are you solving for the values of the variables?
1) r^8 - 1
= (r^4)^2 - (1)^2
= (r^4 - 1)(r^16 + 1r ^4 + 1)
= Not sure what to do from here
2) 4x^2 - 4x - 3 = (x+2)(2x - 1)(2x + 3)
Why do you have to make it 2x and not just have one as 4x?
3) (x^2 + 1) ^ 1/2 + 3(x^2 + 1) ^ -1/2
How do you do this???
Oh sorry no
For number 1 I just need to finish factoring.
#2 I don't understand why when factored 4x^2 - 4x - 3 = (2x - 1)(2x + 3) and not (4x - 2)(x + 3)
#3 I just don't understand how to factor it down.
For 1 I used difference of squares and for 3 I'm supposed to use Sum of Cubes I think.
#3:
If this is how you had it written:
Then we know that anything raised to the power of a half is the same as the square root
and anything raised to the negative a half is the same as one over the square root
Then make it a common denominator
I don't know if you personally want the denominator out of the square root or not but this is it in a nutshell.
If you want to get rid of the square root in the denominator just multiply it by: