Descartes' Rule of Signs Calculator
Our Descartes' rule of signs calculator is here to help you learn and use the famous rule that allows you to find the possible amount of positive roots of any polynomial*, as well as the potential number of its negative roots and nonreal roots.
Have you already learned what Descartes' rule of signs is? If not, don't worry  we explain Descartes' rule of signs below so that you can start using it immediately! To avoid any confusion, we have also included a stepbystep guide on how to use Descartes' rule of signs. We have also provided some examples of Descartes' rule of signs used in practice!
Eager to learn more about polynomials? You may benefit from knowing that Omni has a whole collection of useful calculators:
 Adding and subtracting polynomials;
 Multiplying polynomials;
 Dividing polynomials;
 Synthetic division of polynomials; and
 Rational root test.
Make sure to try every one of them!
What is Descartes' rule of signs?
Descartes' rule of signs is a method of determining the possible number of:
 Positive real zeroes;
 Negative real zeroes; and
 Nonreal zeroes of a polynomial.
This method says that the number of positive zeros is upperbounded by the number of sign changes in the polynomial coefficients and that these two numbers have the same parity. In particular, if the number of sign changes is zero or one, then the polynomial has exactly zero or one positive root, respectively.
This rule was discovered by René Descartes in La Géométrie.
💡 If two numbers have the same parity, they are equal modulo 2 , or have the same remainder when divided by 2 .

How to use Descartes' rule of signs?
To apply Descartes's rule of signs to p(x) = a_{0} + a_{1}x + a_{2}x^{2} + ... + a_{n}x^{n}
follow these steps:
 Count the number of sign changes in
a_{0}, a_{1}, a_{2}, a_{3}, ... , a_{n}
(excluding zeros).  Write down the number of sign changes. Starting from
2
, take away successive even numbers from it until you get1
or0
.  Each number you got from your subtraction gives you the possible number for the positive roots of
p(x)
.  To get the number of negative roots of
p(x)
, repeat Steps 13 forp(x)
.
💡 To get the coefficients of p(x) , just swap the sign of every other coefficient of p(x) , starting from a_{1} . That is, the coefficients of p(x) are a_{0}, a_{1}, a_{2}, a_{3}, ... , a_{n} .

We can express step 4 in words as follows: the number of negative roots of p(x)
is equal to the number of sign changes in the coefficients of p(x)
minus some even number (possibly zero).
How to use this Descartes' rule of signs calculator?
Now that we have explained Descartes' rule of signs, it's time to give brief instruction about Omni's Descartes' rule of signs calculator:
 Enter the coefficients of the polynomial. The fields will appear as you go.
 Our Descartes' rule of signs calculator gives you the answer immediately. It appears beneath the coefficients' fields.
 Turn on the option
Show details?
if you want this Descartes' rule of signs calculator to explain the answer. This way, you can use our calculator to generate examples for Descartes' rule of signs.
Descartes' rule of signs examples
We have learned what Descartes' rule of signs is and how to use it, but it's always best to see some examples. Let's go together through two examples. If you need more, you can use Descartes' rule of signs calculator to generate as many examples as you wish.
Example 1
We consider the polynomial
p(x) = 6x^{5} + 5x^{4}  4x^{3} + 3x^{2} + 2x + 1
.

We see that the degree of
p
is5
. 
The constant term is
1
, so zero is not a root of this polynomial. 
The coefficients of
p(x)
are:6,5,4,3,2,1
. We see that the sign changes twice.Hence, the possible number of positive roots of
p
is:2
or0
. 
We compute
p(x)
:p(x) = 6x^{5} + 5x^{4} + 4x^{3} + 3x^{2}  2x + 1
We see that the coefficients of
p(x)
are:6,5,4,3,2,1
. We see that the sign changes three times.Hence, the possible number of negative roots of
p
is:3
or1
. 
The maximum number of positive roots is
2
, the maximum number of negative roots is3
, the multiplicity of zero as root is0
.Hence, the minimum number of nonreal roots is
5  (0 + 2 + 3) = 0
.
Example 2
Next, let's consider the polynomial
p(x) = x^{3}  2x^{2}  x
.

We see that the degree of
p
is3
. 
The constant term is null, so zero is a root. The smallest power with a nonzero coefficient is
1
, so the multiplicity of zero is1
. 
The coefficients of
p(x)
are:1,2,1
. We see that the sign changes once.Hence, the possible number of positive roots of
p
is1
. 
We compute
p(x)
:p(x) = x^{3}  2x^{2} + x
We see that the coefficients of
p(x)
are:1,2,1
. We see that the sign changes once.Hence, the possible number of negative roots of
p
is1
. 
The number of positive roots is
1
, the number of negative roots is1
, the multiplicity of zero as root is1
.Hence, the minimum number of nonreal roots is
3  (1 + 1 + 1) = 0
.
Note, that we have determined the exact number of positive roots and negative roots using just Descartes' rule of signs and without having to make any serious calculations!
Example 3
Next, let's consider the polynomial
p(x) = x^{3} + x^{2} + 1
.

We see that the degree of
p
is3
. 
The constant term is
1
, so zero is not a root of this polynomial. 
The coefficients of
p(x)
are:1,1,1
. We see that the sign does not change!Hence, the possible number of positive roots of
p
is0
. 
We compute
p(x)
:p(x) = x^{3} + x^{2} + 1
We see that the coefficients of
p(x)
are:1,1,1
. We see that the sign changes once.Hence, the possible number of negative roots of
p
is1
. 
The number of positive roots is
0
, the number of negative roots is1
, the multiplicity of zero as root is0
.Hence, the minimum number of nonreal roots is
3  (0 + 0 + 1) = 2
.
FAQ
Does Descartes' rule of signs always work?
Yes, Descartes' rule of signs works always, but remember that it only states the possible number of positive and negative zeros. It happens very rarely that it gives the exact number of either positive or negative zeros (or both).
How do I determine the multiplicity of zero as a polynomial root?
The multiplicity of zero as root is equal to the smallest power with a nonzero coefficient. For instance:
2x + 4x^{2}
has zero as a root with multiplicity1
.2 + 4x^{2}
does not have zero as a root.2x^{7}  4x^{9}
has zero as a root with multiplicity7
.
How do I determine the number of nonreal roots with Descartes' rule of signs?
To determine the number of nonreal roots, you have to:
 Determine the degree
n
of your polynomial  this is the highest power present in the polynomial.  Work out the multiplicity of zero as the root of your polynomial. Denote it by
k
.  Use Descartes' rule of signs to find the maximum possible number of positive and negative roots. Denote them by
p
andq
, respectively.  Compute
n  (k + p + q)
. This is the minimum number of nonreal roots of your polynomial.
Can Descartes' rule of signs have 0?
Yes, Descartes' rule of signs can give you zero as the possible number of positive or negative roots of a polynomial. In particular, if there are no sign changes in the coefficients, then the rule states that there are exactly zero positive roots!