Question

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. ƒ(x)=−8x4+3x3−6x2+5x−7ƒ(x)=$$-8x^4+3x^3$-6x^2+5x-7$$

Accepted Answer

Identify the degree of the polynomial function. Here, the function is $$f(x) = -8x^4$ + 3x^3$ - 6x^2$ + 5x - 7$$, which is a 4th degree polynomial. Use the Fundamental Theorem of Algebra, which states that a polynomial of degree $$n$$ has exactly $$n$$ roots (zeros) in the complex number system, counting multiplicities. So, there are 4 zeros in total. Apply Descartes' Rule of Signs to determine the possible number of positive real zeros. Count the number of sign changes in $$f(x)$$: from $$-8x^4 to$$ $$+3x^3 ($$change), $$+3x^3 to$$ $$-6x^2 ($$change), $$-6x^2 to$$ $$+5x ($$change), $$+5x to$$ $$-7 ($$change). There are 4 sign changes, so the number of positive real zeros is 4, 2, or 0 (decreasing by even numbers). Apply Descartes' Rule of Signs to $$f(-x) to $$determine the possible number of negative real zeros. Calculate $$f(-x) = -8(-x)^4$ + 3(-x)^3$ - 6(-x)^2$ + 5(-x) - 7$$, simplify it, then count the sign changes in $$f(-x). $$The number of negative real zeros is equal to the number of sign changes or less by an even number. Determine the number of nonreal complex zeros by subtracting the total number of positive and negative real zeros from 4 (the degree). The remaining zeros, if any, are nonreal complex zeros.

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. $ƒ (x) = - 8 x^{4} + 3 x^{3} - 6 x^{2} + 5 x - 7$

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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)=−8x4+3x3−6x2+5x−7ƒ(x)=-8x^4+3x^3-6x^2+5x-7