# Three Factoring Problems

• Jan 29th 2009, 05:49 PM
Dickson
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???
• Jan 29th 2009, 05:57 PM
qzno
are you solving for the values of the variables?
• Jan 29th 2009, 06:02 PM
Dickson
Quote:

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.
• Jan 29th 2009, 06:08 PM
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 : )
• Jan 29th 2009, 06:12 PM
Dickson
Quote:

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.
• Jan 29th 2009, 06:17 PM
qzno
#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:

$\displaystyle \frac{\sqrt{x^2 +1}}{\sqrt{x^2 +1}}$
• Jan 29th 2009, 06:18 PM
Dickson
Quote:

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
• Jan 29th 2009, 06:21 PM
qzno
#1:

$\displaystyle r^8 - 1 = 0$

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

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

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

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

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

: )
• Jan 29th 2009, 06:29 PM
Dickson
Quote:

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