Descartes' Rule of Signs will not tell what is actual value of the roots, but the Rule will tell how many roots are expected. If \(f(x)\) is polynomial, then maximum number of positive roots will be equal to total number of sign changes in \(f(x)\), similarly maximum number of negative roots will be equal to total number of sign changes in \(f(- x)\). 

Example 18: Find the exact number of negative real roots of the equation

\({x^5}-{\rm{ }}{x^4} + {\rm{ }}3{x^3} + {\rm{ }}9{x^2}-{\rm{ }}x{\rm{ }} + {\rm{ }}5{\rm{ }} = {\rm{ }}0\).

Solution: \(f(x) = \underbrace {{x^5} - }_{}\;\underbrace {{x^4} + }_{}3{x^3} + \underbrace {9{x^2} - }_{}\;\underbrace {x\; + }_{}\;5\)

The number of sign changes is 4, hence there are at the most 4 positive real roots.

Now checking the number of sign changes in \(f(-x\)),

=  \(-{x^5}\;-\;{x^4}\;-\;3{x^3} + {\rm{ }}9{x^2} + x + {\rm{ }}5\)

lets look at the signs:

\(f( - x) =  - {x^5} - {x^4} - \underbrace {3{x^3} + }_19{x^2} + x + 5\)

There is only one sign change in this negative case, so there is at the most one negative roots. 

Also the value of \(f(x)\) when x is a large negative number is negative and \(f(0)\) is 5, hence there is  at least one negative root of this equation. 

By the Descartes’ rule there is at the most one negative root, hence there is exactly one negative root of the equation.

There are 4, 2 or 0 positive roots, and exactly 1 negative root.

Example 19: Using Descartes' Rule of Signs, determine the number of real solutions to \(4{x^7} + {\rm{ }}3{x^6} + {x^5} + {\rm{ }}2{x^4}-\;{x^3} + {\rm{ }}9{x^2} + x + {\rm{ }}1{\rm{ }} = {\rm{ }}0.\)

Solution: Lets look first at the polynomial \(f(x)\) (this is the "positive" case):

\(f\left( x \right){\rm{ }} = {\rm{ }} + 4{x^7} + {\rm{ }}3{x^6} + {x^5} + {\rm{ }}2{x^4}\;-\;{x^3} + {\rm{ }}9{x^2} + x + {\rm{ }}1\)

There are two sign changes, so there are two or, counting down in pairs, zero positive solutions. Now lets look at the polynomial \(f (-x)\) (this is the "negative" case):

\(f\left( {-x} \right){\rm{ }} = {\rm{ }}4{\left( {-x} \right)^7} + {\rm{ }}3{\left( {-x} \right)^6} + {\rm{ }}{\left( {-x} \right)^5} + {\rm{ }}2{\left( {-x} \right)^4}-{\rm{ }}{\left( {-x} \right)^3}\;\;\;\; + {\rm{ }}9{\left( {-x} \right)^2} + {\rm{ }}\left( {-x} \right){\rm{ }} + {\rm{ }}1\;\;\;\;\;\;\;\;\;\; = \;-{\rm{ }}4{x^7} + {\rm{ }}3{x^6}-{x^5} + {\rm{ }}2{x^4} + {x^3} + {\rm{ }}9{x^2}-x + {\rm{ }}1\)

There are five sign changes, so there are five or, counting down in pairs, three or one negative solutions. There are two or zero positive solutions, and five, three, or one negative solutions.

In the above example, the maximum number of positive solutions (two) and the maximum number of negative solutions (five) added up to the leading degree (seven). It will always be true that the sum of the possible numbers of positive and negative solutions will be equal to the degree of the polynomial, or two less, or four less, or.... For instance, if we had come up with a maximum answer of "two" for the possible positive solutions in the above example but had come up with only, say, "four" for the possible negative solutions, then it would be known that we had made a mistake somewhere, because 2 + 4 does not equal 7, or 5, or 3, or 1.

Example 20: Use Descartes' Rule of Signs to find the number of real roots of  \(f\left( x \right){\rm{ }} = {x^5} + {x^4} + {\rm{ }}4{x^3} + {\rm{ }}3{x^2} + x + {\rm{ }}1.\)

Solution: Let us look first at \(f(x)\):

\(f\left( x \right){\rm{ }} =  + {x^5} + {x^4} + {\rm{ }}4{x^3} + {\rm{ }}3{x^2} + x + {\rm{ }}1\)

There are no sign changes, so there are no positive roots. Now we look at \(f(-x)\):

\(f\left( {-x} \right){\rm{ }} = {\left( {-x} \right)^5} + {\rm{ }}{\left( {-x} \right)^4} + {\rm{ }}4{\left( {-x} \right)^3} + {\rm{ }}3{\left( {-x} \right)^2} + {\rm{ }}\left( {-x} \right){\rm{ }} + {\rm{ }}1 = \;-{x^5} + {x^4}\;-\;4{x^3} + {\rm{ }}3{x^2}\;-\;x + {\rm{ }}1\)

There are five sign changes, so there are as many as five negative roots.

There are no positive roots, and there are five, three, or one negative roots.

Example 21: Use Descartes' Rule of Signs to determine the possible number of solutions to the equation \(2{x^4}\;-\;{x^3} + {\rm{ }}4{x^2}\;-\;5x + {\rm{ }}3{\rm{ }} = {\rm{ }}0.\)

Solution: Let us look first at \(f(x)\):

\(f\left( x \right){\rm{ }} =  + 2{x^4}\;-\;{x^3} + {\rm{ }}4{x^2}\;-\;5x + {\rm{ }}3\)

There are four sign changes, so there are 4, 2, or 0 positive roots. Now let us look at  \(f (-x)\):

\(f\left( {-x} \right){\rm{ }} = 2{\left( {-x} \right)^4}\;-\;{\left( {-x} \right)^3} + {\rm{ }}4{\left( {-x} \right)^2}\;-\;5\left( {-x} \right){\rm{ }} + {\rm{ }}3\)

\( = {\rm{ }} + 2{x^4} + {x^3} + {\rm{ }}4{x^2} + {\rm{ }}5x + {\rm{ }}3\)

There are no sign changes, so there are no negative roots. There are four, two, or zero positive roots, and zero negative roots. 

Example 22: The harmonic mean of the roots of the equation \((5 + \sqrt 2 ){x^2} - (4 + \sqrt 5 )x + 8 + 2\sqrt 5  = 0\) is:

Solution: Harmonic mean of two numbers \(\alpha \) and \(\beta \) is \(\frac{{2\alpha \beta }}{{\alpha  + \beta }}\). From the given equation \(\alpha  + \beta \) = \(\frac{{4 + \sqrt 5 }}{{5 + \sqrt 2 }}\),

\(\alpha \beta \) = \(\frac{{8 + 2\sqrt 5 }}{{5 + \sqrt 2 }}\).

Harmonic mean = \(\frac{{2\left( {8 + 2\sqrt 5 } \right)}}{{4 + \sqrt 5 }} = 4\)

Example 23: If the roots of the quadratic equation 

\({x^2}-{\rm{ }}bx{\rm{ }} + {\rm{ }}c{\rm{ }} = {\rm{ }}0\) be two consecutive roots, then find the value of \({b^2}-{\rm{ }}4c\)

Solution: Suppose the roots are \(\alpha \) and \(\alpha  + 1\), then

\(\alpha  + \left( {\alpha  + 1} \right) = b\) and \(\alpha (\alpha  + 1) = c\)

Now \({b^2}-{\rm{ }}4c = {\rm{ }}{(2a + {\rm{ }}1)^2}-{\rm{ }}4a(a + {\rm{ }}1){\rm{ }} = {\rm{ }}1\).