Descartes Rule of Signs question???

Question

1. Use Descartes' Rule of Signs to determine the most possible positive, negative, and imaginary roots. x^3 + 2x ^2 - 5x + 5 = 0

Then it gives a chart with p, n, and i, with spaces underneath to fill it out. THANK YOU SO MUCH FOR YOUR HELP!

Answer 1

There are 2 sign changes in the given polynomial. Descartes' Rule of Signs says this means there are at most 2 positive roots.

Now replace x with (-x) in the polynomial:

(-x)³ + 2(-x)² - 5(-x) + 5

= -x³ + 2x² + 5x + 5

There is only one sign change, so the Rule says there is at most one negative root.

For polynomials with real coefficients, imaginary (complex is the preferred term) roots come in conjugate pairs, so there are either 0 or 2 imaginary roots.

So the possibilities are

1 positive root, 0 negative roots, 2 imaginary roots

0 positive roots, 1 negative root, 2 imaginary roots

2 positive roots, 1 negative root, 0 imaginary roots

That's all the Rule can tell us.

As it happens, this polynomial has 1 negative root and 2 imaginary roots. The very similar polynomial equation

x³ + 2x² - 5x + 1 = 0

to which the Rule assigns exactly the same possibilities, has one negative root and 2 positive roots.

Answer 2

Given the equation:

x^3 + 2x^2 - 5x + 5 = 0

Look at the sign + and - in the polynomial on left side

Note where the sign varies from + to - or vice versa.

Count the number of variations of sign.

In this case, x^3 + 2x^2 - 5x + 5 = 0, there are 2 such variations,

the first variation occurs when going from 2nd term (+ sign)  to 3rd term (- sign)

the second variation occurs when going from 3rd term (- sign) to 4th term (+ sign)

According to Descartes'Rule of sign, there are the most possible 2 positive solutions.

To know the number of negative solutions, replace x with -y

(-y)^3 + 2(-y)^2 - 5(-y) + 5 = 0

-y^3 + 2y^2 + 5y + 5 = 0

We see only one variation of sign, when going from the 1st term (- sign)to the 2nd term (+sign). After that there is no more variation, the sign keeps being positive. Thus there is one positive solution for -x, or one negative solution for x

In all the given equation has 2 positive and 1 negative roots. The total (real and imaginary) roots of a polynomial equation of 3rd degree is three. Therefore there is no imaginary root for this equation.

positive root: 2

negative root:1

imaginary root: 0