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anonymous
 5 years ago
does anyone know the way to solve this Polynomial Inequality?
x^2(3+x)(x+4)/((x+5)(x1))>=0
I would like to know how to determine the negative and positive factors of the equation.
anonymous
 5 years ago
does anyone know the way to solve this Polynomial Inequality? x^2(3+x)(x+4)/((x+5)(x1))>=0 I would like to know how to determine the negative and positive factors of the equation.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The way I always learned it was to look for critical points (I think there is a better word for this). Those would be x=0,3,4,5,1 (it comes from equation given). These would be different "critical points" on a number line. Then you can test the points in between these critical points.Then we get that: x<5, 4<=x<=3, x=0, x>1 Note: the reason why some are < while others are <= is because if you were to examine the equation you would realize that it cannot be 5 since that would mean division by zero. On the other hand it could be 4, thus it would have that equation factor as well.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you for the reply. I guess I just do not really understand these types of problems, especially the part about testing the points in between the critical points. I do appreciate your response. Perhaps I can try to find a video to explain it in more graphical terms.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It's because rational functions (rational functions are functions in which the numerator and denominator are both polynomials) are continuous whenever the denominator isn't 0. They are continuous because polynomials are continuous. Continuous functions have something called the "Intermediate Value Property." Basically, suppose that at x = a, the function is negative. Then later at x = b, the function is positive. If the function is continuous, then at some point x = c between a and b, the function must have been 0. That’s why looking for zeros and points of discontinuities is important. It tells you places where the function can change signs from positive to negative. Looking for zeros and points of discontinuities is called looking for critical points. In your function, the critical points are x = 5, 4, 3, 0, and 1. So that means if your function is positive for x = 100000, it follows that the function is positive on the interval 1 < x < infinity. That’s because there are no critical points greater than x = 1. So if f(x) is positive for some random value x where x > 1, then f(x) must be positive throughout the entire interval past x > 1. If f(x) is positive for x = 1000000, then x is positive throughout the entire interval infinity < x < 5 for the same reason. If f(x) is positive for x = 1, then f(x) is positive throughout the entire interval 3 < x < 0, since those are the critical points that surround x = 1. We know that f(x) can’t become negative somewhere in that interval since if it became negative, there should be some other critical point in the interval 3 < x < 0, but there are none. Apply that reasoning to test whether f(x) is nonnegative for the remaining intervals.
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