Could you explain to me decartes Rule of signs?
Here are my questions:
Use Decartes Rule of signs to determine the possible number of postive and negative real zeros for the given function.
1. f(x)=-7x^9+x^5-x^2+6
choices are:
a. 2 or 0 positive zeros,2 or 0 negative zeros
b.2 or 0 positive zeros, 3 or 1 negative zeros
c.3 or 1 positive zeros, 3 or 1 negative zeros
d. 3 or 1 positive zeros, 2 or 0 negative zeros
2. f(x)10x^3-8x^2+x+5
my choices are:
a. 3 or 1 positive zeros, 1 negative zero
b.2 or 0 positive zeros, no negative zeros
c.3 or 1 positive zeros, 2 or 0 negative zeros
d.2 or 0 positive zeros, 1 negative zero
I will be happy to check your answers or thinking.
http://www.purplemath.com/modules/drofsign.htm
Decartes' Rule of Signs is a method used to determine the possible number of positive and negative real zeros of a polynomial function. Here's how it works:
1. Count the number of sign changes in the coefficients of the terms in the polynomial. Ignore any zero coefficients. The number of sign changes represents the maximum number of positive real zeros.
2. Count the number of sign changes in the coefficients of the terms in the polynomial when the exponents are changed from even to odd or odd to even. Ignore any zero coefficients. The number of sign changes represents the maximum number of negative real zeros.
Now, let's apply this rule to the given functions:
1. f(x) = -7x^9 + x^5 - x^2 + 6
- Counting the sign changes in the coefficients, we have one sign change from negative (-7) to positive (6).
- Counting the sign changes when the exponents change, we have one sign change from odd to even (from x^5 to x^2).
Based on the rule, there are either 1 or 3 positive real zeros, and either 1 or 3 negative real zeros. Among the choices provided, the correct answer is (c) 3 or 1 positive zeros, 3 or 1 negative zeros.
2. f(x) = 10x^3 - 8x^2 + x + 5
- Counting the sign changes in the coefficients, we have no sign changes in this case.
- Counting the sign changes when the exponents change, we have one sign change from odd to even (from x^3 to x^2).
Based on the rule, there are either 0 or 2 positive real zeros, and either 1 or 3 negative real zeros. Among the choices provided, the correct answer is (d) 2 or 0 positive zeros, 1 negative zero.
Remember, Decartes' Rule of Signs provides an upper bound on the possible number of positive and negative real zeros. It does not determine the exact number of zeros, but it gives us a range of possibilities.