Answer :
Final answer:
The number of possible positive and negative real zeros of the function f(x)=2x^5+4x^4+9x^3+18x^2-35x-70 can be determined using the polynomial zero-product property.
Explanation:
The number of possible positive and negative real zeros of the function f(x) = 2x^5 + 4x^4 + 9x^3 + 18x^2 - 35x - 70 can be determined using the polynomial zero-product property.
- Count the changes in sign of the coefficients as you write them in a row: +, +, +, +, -, -
- The number of possible positive real zeros is equal to the number of sign changes or less.
- The number of possible negative real zeros is equal to the number of sign changes or less, but with an even number of sign changes.
In this case, there are 2 sign changes, so there can be a maximum of 2 positive real zeros and a maximum of 0 or 2 negative real zeros.
Answer:
Number of possible positive real zeros: 1
Number of possible negative real zeros: 4 or 2 or 0
Step-by-step explanation:
Given polynomial function:
[tex]f(x)=2x^5+4x^4+9x^3+18x^2-35x-70[/tex]
To find the number of possible positive and negative real zeros of the given function, we can use Descartes' Rule of Signs.
Positive Real Zeros
To determine the number of possible positive real zeros, we need to examine the non-zero coefficients of the polynomial in descending order, and count the number of sign changes.
In the given polynomial, there is one sign change between the terms + 18x² and - 35x:
[tex]f(x)=2x^5+4x^4+9x^3\underbrace{+18x^2-35x}_{\sf sign\;change}-70[/tex]
Therefore, there is one possible positive real zero.
Negative Real Zeros
To find the possible number of negative real zeros, begin by replacing each x with -x in the polynomial:
[tex]f(-x)=2(-x)^5+4(-x)^4+9(-x)^3+18(-x)^2-35(-x)-70[/tex]
When (-x) is raised to an odd power, the resulting coefficient is negative, and when (-x) is raised to an even power, the resulting coefficient is positive. Therefore, we negate the sign in front of the terms with odd powers, and leave the signs in front of the terms with even powers unchanged:
[tex]f(-x)=-2x^5+4x^4-9x^3+18x^2+35x-70[/tex]
There are now 4 sign changes, which means that the maximum number of possible negative zeros is 4.
As some of the roots may be complex, we need to count down by 2's to find the complete list of the possible number of zeroes. So, the number of possible negative real zeros is 4 or 2 or 0.
Therefore:
- Number of possible positive real zeros: 1
- Number of possible negative real zeros: 4 or 2 or 0
Additional information
If we graph the function (see attachment), we can see that it crosses the x-axis at (-2, 0), (-√(2.5), 0) and (√(2.5), 0), so there are 2 negative real zeros and one positive real zero.
If we factor the polynomial we get:
[tex]f(x)=(x+2)(2x^2-5)(x^2+7)[/tex]
This gives us:
- 1 positive real zero
- 2 negative real zeros
- 2 complex zeros
