Question 3.3.7: Applying Descartes’ Rule of Signs Determine the different po...
Applying Descartes’ Rule of Signs
Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of
ƒ(x) = x^{4} – 6x³ + 8x² + 2x – 1.
We first consider the possible number of positive zeros by observing that ƒ(x) has three variations in signs.
ƒ(x)=\underbrace{+x^{4}}_{1}\underbrace{-6x^{3}}_{2}+8x^{2} \underbrace{+2x^{2}}_{3} -1
Thus, ƒ(x) has either three or one (because 3 – 2 = 1) positive real zeros.
For negative zeros, consider the variations in signs for ƒ(x) .
ƒ(-x) = (-x)^{4} – 6(-x)³ + 8(-x)²+ 2(-x) – 1
ƒ(-x) = x^{4} + 6x³\underbrace{+8x²}_{1} – 2x – 1
There is only one variation in sign, so ƒ(x) has exactly one negative real zero.
Because ƒ(x) is a fourth-degree polynomial function, it must have four complex zeros, some of which may be repeated. Descartes’ rule of signs has indicated that exactly one of these zeros is a negative real number.
• One possible combination of the zeros is one negative real zero, three positive real zeros, and no nonreal complex zeros.
• Another possible combination of the zeros is one negative real zero, one positive real zero, and two nonreal complex zeros.
By the conjugate zeros theorem, any possible nonreal complex zeros must occur in conjugate pairs because ƒ(x) has real coefficients. The table below summarizes these possibilities.
| Possible Number of Zeros | ||
| Positive | Negative | Nonreal Complex |
| 3 | 1 | 0 |
| 1 | 1 | 2 |
The graph of ƒ(x) in Figure 19 verifies the correct combination of three positive real zeros with one negative real zero, as seen in the first row of the table.*
* The authors would like to thank Mary Hill of College of DuPage for her input into Example 7.
