Guided Example #1 - Polynomial Equations

Guided Example

Use Descartes' rule of signs to find the number of possible positive, negative, and nonreal roots for the following equation. 2x15+3x8-4x7+3x4+6x3+x2+1

a. 0 positive; 0 negative; 15 non-real
b. 1 positive; 0 negative; 14 non-real
c. 0 positive; 1, 3, or 5 negative; 10, 12, 0r 14 non-real
d. 0 or 2 positive; 1 or 3 negative; 10, 12, or 14 non-real

 

The correct answer is D. To see why, view each of the steps below.

 

Step 1

The number of variations of sign in P(x)=2x15+3x8-4x7+3x4+6x3+x2+1

Step 2

is 2. By Descarte's rule, the number of positive real roots is either 2 or 0. Since P(-x)=2(-x)15+3(-x)8-4(-x)7+3(-x)4+6(-x)3+(-x)2+1
=-2x15+3x8+4x7+3x4-6x3+x2+1
there are three variations of sign in P(-x). So there are either 1 or 3 negative real roots.

Step 3

The equation must have 15 roots, because it is a fifteenth- degree polynomial equation. Since there must be 15 roots and there are 0, or two positive and 1 or 3 negative, the other 10,12, or 14 roots must be non-real.

All Steps

The number of variations of sign in P(x)=2x15+3x8-4x7+3x4+6x3+x2+1

is 2. By Descarte's rule, the number of positive real roots is either 2 or 0. Since P(-x)=2(-x)15+3(-x)8-4(-x)7+3(-x)4+6(-x)3+(-x)2+1
=-2x15+3x8+4x7+3x4-6x3+x2+1
there are three variations of sign in P(-x). So there are either 1 or 3 negative real roots.

The equation must have 15 roots, because it is a fifteenth- degree polynomial equation. Since there must be 15 roots and there are 0, or two positive and 1 or 3 negative, the other 10,12, or 14 roots must be non-real.