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How many real and imaginary zeros does f(x) = 3x^5 - 12x^4 + 20x³ - 180x² + 120x+81​

User Jahrel
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Answer:

Explanation:

To find the number of real zeros of a polynomial, we can use Descartes' rule of signs. According to this rule, the number of positive real zeros of a polynomial is equal to the number of sign changes in the coefficients of the polynomial, or is less than that by a multiple of 2. Similarly, the number of negative real zeros is equal to the number of sign changes in the coefficients of f(-x), or is less than that by a multiple of 2.

For f(x) = 3x^5 - 12x^4 + 20x³ - 180x² + 120x+81​, there are 2 sign changes in the coefficients, so the number of positive real zeros is either 2 or 0. To find the number of negative real zeros, we can substitute -x for x in f(x) and simplify:

f(-x) = 3(-x)^5 - 12(-x)^4 + 20(-x)³ - 180(-x)² + 120(-x)+81

= -3x^5 - 12x^4 - 20x³ - 180x² - 120x + 81

There are 3 sign changes in the coefficients of f(-x), so the number of negative real zeros is either 3 or 1. Therefore, f(x) has either 2 or 0 positive real zeros, and either 3 or 1 negative real zeros.

To find the number of imaginary zeros, we can use the complex conjugate root theorem, which states that if a polynomial with real coefficients has a+bi as a root (where a and b are real numbers and i is the imaginary unit), then its conjugate a-bi is also a root. Since the coefficients of f(x) are all real, any non-real roots must occur in conjugate pairs.

Since f(x) has degree 5, it has 5 complex roots (counting multiplicities). If all the real zeros are distinct, then there are 5 distinct complex roots, and hence 5 imaginary zeros (again, counting multiplicities). If there are any repeated real zeros, then there are fewer than 5 distinct complex roots, and hence fewer than 5 imaginary zeros.

In summary, the number of real zeros of f(x) is either 2 or 0 positive zeros and either 3 or 1 negative zeros. The number of imaginary zeros is at most 5 (counting multiplicities).

User Dalc
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