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# Descartes Rule of Signs Help (page 3)

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By McGraw-Hill Professional
Updated on Oct 4, 2011

## Descartes Rule of Signs Practice Problems

#### Practice

1. List the candidates for rational zeros. Do not try to find the zeros. f(x) = 3 x 4 + 8 x 3 −11 x 2 + 3 x + 4

2. List the candidates for rational zeros. Do not try to find the zeros. P(x) = 6 x 4 − 24

3. Completely factor h(x) = 2 x 3 + 5 x 2 − 23 x + 10.

4. Completely factor P(x) = 7 x 3 + 26 x 2 − 15 x + 2.

5. Use Descartes’ Rule of Signs to count the possible number of positive zeros and the possible number of negative zeros of f(x) = 2 x 4 − 6 x 3x 2 + 4 x − 8.

6. Use Descartes’ Rule of Signs to count the possible number of positive zeros and the possible number of negative zeros of f(x) = − x 3x 2 + x + 1.

7. Show that the given values for a and b are lower and upper, respectively, bounds for the zeros of f(x) = x 3 − 6 x 2 + x + 5; a = −3, b = 7.

8. Show that the given values for a and b are lower and upper, respectively, bounds for the zeros of f(x) = x 4x 2 − 2; a = −2, b = 2.

9. Sketch the graph for g(x) = x 3x 2 − 17 x − 15.

#### Solutions

1. Possible numerators: ±1, ±2, ±4

Possible denominators: 1 and 3

Possible rational zeros:

2. Possible numerators: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24

Possible denominators: 1, 2, 3, 6

Possible rational zeros (with duplicates omitted): ±1, ±2, ±3, ±4, ±6,

3. The possible rational zeros are ±1, ±2, ±5, ±10, , and . Because h (2) = 0, x = 2 is a zero of h(x) .

h(x) = (x − 2)(2 x 2 + 9 x − 5)

h(x) = ( x − 2)(2 x − 1)( x + 5)

4. The possible rational zeros are ±1, ±2, , and . Because , is a zero for P(x)

We will use the quadratic formula to find the zeros for x 2 + 4 x − 1.

5. There are 3 sign changes in f(x) , so there are 3 or 1 positive zeros.

f(−x) = 2(−x) 4 − 6(−x) 3 − (−x) 2 + 4(− x )− 8

= 2 x 4 + 6 x 3x 2 − 4 x − 8

There is 1 sign change in f(−x) , so there is exactly 1 negative zero.

6. There is 1 sign change in f(x) , so there is exactly 1 positive zero.

f(−x) = −(− x ) 3 − (− x ) 2 + (− x ) + 1

= x 3x 2x + 1

There are 2 sign changes in f(−x) , so there are 2 or 0 negative zeros.

7.

The entries on the bottom row alternate between positive and negative (or nonnegative and nonpositive), so a = −3 is a lower bound for the zeros of f(x) .

The entries on the bottom are positive (nonnegative), so b = 7 is an upper bound for the positive zeros of f(x) .

8.

The entries on the bottom row alternate between positive and negative, so a = −2 is a lower bound for the negative zeros of f(x) .

The entries on the bottom row are all positive, so b = 2 is an upper bound for the positive zeros of f(x) .

9. The x -intercepts are − 3, − 1, and 5. We will plot points for x = −3.5, x = −2, x = 0, x = 3, and x = 5.5.

Fig. 7.14

Practice problems for this concept can be found at: Polynomial Functions Practice Test.

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