There is a multiple zero at 0 (which means that it touches there), and there are single zeros at -2 and 2 (which means that they cross). There is also 2 imaginary zeros at i and -i.
You can find this by factoring. Start by pulling out the greatest common factor, which in this case is -x^2.
-x^6 + 3x^4 + 4x^2
-x^2(x^4 - 3x^2 - 4)
Now we can factor the inside of the parenthesis. You do this by finding factors of the last number that add up to the middle number.
-x^2(x^4 - 3x^2 - 4)
-x^2(x^2 - 4)(x^2 + 1)
Now we can use the factors of two perfect squares rule to factor the middle parenthesis.
-x^2(x^2 - 4)(x^2 + 1)
-x^2(x - 2)(x + 2)(x^2 + 1)
We would also want to split the term in the front.
-x^2(x - 2)(x + 2)(x^2 + 1)
(x)(-x)(x - 2)(x + 2)(x^2 + 1)
Now we would set each portion equal to 0 and solve.
First root
x = 0 ---> no work needed
Second root
-x = 0 ---> divide by -1
x = 0
Third root
x - 2 = 0
x = 2
Forth root
x + 2 = 0
x = -2
Fifth and Sixth roots
x^2 + 1 = 0
x^2 = -1
x = +/- 
x = +/- i
3.5, it is the coefficient and describes how many n there are
140 is a composite number. Factor pairs: 140 = 1 x 140, 2 x 70, 4 x 35, 5 x 28, 7 x 20, 10 x 14. Factors of 140: 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, 140. Prime factorization: 140 = 2 x 2 x 5 x 7, which can also be written 140 = 2² x 5 x 7
Hey there!
The answer to your question is 
Given:

First, we distribute the negative sign to both terms (
) This would make the
negative and the
positive, so we have:

Add them together and we get:

Have a nice day!