*Monday, November 2nd, 2020*

Descartes’ rule of signs is used to get information on the number of real zeros of a polynomial. This is useful in the cases where the graph is not provided for a polynomial.

You need to know that for equations with real coefficients, complex roots occur in pairs. As a result, any equation with an odd degree must have at least 1 real root. This combined with Descartes rule generates some interesting insights, as we will discuss in following examples.

Let us understand the rule with an example:

This rule doesn’t help us actually find the roots, but helps us identify the number and type of roots that we can expect for a given polynomial.

Follow these steps:

**1**. Look at signs of the coefficients of f(x). Value of coefficients do not hold any importance in the process.

Count the number of times the sign of the coefficient is getting changed from positive to negative or negative to positive.

There are 4 sign changes in this case. Hence, there can be a **maximum **of 4 positive roots for the given polynomial.

But, some of the roots may be generated by the Quadratic Formula , and these pairs of roots may be complex and thus not graphable as x-intercepts. So, we have to count down by two’s in order to find the complete list of the possible number of zeroes.

Hence, possible values are 4,2,0 for the number of roots.

**2**. Now, look at signs of the coefficients of f(-x). Value of coefficients do not hold any importance in the process. Replace x by -x in the given polynomial.

Count the number of times the sign changes:

So, there is only 1 negative root. There is no need to subtract 2s from this number of roots because it will generate a negative number.

Let us consider another example:

There are again 4 sign changes in f(x). Thus, number of positive roots can be 4, 2 or 0.

Now, consider f(-x).

There is only 1 sign change. Hence, the number of negative roots is one.

Hence, 2 scenarios are possible:

Negative | Positive | Complex |

1 | 4 | 0 |

1 | 2 | 2 |

1 | 0 | 4 |

Now that we have discussed some examples in detail, here is a problem for your practice followed by the solution:

**Problem 1:** How many non-real roots does the equation x^{4} – 2x^{2} +3 x – 2=0 have?

**Solution**: Let f(x) = x^{4} – 2x^{2} +3 x – 2

f(x) has 3 sign changes. Hence, it has 3 or 1 positive roots.

f(-x) has 1 sign change. Hence, it has exactly 1 negative root.

As the sum of coefficients of f(x)=0, x=1 is a root of the equation.

f(x)= (x-1) (x^{3} + x^{2} – x + 2)

Let g(x) = x^{3} + x^{2} – x + 2

By hit and trial,

g(-2)=0

g(x)= (x+2)(x^{2} – x + 1)

x^{2} – x + 1 has 2 non real roots.

Thus, g(x) has 1 positive, 1 negative and 2 non real roots.

**Problem 2: **Find the number of positive roots for the polynomial : 3x^{2} – x – 2**–**x^{4} + x^{3} .

**Solution**: Firstly, arrange the terms in the descending order of exponents and then proceed with the steps mentioned earlier.

f(x)= -x^{4} + x^{3}+3x^{2} – x – 2

It has 2 sign changes. Hence, 2 or 0 positive roots.

f(-x)= -x^{4} – x^{3}+3x^{2} + x – 2

This again has 2 sign changes. Hence, number of negative roots is 2 or 0.

Total possibilities:

Real positive zeroes | Real negative zeroes | Complex zeros |

2 | 2 | 0 |

2 | 0 | 2 |

0 | 2 | 2 |

0 | 0 | 4 |

**Problem 3:** Find the possible number of real roots of the polynomial x^{3}-x^{2} -14 x +24. Verify.

**Solution**: f(x) has 2 sign changes. Hence, positive roots are 2 or 0.

f(-x) has 1 sign change. Hence, number of negative roots is 1.

Let us find out the roots and verify this:

The polynomial can be rewritten as (x-2) (x-3) (x+4).

So, we can see that there are 2 positive and 1 negative roots.

**Note : repeated roots are counted separately.**

For example: (x-3)^2 = + x^{2} -18 x +9. There are 2 sign changes, hence two positive roots.

Check out the following links for quadratic equations and other related topics:

Quadratic Equations – Concepts and Questions with Solutions for CAT Exam

Quantitative Aptitude – Algebra – Polynomials

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