Descartes' Rule of Signs

[nycmathguy]

Section 2.5
Question 80

Can you show me how to do 80?

 

[MathLover1]

f(x)=4x^2+8x+3

The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number.

So, the coefficients are 4,-8,3.

As can be seen, there are 2 changes.

This means that there are 2 or 0 positive real roots.

To find the number of negative real roots, substitute x with -x in the given polynomial:
f(x)=4x^2-8x+3 becomes f(x)=4x^2+8x+3.

The coefficients are 4,8,3.

As can be seen, there are 0 changes.

This means that there are 0 negative real roots.

 

[nycmathguy]

f(x)=4x^2+8x+3

The rule states that if the terms of a single-variable polynomial with real coefficients are ordered by descending variable exponent, then the number of positive roots of the polynomial is either equal to the number of sign differences between consecutive nonzero coefficients, or is less than it by an even number.

So, the coefficients are 4,-8,3.

As can be seen, there are 2 changes.

This means that there are 2 or 0 positive real roots.

To find the number of negative real roots, substitute x with -x in the given polynomial:
f(x)=4x^2-8x+3 becomes f(x)=4x^2+8x+3.

The coefficients are 4,8,3.

As can be seen, there are 0 changes.

This means that there are 0 negative real roots.

Thank you. I will use this reply to answer a few on paper.