College Algebra [HELP]

**magnum** · September 8th 2008, 06:42 PM

Okay so I'm not the smartest tool in the shed. Need someone to come around and do this.

SIMPLIFY BY FACTORING:

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.

**11rdc11** · September 8th 2008, 06:47 PM

1st Factor out

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

you will get this

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

**magnum** · September 8th 2008, 06:54 PM

Originally Posted by 11rdc11

1st Factor out

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

you will get this

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

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)
$1. (x+1)(\sqrt{2x-1})(3(x+1) -2(2x-1))$
$2. (x+1)(\sqrt{2x-1})(3x+1-4x+2)$
$3. (x+1)(\sqrt{2x-1})(3-x)$
Am I wrong? If so, what am I doing wrong?

**11rdc11** · September 8th 2008, 07:20 PM

Why are you dividing?

**magnum** · September 8th 2008, 07:24 PM

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?

**Legendsn3verdie** · September 8th 2008, 08:39 PM

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.

**11rdc11** · September 8th 2008, 09:03 PM

I'm want to help but just not sure how to explain it. Umm lets try again.

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

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

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

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

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

which now you simply

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

simplfied more

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

Hope this helps and fixed a mistake in my original post

**magnum** · September 8th 2008, 09:04 PM

thanks but I think I got it....
I distributed wrong on my earlier post...
Final Answer...
$(x+1)(\sqrt{2x-1})(5-x)$

I just switch it back to the square root so I would be less confused with the fractional powers....

**Legendsn3verdie** · September 8th 2008, 09:09 PM

Originally Posted by 11rdc11

I'm want to help but just not sure how to explain it. Umm lets try again.

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

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

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

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

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

which now you simply

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

simplfied more

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

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.

**11rdc11** · September 8th 2008, 09:10 PM

No problem

**magnum** · September 8th 2008, 11:10 PM

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...

Oh by the way is there an FAQ or guide for the math code on the forum?

**11rdc11** · September 8th 2008, 11:15 PM

http://www.mathhelpforum.com/math-he...-tutorial.html

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