Question

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. See Example 7. ƒ(x)=9x6-7x4+8x2+x+6

Accepted Answer

Identify the degree of the polynomial function $$f(x) = 9x^6$ - 7x^4$ + 8x^2$ + x + 6. $$The degree is the highest power of $$x$$, which is 6 in this case. Use Descartes' Rule of Signs to determine the possible number of positive real zeros. Count the number of sign changes in $$f(x)$$: $$9x^6 ($$positive), $$-7x^4 ($$negative), $$+8x^2 ($$positive), $$+x ($$positive), $$+6 ($$positive). 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 function and simplify the signs of each term, then count the sign changes in $$f(-x). $$The number of negative zeros is either equal to the number of sign changes or less than it by an even number. Recall that the total number of zeros (counting multiplicities) is equal to the degree of the polynomial, which is 6. Use this to find the number of nonreal complex zeros by subtracting the possible numbers of positive and negative real zeros from 6. Summarize the possible combinations of positive, negative, and nonreal complex zeros based on the results from the previous steps, keeping in mind that the number of zeros must add up to 6 and that the number of positive and negative zeros can decrease by even numbers from the counts found by Descartes' Rule of Signs.

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. See Example 7. ƒ(x)=9x⁶-7x⁴+8x²+x+6

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Key Concepts

Fundamental Theorem of Algebra

Descartes' Rule of Signs

Complex Conjugate Root Theorem