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Consider the given function.
P(x)=-0.006x^4+0.15x^3-0.05x^2-1.8x
We see the degree of the polynomial is 4. According to the Fundamental Theorem of Algebra, the function has four zeros. This is including any repeated zeros. We should also consider the zero at the origin. Notice that we can factor out x.
Let's now determine the type of zeros. To do so, we will examine the number of sign changes for P(x).
There are two sign changes for the coefficients of P(x). According to Descartes' Rule of Signs, the function has zero or two positive real zeros. Next, let's write and simplify P(- x). P(- x)=-0.006x^4-0.15x^3-0.05x^2+1.8x As we did for P(x), we will examine the number of sign changes for P(- x).
There is only one sign change for the coefficients of P(- x). Once again, according to Descartes' Rule of Signs, the function has one negative real zero. Therefore, there are two options for the types of zeros of P(x). Remember, we know that there are four zeros in total.
| Option 1 | Option 2 | ||
|---|---|---|---|
| Real Zeros | 0 positive | Real Zeros | 2 positive |
| 1 negative | 1 negative | ||
| Zeros at the Origin | 1 | Zeros at the Origin | 1 |
| Imaginary Zeros | 2 | Imaginary Zeros | |
| Number of Zeros | 0+ 1+ 1+2=4 | Number of Zeros | 2+ 1+ 1+ =4 |