Descartes’ rule of signs is a way to determine the nature of roots to an equation. What this means in simple terms is if there is an equation that is set equal to zero, use Descartes’ Rule of Signs to see if the solutions, or roots, are positive, negative, or imaginary. The number of solutions to an equation will always equal the degree. Therefore if the equation is of degree 5 then there will be 5 solutions, or also called roots. The type of roots whether positive, negative, or imaginary and the number of each can be determined by counting the number of sign changes in the polynomial. Here is an example.

Example

Determine the nature of the roots of f(x) = 3x³ –2x² +4x – 7

 

Look at the polynomial from one term to the next and count the number of times that the signs change, either from positive to negative or negative to positive.

 

 

 

f(x) = 3x³ –2x² +4x – 7

 

 

Notice from the lines drawn that there are 3 changes in sign. That is, it goes from +3 to –2, -2 to +4, and +4 to –7. This means there are 3 possible positive roots. Now anytime there is the possibility of 2 or more roots of one type, like in this case, where there are 3 positive, there is also the possibility of 2 less roots than that. This means that there are either 3 or 1 positive roots. If there had been 5 sign changes in the original polynomial then there could have been 5, 3, or 1 possible positive roots. Take the number of possible roots and subtract 2 from it unless it would get a negative number. If there were 6 sign changes, then one could have 6,4,2, or 0 possible positive roots.

 

This case is of degree 3, since the highest exponent is 3. That means that there are 3 roots and either all 3 are positive or just 1 is, because of the 3 sign changes. Now see how many negative roots there are.

 

To do this, look at the f(-x) function. Plug in –x everywhere there is an x and simplify.

 

f(-x) = 3(-x)³ –2(-x)² +4(-x) – 7

 

f(-x) = -3x³ –2x² -4x – 7

 

Now that it is simplified, count the sign changes. Since they are all the same sign, negative, there are no sign changes. Therefore, there are no negative roots.

 

Since there are not any negative roots and the only other roots possible to look at are imaginary, just use them to fill in the blanks. If there are 3 positive roots and only 3 possible roots, that means that there are 0 negative, and 0 imaginary. If the case is true where there is only 1 positive root, and there are no negative roots, then that means there would have to be 2 imaginary roots since there has to be a total of three roots. Using Descartes’ Rule of Signs one might not be able state the specific nature of roots but it can narrow it down. In this case it is narrowed down to 2 possibilities.

 

3 positive		1 positive
0 negative	or	0 negative
0 imaginary		2 imaginary

 

In each of these possibilities there are a total of 3 roots, which is because the problem is of degree 3. The number of roots in each possible case must equal the degree of the original polynomial.

 

Note- Imaginary roots will always come in groups of two if they exist. Thus the only possible number of imaginary roots one can have are 0, 2, 4, 6, etc….

 

Example

Describe the nature of the roots of f(x) = 4x⁵ –3x⁴ +3x³ +2x² +7x + 4

 

Count the number of sign changes to find out how many possible positive roots there are.

 

f(x) = 4x⁵ –3x⁴ +3x³ +2x² +7x + 4

 

There are 2. That means there are either, 2 or 0 positive roots. Now find f(-x).

 

f(-x) = -4x⁵ –3x⁴ -3x³ +2x² -7x + 4

 

Count the number of sign changes to find out the number of negative roots.

 

f(-x) = -4x⁵ –3x⁴ -3x³ +2x² -7x + 4 There are 3. That means that there are either 3 or 1 negative roots.

 

List all the possible outcomes starting with 2 positive roots. Remember that there are a total of 5 roots since it is of degree 5.

2 positive

3 negative

0 imaginary

 

or

 

2 positive

1 negative

2 imaginary

List all the possible roots with 0 positive roots.

0 positive

3 negative

2 imaginary

 

or

 

0 positive

1 negative

4 imaginary

 

 

These are all the possible outcomes for the nature of the roots, list them all together.

 

2 positive 3 negative 0 imaginary

or

2 positive 1 negative 2 imaginary

or

0 positive 3 negative 2 imaginary

or

0 positive 1 negative 4 imaginary

 

 

Answer!

 

 

 

 

Example

 

Describe the nature of the roots for f(x) = 7x⁴ -3x³ +2x² -5x + 1

Count the number of sign changes.

 

 

f(x) = 7x⁴ -3x³ +2x² -5x + 1

 

There are 4. So, there are 4,2,or 0 positive roots. Find f(-x).

 

f(-x) = 7x⁴ +3x³ +2x² +5x + 1

 

Count the sign changes.

 

f(x) = 7x⁴ -3x³ +2x² -5x + 1

 

There are 0. There are 0 negative roots. List all the possible roots.

4 positive 0 negative 0 imaginary

or

2 positive 0 negative 2 imaginary

or

0 positive 0 negative 4 imaginary

 

 

 

 

Answer!

 

Special Cases- In working problems like this there are 2 main kinds of problems

to look out for. There are those with missing terms and those with x in every term.

 

Example

Find the nature of the roots of f(x)=3x³ –1

 

This is a case with missing terms, x² and x. Treat it just like the others by first counting the number of sign changes.

 

 

f(x) = 3x³ –1

 

There is 1 sign change so 1 positive root. Find f(-x).

 

f(-x) = -3x³ –1

 

Count the sign changes in f(-x) to find out how many negative roots there are.

 

f(-x) = -3x³ –1

 

There are no sign changes so there are 0 negative roots. List all possibilities. Remember that the highest exponent is 3 so there are 3 roots.

 

1 positive

0 negative

2 imaginary

Answer!

 

The other special case to look out for is when there is at least one x in every term. When this happens, there will be at least one root at zero.

 

 

 

Example

 

Find the nature of the roots of f(x) = 5x⁵ –3x⁴ +3x³

 

Notice that this is of degree 5, which means there are 5 roots. Not only is there an x in every term, but every term has 3 of them. The first thing to do is to factor out all possible x’s, which is 3 in this case.

f(x) = x³(4x² –3x +3)

 

Since there were 3 x’s factored out, that means that 3 of the roots are at 0, they are neither positive, negative, nor imaginary. The rest of the equation, 4x² –3x +3, is all that is left. Count the number of sign changes.

 

 

f(x) = 4x² –3x +3

 

There are two sign changes, which means there are either 2 or 0 positive roots. Find f(-x) and count the number of sign changes.

 

f(-x) = 4x² +3x +3

 

There are no sign changes so there are no negative roots. Write out all the answers keeping in mind that 3 are at zero.

 

3 at zero 2 positive 0 negative 0 imaginary

or

3 at zero 0 positive 0 negative 2 imaginary

 

 

 

Answer!