#1**+5 **

(16)((x^2+1)^3))(1-3x^2)

The only thing that makes this 0 is found in the last term

We have

1 - 3x^2 = 0 add 3x^2 to both sides

1 = 3x^2 divide bpth sides by 3

1/3 = x^2 take the square root of both sides

±√(1/3) = x

And these are the two values that make this function = 0

CPhill
Mar 2, 2015

#3**+5 **

Notice that (x^2 +1) can never equal 0

Here's a graph.........https://www.desmos.com/calculator/1kvavvdo1i

Notice that there are only two zeroes at x = ±√(1/3)

CPhill
Mar 2, 2015