 x + a 0 . The possible number of positive zeros is the number of sign changes minus an even whole number. For example, if there are 5 sign changes, there are 5 or 3 or 1 positive zeros. If there are 6 sign changes, there are 6 or 4 or 2 or 0 positive zeros. The polynomial function f(x) = 3 x 4 − 2 x 3 + 7 x 2 + 5 x − 8 has 3 sign changes: from 3 to −2, from −2 to 7, and from 5 to −8. There are either 3 or 1 positive zeros. The maximum number of negative zeros is the number of sign changes in the polynomial f(− x ). The possible number of negative zeros is the number of sign changes in f(− x ) minus an even whole number.

#### Examples

Use Descartes’ Rule of Signs to count the possible number of positive zeros and negative zeros for the polynomial functions.

• f(x) = 5 x 3 − 6 x 2 − 10 x + 4

There are 2 sign changes: from 5 to −6 and from −10 to 4. This means that there are either 2 or 0 positive zeros. Before we count the possible number of negative zeros, remember from earlier in the book that for a number a, a (−x) even power = ax even power and − a ( x ) odd power = − ax odd power.

f(−x) = 5(− x ) 3 − 6(− x ) 2 − 10(−x) + 4

= −5 x 3 − 6 x 2 + 10 x + 4

There is 1 sign change, from −6 to 10, so there is exactly 1 negative zero.

• P(x) = x 5 + x 3 + x + 4

There are no sign changes, so there are no positive zeros.

P (− x ) = (− x ) 5 + (− x ) 3 + (− x ) + 4

= − x 5x 3x + 4

There is 1 sign change, so there is exactly 1 negative zero.

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