Use Descartes Rules of Signs to determine the possible numbers of positive, negative and imaginary zeros of the function.

Question

Use Descartes Rules of Signs to determine the possible numbers of positive, negative and imaginary zeros of the function.

𝑓(𝑥)=4𝑥3 −3𝑥2 +2𝑥−1

(HINT: Use a table format as done in the notes when writing your answer)

Answer 1

4x^3 -3x^2 +2x -1 has 3 changes in sign: +, -, +, - which means 3 potential positive zeros

change x to -x

-4x^3 -3x^2 -2x -1 has no changes in sign - - - - which means zero potential negative zeros

there's either 3 positive zeroes

or 1 positive and 2 imaginary. The imaginary zeros come in pairs.

Two possibilities

number of zeros

Positive, negative, imaginary

3 0 0

1 0 2

one positive zero is between x=1/2 and x=2/3, unless there's a 2nd positive zero in that range, the other 2 zeros are imaginary

take the derivative and solve for x, that gives two imaginary turning points. They would be real if there were 3 real zeros.

So, conclusion: there is one positive zero, zero negative zeros, and two imaginary zeroes.