# College Algebra [HELP]

• Sep 8th 2008, 06:42 PM
magnum
College Algebra [HELP]
Okay so I'm not the smartest tool in the shed. Need someone to come around and do this. (Headbang)

SIMPLIFY BY FACTORING:
http://i110.photobucket.com/albums/n...kproblem96.jpg

I came up with (2x-1)^1/2(x+1)(5-x)
Don't ask me how I got it LOL

A little explanation would be nice too.
thanks in advance! I really don't want to fail this course.
• Sep 8th 2008, 06:47 PM
11rdc11
1st Factor out

$\displaystyle (x + 1)(\sqrt{2x-1})$

you will get this

$\displaystyle (x + 1)(\sqrt{2x-1})[3(x+1) -2(2x-1)]$
• Sep 8th 2008, 06:54 PM
magnum
Quote:

Originally Posted by 11rdc11
1st Factor out

$\displaystyle (x + 1)(\sqrt{2x-1})$

you will get this

$\displaystyle (x + 1)(\sqrt{2x-1})[3(x+1) -2\sqrt{2x-1}]$

Can you do it step by step? I keep getting lost after the 1st grouping. That 3/2 power keeps messing me up.

EDIT (MY STEPS)
$\displaystyle 1. (x+1)(\sqrt{2x-1})(3(x+1) -2(2x-1))$
$\displaystyle 2. (x+1)(\sqrt{2x-1})(3x+1-4x+2)$
$\displaystyle 3. (x+1)(\sqrt{2x-1})(3-x)$
Am I wrong? If so, what am I doing wrong?
• Sep 8th 2008, 07:20 PM
11rdc11
Why are you dividing?
• Sep 8th 2008, 07:24 PM
magnum
Quote:

Originally Posted by 11rdc11
Why are you dividing?

I'm not I'm trying to get the 1/2 power up there but it doesn't work.

Edit:
Does that look a little clearer on what I did?
• Sep 8th 2008, 08:39 PM
Legendsn3verdie
warning i may be wrong here is what i did:::::

(x+1)[3(2x-1)^(1/2) (x+1) - 2(2x-1)^(3/2)]

(x+1)(2x-1) [ 3(2x-1)^(-1/2)(x+1) - 2(2x-1)^(1/2)

(x+1)(2x-1) [ {3(x+1) / (2x-1)^(1/2)} - 2(2x-1)^(1/2)

(x+1)(2x-1) [ {3(x+1) / (2x-1)^(1/2)} - 2(2x-1)^(1/2) <--now u gotta subtract these fraction by common denominator =

(x+1)(2x-1) [ {3(x+1) - 2(2x-1) /(2x-1)^(1/2)}
haha now here comes my long shot this is prolly wrong (just tryin to help u man since no1 else will)

(x+1)(2x-1) [3x+3 -4x+2 / (2x-1)^1/2)]

i duno bro i treid for ya.
• Sep 8th 2008, 09:03 PM
11rdc11
I'm want to help but just not sure how to explain it. Umm lets try again.

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

Ok now lets see what can be factored out of it. You can take out an (x+1)

$\displaystyle (x+1)[3(2x-1)^{\frac{1}{2}}(x + 1)-2(2x-1)^{\frac{3}{2}}]$

It can still be factored some more by factoring out $\displaystyle (2x-1)^{\frac{1}{2}}$

$\displaystyle (x+1)(2x-1)^{\frac{1}{2}}[3(x + 1)-2(2x-1)]$

which now you simply

$\displaystyle (x+1)(2x-1)^{\frac{1}{2}}[3x + 3-4x+2)]$

simplfied more

$\displaystyle (x+1)(2x-1)^{\frac{1}{2}}(-x + 5)$

Hope this helps and fixed a mistake in my original post
• Sep 8th 2008, 09:04 PM
magnum
thanks but I think I got it....
I distributed wrong on my earlier post...
$\displaystyle (x+1)(\sqrt{2x-1})(5-x)$

I just switch it back to the square root so I would be less confused with the fractional powers....
• Sep 8th 2008, 09:09 PM
Legendsn3verdie
Quote:

Originally Posted by 11rdc11
I'm want to help but just not sure how to explain it. Umm lets try again.

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

Ok now lets see what can be factored out of it. You can take out an (x+1)

$\displaystyle (x+1)[3(2x-1)^{\frac{1}{2}}(x + 1)-2(2x-1)^{\frac{3}{2}}]$

It can still be factored some more by factoring out $\displaystyle (2x-1)^{\frac{1}{2}}$

$\displaystyle (x+1)(2x-1)^{\frac{1}{2}}[3(x + 1)-2(2x-1)]$

which now you simply

$\displaystyle (x+1)(2x-1)^{\frac{1}{2}}[3x + 3-4x+2)]$

simplfied more

$\displaystyle (x+1)(2x-1)^{\frac{1}{2}}(-x + 5)$

Hope this helps and fixed a mistake in my original post

believe it or not we got the same answer.. if i factor out the denominator in the answer i got i can get that.. guess i did it right.
• Sep 8th 2008, 09:10 PM
11rdc11
No problem
• Sep 8th 2008, 11:10 PM
magnum
just wanted to make sure...

Thanks all.

I froze up during class and forgot how to factor and forgot how fractional exponents were handled. I haven't done math in 2 years...There's bound to be some holes in my memory...Now I can turn it in on thursday... (Rofl)

Oh by the way is there an FAQ or guide for the math code on the forum?
• Sep 8th 2008, 11:15 PM
11rdc11