Question

i guess no one wants to help me
For the polynomial, list each real zero and its multiplicity. Determine whether the graph crosses or touches the x-axis at each x -intercept. f(x) = 2(x 2 + 5)(x 2 + 1) 2
Determine whether the graph crosses or touches the x-axis at each x -intercept.

f(x) = 2(x2 + 5)(x2 + 1)2
A) -5, multiplicity 1, touches x-axis; -1, multiplicity 2, crosses x-axis
B) -5, multiplicity 1, crosses x-axis; -1, multiplicity 2, touches x-axis
C) No real zeros
D) , multiplicity 1, crosses x-axis; -, multiplicity 1, c

Answer 1

I'd love to help you understand how to do these all by yourself! Which parts don't you understand how to do?

Answer 2

Is your polynomial\[f(x) = 2(x^2+5)(x^2+1)^2\]
?

Answer 3

yes it is

Answer 4

I'll assume that it is. The zeros of a polynomial f(x) are just the places where f(x) = 0. How do we find them? Well, that can be tricky in the general case, but in this case, we've been given the polynomial in factored form, which makes it very easy. Because it is factored, it is written as a product. How do we make the product of some numbers = 0? There's only one way: make one or more of the numbers be 0! Doesn't matter how complicated the expression is, if it is a product, we just need to make the various quantities = 0 and the product will be 0.

Now, how do we make \[f(x) = 2(x^2+5)(x^2+1)^2 = 0\]\[2(x^2+5)(x^2+1)(x^2+1) = 0\]Now we can set any of the things in parentheses equal to 0 and solve them; the results will be our zeros, because if any of those = 0, the whole product will be 0.
\[x^2+5 = 0\]\[x^2 = -5\]\[x = \pm \sqrt{-5} = \pm \sqrt{-1*5} = \pm i\sqrt{5}\]
\[x^2+1 = 0\]\[x^2 = -1\]\[x= \pm \sqrt{-1} = \pm i\]

So there are our roots! the +- signs are due to the fact that -a*-a = a^2 = a*a, so both the negative and positive values work.

Answer 5

How many of those are real numbers, not complex numbers?