Question

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. ƒ(x)=2x5−7x3+6x+8ƒ(x)=$$2x^5$-7x^3+6x+8$$

Accepted Answer

Identify the degree of the polynomial function $$f(x) = 2x^5$ - 7x^3$ + 6x + 8. $$The degree is 5, which means there are 5 zeros in total (counting multiplicities and including complex zeros). Use Descartes' Rule of Signs to determine the possible number of positive real zeros. Count the number of sign changes in $$f(x) = 2x^5$ - 7x^3$ + 6x + 8. $$Each sign change corresponds to a possible positive zero, and the actual number of positive zeros is either equal to the number of sign changes or less than it by an even number. Apply Descartes' Rule of Signs to $$f(-x) to $$find the possible number of negative real zeros. Substitute $$-x$$ into the function to get $$f(-x) = 2(-x)^5$ - 7(-x)^3$ + 6(-x) + 8$$, simplify it, and then count the sign changes. The number of negative zeros is either equal to the number of sign changes or less than it by an even number. Determine the possible number of nonreal complex zeros by subtracting the possible numbers of positive and negative real zeros from the total degree (5). Remember that complex zeros come in conjugate pairs, so the number of nonreal complex zeros must be even. Summarize the possible combinations of positive, negative, and nonreal complex zeros based on the results from the previous steps, ensuring the total number of zeros adds up to 5.

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. $ƒ (x) = 2 x^{5} - 7 x^{3} + 6 x + 8$

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Fundamental Theorem of Algebra

Descartes' Rule of Signs

Complex Conjugate Root Theorem

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. ƒ(x)=2x5−7x3+6x+8ƒ(x)=2x^5-7x^3+6x+8