Descartes' Rule of Signs

To justify this theorem, the graph of a polynomial function will be examined for non-negative values of $x.$ Consider an arbitrary polynomial function $p(x)$ of degree $n.$ Here, $a_n$ is the leading coefficient and $a_0$ the constant term. Consider the case where both $a_n$ and $a_0$ are positive. Because $a_0>0,$ the graph intersects the $y\text{-}$axis above the $x\text{-}$axis. Also, since $a_n>0,$ the end behavior of the graph for positive values of $x$ is up.

In this case, the number of sign changes must be even. This is because every time a positive sign changes to a negative sign, it must return to positive again, making the number of sign changes a multiple of $2.$ The number of times the graph crosses the $x\text{-}$axis must also be even, since the graph begins and ends above the $x\text{-axis}.$

Without sign changes, a polynomial function only increases or decreases as $x$ increases. Therefore, each change in direction of the graph is related to a sign change in the polynomial. To explore this idea, consider the following polynomial function. It has already been said that only non-negative values of $x$ are considered, so only non-negative values of $x$ will be considered for the graph of $f(x).$ The $y\text{-}$intercept of the graph is $24.$ Then, the graph goes down from $24$ for a while before going up. This happens because in the first interval, the term $\text{-}8x^2$ directs the graph before the leading term $x^3$ takes direction. For the values of this first interval, $x^3$ is less than $8x^2$ and, therefore, $x$ is less than $8.$ This value can be substituted into the function $f(x)$ to find its corresponding output. When $x=8,$ the value of $f$ returns to the original value of $24$ before continuously going up as $x$ increases. In the interval that goes from $0$ to $8,$ the term $\text{-}8x^2$ drives the graph down. Then the value of $x^3$ catches up and directs the graph upwards. Now consider a variation of the function. The same will be done to find the values where $x^2$ is greater than $8x^3.$ This time the interval is significantly shorter than in the previous case. This can also be seen in the graph.

Here, the effect of the term $\text{-}{x}^2$ is negligible compared to the effect of the term $8x^3.$ This indicates that a change in direction of the graph of a function is caused by a sign change, but a sign change does not always result on a change in direction of the graph. Consider the graph of the initial polynomial function again.

The function starts above the $x\text{-}$axis and has $4$ significant direction changes. Therefore, there are at least $4$ sign changes in the equation. Also, each time the graph moves in a different direction, it crosses the $x\text{-}$axis, which suggests that the number of zeros can be the same as the number of sign changes. Consider now a different graph.

Here, the number of zeros is less than the number of times the graph changes direction. Since sign changes indicate changes in direction, it is possible to have fewer zeros than sign changes. Also, since there cannot be more zeros without changing direction, there cannot be more zeros than sign changes. Two conclusions can be made from this reasoning.

1. The number of sign changes and the number of zeros are both even.
2. The number of zeros has to be less than or equal to the number of sign changes.

This is equivalent to what is written in the rule. The other cases can be examined on a similar way. Also, negative zeros are found by considering $p(\text{-}x)$ because this is a reflection in the $y\text{-}$axis. Note that this is an informal justification and cannot be taken as a formal proof.