# Help with factoring

• Oct 16th 2006, 04:33 PM
Taylor
Help with factoring
I need to factor these four problems:
2
3x -1/3

2
(y-5) -36

2
(2y-7) -1

16
x -1

• Oct 16th 2006, 04:40 PM
ThePerfectHacker
Quote:

Originally Posted by Taylor
I need to factor these four problems:
2
3x -1/3

2
(y-5) -36

2
(2y-7) -1

16
x -1

What are those 2's on top?
• Oct 16th 2006, 04:55 PM
Taylor
The twos on top are exponents.
• Oct 16th 2006, 04:59 PM
ThePerfectHacker
Quote:

Originally Posted by Taylor
I need to factor these four problems:
2
3x -1/3

It seems factored to my divine eye.

Quote:

2
(y-5) -36
First 36=6^2
Thus,
Difference of two squares
We have,
(y-5-6)(y-5+6)=(y-11)(y+1)

Quote:

2
(2y-7) -1
Same idea since 1=1^2
We have,
(2y-7-1)(2y-7+1)=(2y-8)(2y-6)
Factor the twos,
2(y-4)*2(y-3)
Thus,
4(y-4)(y-3)

Quote:

16
x -1
Difference of two squaures on 1=1^2 we have,
(x^8+1)(x^8-1)
Thus,
(x^8+1)(x^4+1)(x^4-1)
Thus,
(x^8+1)(x^4+1)(x^2+1)(x^2-1)
Thus,
(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)
• Oct 16th 2006, 05:08 PM
Taylor
Thank you sooooo much for helping me with the problems.
• Oct 17th 2006, 03:36 AM
CaptainBlack
Quote:

Originally Posted by Taylor
I need to factor these four problems:
2
3x -1/3

Factor:

3x^2 - 1.3 = 3(x^2 - 1.9)

the bracket is the difference of two squares, so:

3x^2 - 1.3 = 3(x^2 - 1.9) = 3 (x-1/3) (x+1/3)

,,,,,,,,,,,,,,,= (3x-1)(x+1/3).

RonL