Question

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

Accepted Answer

Identify the degree of the polynomial function $$f(x) = 6x^4$ + 2x^3$ + 9x^2$ + x + 5. $$The degree is 4, so there are 4 zeros in total (counting multiplicities and complex zeros). Use Descartes' Rule of Signs to determine the possible number of positive real zeros by counting the sign changes in $$f(x). $$Write down the signs of the coefficients: $$6, +2, +9, +1, +5$$ and count how many times the sign changes from positive to negative or vice versa. Apply Descartes' Rule of Signs to $$f(-x) to $$find the possible number of negative real zeros. Substitute $$-x$$ into the function and simplify to get $$f(-x) = 6x^4$ - 2x^3$ + 9x^2$ - x + 5. $$Then count the sign changes in this new polynomial. List the possible numbers of positive and negative real zeros based on the counts from steps 2 and 3. Remember that the number of positive or negative zeros can be the number of sign changes or less than that by an even number (e.g., if there are 2 sign changes, possible zeros are 2 or 0). Determine the number of nonreal complex zeros by subtracting the total number of positive and negative real zeros from the degree of the polynomial (4). The remaining zeros must be nonreal complex zeros, which come in conjugate pairs.

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

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