Question

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. ƒ(x)=7x5+6x4+2x3+9x2+x+5ƒ(x)=$$7x^5+6x^4$+2x^3$+9x^2+x+5$$

Accepted Answer

Identify the degree of the polynomial function $$f(x) = 7x^5$ + 6x^4$ + 2x^3$ + 9x^2$ + x + 5. $$The degree is the highest power of $$x$$, which is 5 in this case. Use Descartes' Rule of Signs to determine the possible number of positive real zeros. Count the number of sign changes in the coefficients of $$f(x)$$: $$7, +6, +2, +9, +1, +5. $$Since all coefficients are positive, there are 0 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. Substitute $$-x$$ into $$f(x)$$ and simplify the signs of each term: $$7(-x)^5$$ + $$6(-x)^4$$ + $$2(-x)^3$$ + $$9(-x)^2 + (-x) + 5$$ = $$-7x^5$$ + $$6x^4$$ - $$2x^3$$ + $$9x^2 - x + 5$. Count the sign changes in the sequence -7$$, +6, -2, +9, -1, +5$$. Determine the possible number of negative real zeros by counting the sign changes in f(-x$$)$$. Each sign change corresponds to a possible negative zero, and the actual number of negative zeros is either equal to the number of sign changes or less than it by an even number. Use the Fundamental Theorem of Algebra to find the total number of zeros (counting multiplicities), which is equal to the degree (5). Subtract the possible number of positive and negative real zeros from 5 to find the possible number of nonreal complex zeros.

Determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of each function. $ƒ (x) = 7 x^{5} + 6 x^{4} + 2 x^{3} + 9 x^{2} + x + 5$

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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)=7x5+6x4+2x3+9x2+x+5ƒ(x)=7x^5+6x^4+2x^3+9x^2+x+5