Using the descartes rule of signs, describe the real zeroes of the function. 2x^5-x^4-2x^3+4x^2+x-2 i am pretty sure there is only one positive root but i am having trouble finding the number of negative roots.

This is about understanding Descartes rule of signs.

Maximum number of real positive zeros = 3

Maximum number of real negative zeros = 2

While substituting -x for x in the polynomial, simplifying and looking at the amount of changes will give us the maximum number of negative zeros.

f(x) = 2x⁵ - x⁴- 2x³ + 4x² + x - 2

Looking at the polynomial, there are only 3 sign changes from term to term.

Thus,

Maximum number of positive real zeros = 3

Let's find maximum number of negative real zeros by putting -x for x to get;

f(x) = 2(-x)⁵ - (-x)⁴- 2(-x)³ + 4(-x)² + (-x) - 2

f(x) = -2x⁵ - x⁴ + 2x³ + 4x² - x - 2

Looking at this, we can see that there are only 2 sign changes. Thus;

Maximum number of real negative zeros = 2

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as you already know, Descartes rule of signs check the times the sign changes or f(x) and for f(-x)

by the fundamental theorem of algebra, the polynomial has a degree of 5, so it has to have at most 5 zeros/solutions/roots.

f(x) has 3 sign changes, notice, that means, it has either 3, or (3-2), 1 positive zeros.

for f(-x), recall that x³, will be (-x)³ = (-x)(-x)(-x) = -x³, so in short, if the exponent is ODD, the sign changes for that term, if it's EVEN, it doesn't change.

so for f(-x), we have 2 sign changes, meaning, it has either 2 or (2-2), 0 negative roots.

the slack is picked up by the complex roots.

so

3 positive, 2 negative, 0 complex

or

1 positive, 2 negative, 2 complex  *recall complex always come in pairs*

or

3 positive, 0 negative, 2 complex

or

1 positive, 0 negative, 4 complex.

f

step-by-step explanation:

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question is incomplete.