Question

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. ƒ(x)=3x4+2x3−8x2−10x−1ƒ(x)=$$3x^4+2x^3$-8x^2-10x-1$$

Accepted Answer

Identify the degree of the polynomial function $$f(x) = 3x^4$ + 2x^3$ - 8x^2$ - 10x - 1. $$Since the highest power of $$x is 4$$, the polynomial is of degree 4, so it has 4 zeros in total (counting multiplicities and complex zeros). Use Descartes' Rule of Signs to determine the possible number of positive real zeros. Count the number of sign changes in $$f(x) = 3x^4$ + 2x^3$ - 8x^2$ - 10x - 1. $$Each sign change indicates 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 polynomial to get $$f(-x) = 3(-x)^4$ + 2(-x)^3$ - 8(-x)^2$ - 10(-x) - 1$$, 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 (4). Since complex zeros come in conjugate pairs, the number of nonreal zeros must be even. List all possible combinations of positive, negative, and nonreal zeros based on the above calculations, ensuring the total adds up to 4 and the number of nonreal zeros is even.

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. $ƒ (x) = 3 x^{4} + 2 x^{3} - 8 x^{2} - 10 x - 1$

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Key Concepts

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)=3x4+2x3−8x2−10x−1ƒ(x)=3x^4+2x^3-8x^2-10x-1