Question 1135039: Consider the following f(x)=x^5+x^3-6x
a. Find all the real zeros of the polynomial function.
b. Determine the end behavior of the polymial.
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! Factor.
There's an x in common with every one of those terms so we can definitely factor out x.
x(x^4 + x^2 - 6)
Think of this as:
x((x^2)^2 + (x)^2 - 6)
Let's take care of the easy one first. When x is a factor, this would just mean that x = 0. But since the polynomial has degree 5, we must check to make sure there aren't others (up to a maximum of 5).
So by any quadratic factoring/solving method you want to use, you should get to this result:
x^2 = (-1 +- sqrt(1^2 - (-24)))/2
x^2 = -3 or 2
Clearly x^2 = -3 does not yield any real zeroes, so look at x^2 = 2. While not a perfect square, this still has two answers. Those are +-sqrt(2).
b) The degree is 5, hence this function has odd behavior. The leading coefficient is greater than 0, so we know the end behavior is that of any positive odd function, namely from the left it is decreasing and from the right, it is increasing.
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