1. Three Factoring Problems

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???

2. are you solving for the values of the variables?

3. Originally Posted by qzno
are you solving for the values of the variables?
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.

4. #2:

$\displaystyle 4x^2 - 4x - 3 = 0$
where there is a 4 in front of the x squared, you take this and multiply it by the last number
so 4 x -3 = -12 meaning
what multiplies to give you -12 and adds to give you -4?
(-6 and +2)

$\displaystyle 4x^2 + 2x - 6x - 3 = 0$
$\displaystyle 2x(2x + 1) - 3(2x + 1)$

Factors:
$\displaystyle (2x - 3)(2x + 1)$

hope this helps with #2 : )

5. Originally Posted by qzno
#2:

$\displaystyle 4x^2 - 4x - 3 = 0$
where there is a 4 in front of the x squared, you take this and multiply it by the last number
so 4 x -3 = -12 meaning
what multiplies to give you -12 and adds to give you -4?
(-6 and +2)

$\displaystyle 4x^2 + 2x - 6x - 3 = 0$
$\displaystyle 2x(2x + 1) - 3(2x + 1)$

Factors:
$\displaystyle (2x - 3)(2x + 1)$

hope this helps with #2 : )
Oh geez yeah I forgot to put the -6 and 2 back into the equation...thanks a lot.

6. #3:
If this is how you had it written:

$\displaystyle (x^2 + 1)^{\frac{1}{2}} + 3(x^2 + 1)^{\frac{-1}{2}} = 0$

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

$\displaystyle \sqrt{x^2 +1} + \frac{3}{\sqrt{x^2 + 1}} = 0$

Then make it a common denominator

$\displaystyle \frac{x^2 + 1 + 3}{\sqrt{x^2 + 1}} = 0$

$\displaystyle \frac{x^2 + 4}{\sqrt{x^2 + 1}} = 0$

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:

$\displaystyle \frac{\sqrt{x^2 +1}}{\sqrt{x^2 +1}}$

7. Originally Posted by Dickson
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???
Nevermind about #1 just made sense to me...not sure why I didn't get it haha

8. #1:

$\displaystyle r^8 - 1 = 0$

$\displaystyle (r^4)^2 - 1 = 0$

$\displaystyle (r^4 + 1)(r^4 - 1)$

$\displaystyle (r^4 + 1)((r^2)^2 -1)$

$\displaystyle (r^4 + 1)(r^2 + 1)(r^2 - 1)$

$\displaystyle (r^4 + 1)(r^2 + 1)(r + 1)(r - 1)$

: )

9. Originally Posted by qzno
#1:

$\displaystyle r^8 - 1 = 0$

$\displaystyle (r^4)^2 - 1 = 0$

$\displaystyle (r^4 + 1)(r^4 - 1)$

$\displaystyle (r^4 + 1)((r^2)^2 -1)$

$\displaystyle (r^4 + 1)(r^2 + 1)(r^2 - 1)$

$\displaystyle (r^4 + 1)(r^2 + 1)(r + 1)(r - 1)$

: )
Haha thanks...yeah I got it, don't know how I didn't before