how do you find the zeros of (16)((x^2+1)^3))(1-3x^2)

I like seeing your question anon  

Sometimes we wonder if we are teaching thin air.    LOL

(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

How come (1-3x^2) is the only thing that makes this zero?

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)