Question

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. See Example 7. ƒ(x)=x3+2x2+x-10

Accepted Answer

Identify the degree of the polynomial function $$f(x) = x^3$ + 2x^2$ + x - 10. $$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 $$f(x) = x^3$ + 2x^2$ + x - 10. $$The signs of the coefficients are $$+, +, +, -$$, so there is 1 sign change (from $$+ to$$ $$-). $$This means there is exactly 1 positive real zero or fewer by an even number (which in this case can only be 1). Apply Descartes' Rule of Signs to $$f(-x) to $$find the possible number of negative real zeros. Compute $$f(-x) = (-x)^3$ + 2(-x)^2$ + (-x) - 10 = -x^3$ + 2x^2$ - x - 10. $$The signs of the coefficients are $$-, +, -, -. $$Count the sign changes: from $$- to$$ $$+ (1$$), $$+ to$$ $$- (2$$), and $$- to$$ $$- (no $$change). So there are 2 sign changes, meaning there could be 2 or 0 negative real zeros (subtracting even numbers). Summarize the possible combinations of positive and negative real zeros based on the above: positive zeros could be 1, negative zeros could be 2 or 0. Since the total number of zeros is 3, the remaining zeros (if any) must be nonreal complex zeros. Conclude the possible distributions of zeros: (1 positive, 2 negative, 0 nonreal), or (1 positive, 0 negative, 2 nonreal). These are the different possibilities for the numbers of positive, negative, and nonreal complex zeros of the function.

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. See Example 7. ƒ(x)=x³+2x²+x-10

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Key Concepts

Fundamental Theorem of Algebra

Descartes' Rule of Signs

Complex Conjugate Root Theorem