Descartes' Rule

Descartes' Rule - How it Works - Video

Descartes' Rules

Descarte's Rule:

Descarte's Rule has two statutes. The first one is for the positive real zeros. The second one is for the negative real zeros.

For instance if we find that our function has 4 possible positive real zeros. We could also have 2 possible positive real zeros. We could also have 0 possible positive real zeros. That is what it means by less than that number be even integer. So we subtract 2 until we reach 1 or 0 real zeros whether it is either positive or negative.

Example 1

Example 1:

We have our function f(x) = 4x5 - 5x4 + x3 - 2x2 - 3x + 8.

We can use Descartes' rule to find the number of possible positive real zeros, possible negative real zeros, and complex real zeros. We can use this information to narrow down our search for the zeros.

The first step that we need to is add positive sign in front of the first coefficient if need be. This will helps us determine how times the sign changes. Now we have f(x) = + 4x5 - 5x4 + x3 - 2x2 - 3x + 8. After drawing arcs from each change of sign, we found that there are 4 in our function. Since the input of f(x) is positive, we found the number of possible positive real zeros, 4. We could also have 2 positive real zeros or 0 positive real zeros.

Now we have to find the number of possible negative real zeros. In order to do that our input must be negative. So let's substitute -x into f(x). Now we have f(-x) = + 4(-x)5 - 5(-x)4 + (-x)3 - 2(-x)2 - 3(-x) + 8. Let's simplify and our result is f(-x) = - 4x5 - 5x4 - x3 - 2x2 + 3x + 8. After drawing arcs from each change of sign, we found that there was only 1  in our function. So the number of possible number of negative real zeros is 1. We don't subtract 2 anymore because we reached 0 or 1. 

Now we can make our table.

+ Real Zeroes

4

2

0

- Real Zeroes

1

1

1

i Real Zeroes

5

2

4

Total

5

5

5

Our total has to be the highest degree, which our function is 5.

Example 2

Example 2:

We have our function f(x) = -2x6 + x4 + 4x2 + 9x - 1.

We can use Descartes' rule to find the number of possible positive real zeros, possible negative real zeros, and complex real zeros. We can use this information to narrow down our search for the zeros.

Since we have a negative sign out in front, we don't need to add any signs. We also don't have to worry about the "missing terms" like 0x5 or 0x3. So we have f(x) = -2x6 + x4 + 4x2 + 9x - 1. After drawing arcs from each change of sign, we found that there are 2 in our function. Since the input of f(x) is positive, we found the number of possible positive real zeros, 2. We could also have 0 positive real zeros.

Now we have to find the number of possible negative real zeros. In order to do that our input must be negative. So let's substitute -x into f(x). Now we have f(-x) = -2(-x)6 + (-x)4 + 4(-x)2 + 9(-x) - 1. Let's simplify and our result is f(-x) = -2x6 + x4 + 4x2 - 9x - 1. After drawing arcs from each change of sign, we found that there are 2 in our function. So the number of possible number of negative real zeros is 2. We could also have 0 negative real zeros.

Now we can make our table.

+ Real Zeroes

2

0

2

0

- Real Zeroes

2

2

0

0

i Real Zeroes

2

4

4

6

Total

6

6

6

6

Our total has to be the highest degree, which our function is 6.

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