SOLUTION: Hello: I am having some trouble with this problem: x^3+8x^2+11x-20 I am to find all real zeros using DeCartes' Rule of Signs and Rational Zeros Theorem. My instructor nev

Question 31012This question is from textbook College Algebra
: Hello:
I am having some trouble with this problem:
x^3+8x^2+11x-20
I am to find all real zeros using DeCartes' Rule of Signs and Rational Zeros Theorem. My instructor never explained the Rule of Signs, and I don't understand it very well in the book. I have figured out p/q, but is there a way to find the zeros without plugging them into the f(x) individually?
Thanks,
Brad Olson This question is from textbook College Algebra

Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
I am having some trouble with this problem:
x^3+8x^2+11x-20
I am to find all real zeros using DeCartes' Rule of Signs and Rational Zeros Theorem. My instructor never explained the Rule of Signs, and I don't understand it very well in the book
SEE BELOW THE EXAMPLES AND THEOREM .IF NOT CLEAR COME BACK
Descartes's Rule of Signs
Let F(X) be a polynomial where are real coefficients.
The number of POSITIVE REAL ZEROS of f(X) is either equal to the number of sign changes of successive terms of f(x) or is less than that number by an even number (until 1 or 0 is reached).
The number of NEGATIVE REAL ZEROS of f is either equal to the number of sign changes of successive terms of f(-x) or is less than that number by an even integer (until 1 or 0 is reached).

This can help narrow down your possibilities when you do go on to find the zeros.
Example 5: Find the possible number of positive and negative real zeros of using Descartes�s Rule of Signs.
F(X)=3X^4-5X^3+2X^2-X+10
.........S....S....S..S
In this problem it isn�t asking for the zeros themselves, but what are the possible number of them. This can help narrow down your possibilities when you do go on to find the zeros.
Possible number of positive real zeros:

The S are showing where there are sign changes between successive terms, going left to right. The first S on the left shows a sign change from positive 3 to negative 5. The 2nd S shows a sign change from negative 5 to positive 2. The third S shows a sign change from positive 2 to negative 1. And the last S shows a sign change from negative 1 to positive 10.
There are 4 sign changes between successive terms, which means that is the highest possible number of positive real zeros. To find the other possible number of positive real zeros from these sign changes, you start with the number of changes, which in this case is 4, and then go down by even integers from that number until you get to 1 or 0.
Since we have 4 sign changes with f(x), then there is a possibility of 4 or 4 - 2 = 2 or 4 - 4 = 0 positive real zeros.

Possible number of negative real zeros:
F(X)=3(-X)^4-5(-X)^3+2(-X)^2-(X)+10
=3X^4+5X^3+2X^2+X+10
Note how there are no sign changes between successive terms.
This means there are no negative real zeros.
Since we are counting the number of possible real zeros, 0 is the lowest number that we can have. This piece of information would be helpful when determining real zeros for this polynomial.
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I have figured out p/q,
GOOD...SO I AM NOT ELABORATING IT
but is there a way to find the zeros without plugging them into the f(x) individually?
NO THERE IS NO WAY YOU HAVE TO SUBSTITUTE TO FIND IT IS A ZERO OR NOT/

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