Question

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

Accepted Answer

Identify the degree of the polynomial function $$f(x) = 3x^3$ + 6x^2$ + x + 7. $$Since the highest power of $$x is 3$$, the polynomial is cubic and has exactly 3 zeros (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 $$f(x) = 3x^3$ + 6x^2$ + x + 7. $$Since all coefficients are positive, there are no sign changes, so there are 0 positive real zeros. Apply Descartes' Rule of Signs to $$f(-x) to $$find the possible number of negative real zeros. Compute $$f(-x) = 3(-x)^3$ + 6(-x)^2$ + (-x) + 7 = -3x^3$ + 6x^2$ - x + 7. $$Count the sign changes in this expression to determine the possible number of negative real zeros. Based on the number of positive and negative real zeros found, use the Fundamental Theorem of Algebra to determine the number of nonreal complex zeros. Remember that the total number of zeros (real and nonreal) must be 3. Summarize the possible combinations of positive, negative, and nonreal zeros based on the above analysis, considering that the number of positive and negative zeros can decrease by even numbers according to Descartes' Rule of Signs.

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

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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)=3x3+6x2+x+7ƒ(x)=3x^3+6x^2+x+7