Descartes’ Rule of Signs With Examples

In this section we shall examine the number and approximate location of real roots of a polynomial equation with real coefficients using Descartes’ rule of signs.

When two consecutive coefficients of a polynomial f(x) have same signs, we say that there is a continuation of signs; but if they have opposite signs, they present a variation of signs.

In the polynomial 7x¹² + 2x¹¹ + 4x¹⁰ + x⁹ – 3x⁶ – 2x⁴ + 3x³ – 5x² + 9x – 6 there are five variations and four continuations:

+  +  +  +  –  –  +  –  +  –

In a complete polynomial f(x), if x is replaced by (-x), then a variation becomes a continuation and vice versa.

Descartes’ Rule of Signs: Statement

If f(x) be a polynomial with real coefficients, the number of positive roots of the equation f(x) = 0 is either equal to the number of changes of signs in the successive coefficients of f(x) or less than that number of changes of sign by an even number.

The number of negative roots of the equation f(x) = 0 is either equal to the number of changes of sign of the coefficients of f(-x) or less than that number of changes of sign by an even number.

Observation

Descartes’ Rule of Signs do not determine actual number of real positive or real negative roots of an algebraic equation, but it indicates only the maximum limit of the number of real positive or negative roots of an equation. This rule can also indicate the existence and minimum number of imaginary roots for equations with real coefficients. If the degree of an equation f(x) = 0 be n and f(x) has p changes of sigh while f(-x) has q changes of sign, the number of positive real roots of f(x) = 0 is either equal to p or less than p by an even number. The number of negative real roots of f(x) = 0 is either equal to q or less than q by an even number.

And the number of imaginary roots of the equation f(x) = 0 is either equal to (n – p – q) or greater than (n – p – q) by an even number.

Read the post how to find the nature of the roots of an equation

Apply Descartes’ rule of signs to determine the nature of the roots of the following equations:

\[(i){{x}^{4}}+{{x}^{2}}+x-1=0\]

\[(ii){{x}^{4}}+16{{x}^{2}}+7x-10=0\]

\[(iii){{x}^{n}}-1=0\]

Solution:

(i) Let the given equation be

\[f(x)={{x}^{4}}+{{x}^{2}}+x-1=0\]

\[i.e.,f(x)={{x}^{4}}+{{x}^{2}}+x-1\]

And

\[f(-x)={{x}^{4}}+{{x}^{2}}-x-1\]

The pattern in sign of the coefficients of f(x) and f(-x) are respectively

+  +  +  –

and       +  +  –  –

Since the number of changes of sign of f(x) is only one, the number of real positive root is ≤ 1.

Again, since f(0) = -1 and f(∞) → ∞, there must be at least one real positive root of the equation. So the number of real positive root is ≥ 1.

Combining these two, we find that the number of positive real root is exactly 1.

Also, there is only one change of sign in f(-x). By similar arguments as above, number of negative real root is exactly 1.

Now, the degree of the equation is 4; so it has four roots of which one is positive real, one is negative real and the remaining two are imaginary.

(ii) Denoting the given equation by f(x) = 0, we have

\[f(x)={{x}^{4}}+16{{x}^{2}}+7x-10\]

The pattern of sign of the coefficients of f(x) is

+  +  +  –

There is only one change of sign, so, the number of positive real roots is ≤ 1.

Again since f(0) = -10 < 0 and f(∞) > 0, there is at least one positive root, i.e., number of positive real root is ≥ 1.

From these, we conclude that the number of positive real roots is exactly 1.

Since

\[f(-x)={{x}^{4}}+16{{x}^{2}}-7x-10\]

The pattern of sign of coefficients in f(-x) is

+  +  –  –

As there is only one change in sign, the number of negative real root is ≤ 1. By above arguments similar to above, the number of negative real root is exactly one.

The given equation is of fourth degree. So it has four roots of which one is positive real, one is negative real and the remaining two are imaginary.

(iii) Let the given equation be f(x) = 0, where

\[f(x)={{x}^{n}}-1\]

The pattern of sign of the coefficients in f(x) is

+  –

There is one change in sign. By the rule number of positive real root is ≤ 1.

Now since f(0) = -1 < 0 and f(∞) > 0, the number of positive roots is ≥ 1.

From these two we conclude that the number of positive real root is exactly one.

Now since

\[f(-x)={{\left( -x \right)}^{n}}-1\]

If n is even,

\[f(-x)={{x}^{n}}-1\]

As there is only one change of sign of the coefficients in f(-x), the number of negative real root is ≤ 1. Again, since f(0) = -1 < 0 and f(∞) > 0, the number of negative roots is ≥ 1. From these two, we conclude that the number of negative real root is exactly one. The degree of the equation is n, so it has altogether n roots of which one is positive real, one is negative real and the remaining (n – 2) are imaginary.

Again if n be odd,

\[f(-x)=-{{x}^{n}}-1\]

There is no change of sign of the coefficients in f(-x). So the equation has no negative real root. In this case, the given equation has one positive real root and (n – 1) imaginary roots.

Prove that the equation x⁷ – 2x⁴ + 3x³ – 1 = 0 has at least four imaginary roots.

Solution

Let the equation be f(x) = 0, where

\[f(x)={{\text{x}}^{\text{7}}}-\text{ 2}{{\text{x}}^{\text{4}}}+\text{ 3}{{\text{x}}^{\text{3}}}\text{ 1}\]

The pattern of signs of the coefficients in f(x) is

+  –  +  –

There are three changes in the signs.

According to the Rule of signs, the number of positive real roots is ≤ 3.

\[f(-x)=-{{\text{x}}^{\text{7}}}-\text{ 2}{{\text{x}}^{\text{4}}}-\text{3}{{\text{x}}^{\text{3}}}\text{ 1}\]

There is no change in sign. So, there is no negative real root of the equation.

The degree of the equation is 7, so it has 7 roots.

We have two possibilities regarding the number of roots of the equation.

So the equation has at least four imaginary roots.;