## Tuesday, December 25, 2012

We learned how to find a zero of a function using synthetic division. Now we will learn how to determine the number and nature of zeroes, (also known as roots) that a polynomial function has. To determine the number of positive and negative real roots, we use a technique founded by Rene Descartes, thus named Decartes' Rule of Signs.

To find the number of positive real roots, start with the sign of the coefficient of the term with the highest power of the variable. Count the sign changes as you proceed through the polynomial. The number of sign changes is the number of positive real roots or less than it by a multiple of two. For example, if there are 2 sign changes there are either 2 or (2 – 2) =
0 positive real roots.

Example: Find the number of positive real roots of f(x) = -4x^5 – 11x^4 + 2x^3 + 9x^2 - x + 3.
Starting with -4x^5 , there is a sign change at 2x^3, another at -x and a third at 3. Therefore there are 3 sign changes and either 3 or (3 – 2) = 1 positive real roots.

To find the number of negative real roots of f(x), substitute -x into f(x) to get f(-x). Then proceed in the same manner as you would when finding the number of positive real roots. In the above example,

f(-x) = -4(-x)^5 – 11(-x)^4 + 2(-x)^3 + 9(-x)^2 - (-x) + 3
f(-x) = 4x^5 - 11x^4 – 2x^3 + 9x^2 + x + 3

Starting with 4x^5 , there is a sign change at 11x^4 and another at 9x^2 . Therefore, there are 2 or (2 – 2) = 0 negative real roots.

• Note that the number of roots in a polynomial function equals the highest power. In the above example, there are 5 roots. The sum of the positive and negative roots will not always equal the highest power. In such cases, those polynomial functions have imaginary roots, which we will deal with in a later chapter.

Example: Find the number of positive and negative real roots of f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5.
For positive real roots f(x) = 6x^4 – 9x^3 – 2x^2 + 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -9x^3, another at 5x and a third at -5. Therefore there are 3 or (3 – 2) = 1 positive real roots.

For negative real roots f(-x) = 6(-x)^4 – 9(-x)^3 – 2(-x)^2 + 5(-x) – 5.

f(-x) = 6x^4 + 9x^3 – 2x^2 – 5x – 5. Start with 6x4 and count the sign changes. Notice a change at -2x^2 and no other changes. Therefore there is 1 negative real root.