Question

Solve each problem. Use Descartes' rule of signs to determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of ƒ(x)=x3+3x2−4x−2ƒ(x)=$$x^3$+3x^2-4x-2.$$

Accepted Answer

Write down the polynomial function: $$f(x) = x^3$ + 3x^2$ - 4x - 2. $$Apply Descartes' Rule of Signs to find the possible number of positive real zeros by counting the sign changes in $$f(x). $$Look at the coefficients of $$f(x) in $$order: $$+1, +3, -4, -2. $$Count how many times the sign changes from one term to the next. To find the possible number of negative real zeros, evaluate $$f(-x) by $$substituting $$-x$$ into the polynomial: $$f(-x) = (-x)^3$ + 3(-x)^2$ - 4(-x) - 2. $$Simplify this expression and then count the sign changes in the coefficients of $$f(-x). $$Use the counts of sign changes from steps 2 and 3 to list the possible numbers of positive and negative real zeros. Remember, the number of positive or negative real zeros is either equal to the number of sign changes or less than that by an even number (e.g., if there are 3 sign changes, possible zeros are 3 or 1). Determine the number of nonreal complex zeros by using the fact that the total number of zeros (counting multiplicities) equals the degree of the polynomial (which is 3). Subtract the possible positive and negative zeros from 3 to find the possible number of nonreal complex zeros.

Solve each problem. Use Descartes' rule of signs to determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of $ƒ (x) = x^{3} + 3 x^{2} - 4 x - 2$ .

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Key Concepts

Descartes' Rule of Signs

Polynomial Zeros and Their Types

Evaluating f(-x) to Find Negative Zeros

Solve each problem. Use Descartes' rule of signs to determine the different possibilities for the numbers of positive, negative, and nonreal complex zeros of ƒ(x)=x3+3x2−4x−2ƒ(x)=x^3+3x^2-4x-2.