# Use Descartes’ Rule of Signs

There is a straightforward way to determine the possible numbers of positive and negative real zeros for any polynomial function. If the polynomial is written in descending order,** Descartes’ Rule of Signs** tells us of a relationship between the number of sign changes in [latex]f\left(x\right)[/latex] and the number of positive real zeros. For example, the polynomial function below has one sign change.

There is a similar relationship between the number of sign changes in [latex]f\left(-x\right)[/latex] and the number of negative real zeros.

In this case, [latex]f\left(\mathrm{-x}\right)[/latex] has 3 sign changes. This tells us that [latex]f\left(x\right)[/latex] could have 3 or 1 negative real zeros.### A General Note: Descartes’ Rule of Signs

According to **Descartes’ Rule of Signs**, if we let [latex]f\left(x\right)={a}_{n}{x}^{n}+{a}_{n - 1}{x}^{n - 1}+...+{a}_{1}x+{a}_{0}[/latex] be a polynomial function with real coefficients:

- The number of positive real zeros is either equal to the number of sign changes of [latex]f\left(x\right)[/latex] or is less than the number of sign changes by an even integer.
- The number of negative real zeros is either equal to the number of sign changes of [latex]f\left(-x\right)[/latex] or is less than the number of sign changes by an even integer.

### Example 7: Using Descartes’ Rule of Signs

Use Descartes’ Rule of Signs to determine the possible numbers of positive and negative real zeros for [latex]f\left(x\right)=-{x}^{4}-3{x}^{3}+6{x}^{2}-4x - 12[/latex].

### Solution

Begin by determining the number of sign changes.

There are two sign changes, so there are either 2 or 0 positive real roots. Next, we examine [latex]f\left(-x\right)[/latex] to determine the number of negative real roots.[latex]\begin{cases}f\left(-x\right)=-{\left(-x\right)}^{4}-3{\left(-x\right)}^{3}+6{\left(-x\right)}^{2}-4\left(-x\right)-12\hfill \\ f\left(-x\right)=-{x}^{4}+3{x}^{3}+6{x}^{2}+4x - 12\hfill \end{cases}[/latex]

Again, there are two sign changes, so there are either 2 or 0 negative real roots.There are four possibilities, as we can see below.

Positive Real Zeros | Negative Real Zeros | Complex Zeros | Total Zeros |
---|---|---|---|

2 | 2 | 0 | 4 |

2 | 0 | 2 | 4 |

0 | 2 | 2 | 4 |

0 | 0 | 4 | 4 |

### Try It 6

Use Descartes’ Rule of Signs to determine the maximum possible numbers of positive and negative real zeros for [latex]f\left(x\right)=2{x}^{4}-10{x}^{3}+11{x}^{2}-15x+12[/latex]. Use a graph to verify the numbers of positive and negative real zeros for the function.

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- Precalculus.
**Provided by:**OpenStax**Authored by:**Jay Abramson, et al..**Located at:**https://openstax.org/books/precalculus/pages/1-introduction-to-functions.**License:**CC BY: Attribution.**License terms:**Download For Free at : http://cnx.org/contents/[email protected]..

## Analysis of the Solution

We can confirm the numbers of positive and negative real roots by examining a graph of the function. We can see from the graph in Figure 3 that the function has 0 positive real roots and 2 negative real roots.

Figure 3