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A little help...

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A little help...

by adsuper7 » Fri Oct 14, 2011 8:47 pm
Having a bit of trouble solving this problem.

xy+2xy(1+y)2y

A. xy(2y+1)^2
B. (2xy+1)(2xy-1)
C. (2x+y)^2
D. (x+2y)^2
E. y(x-2y)^2
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by Anurag@Gurome » Fri Oct 14, 2011 8:55 pm
adsuper7 wrote:xy+2xy(1+y)2y
I guess the problem is to find the equivalent expression or the factorized form of the given expression. Anyways, here is the solution...

# xy + 2xy(1 + y)2y
= xy[1 + 2(1 + y)2y]
= xy[1 + 4y + 4y²]
= xy(2y + 1)²

The correct answer is A.
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by zooki » Sat Oct 15, 2011 5:59 am
in the original expression the power of x is 1. Only Answer A satisfies this.
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by sam2304 » Sat Oct 15, 2011 6:01 am
That's awesome. Easier than solving. Thanks zooki.
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by adsuper7 » Sat Oct 15, 2011 9:18 am
Thanks all!
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by HeyArnold » Sun Oct 16, 2011 7:31 am
It would not take too long to just plug in #'s here either.

x = 2
y = 1

(2x1) + 2(2x1)(1+1)2(1) = 18

only A. = 18

(2x1)(2x1 + 1)^2 = 2(3)^2 = 18
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by HeyArnold » Sun Oct 16, 2011 7:33 am
It would not take too long to just plug in #'s here either.

x = 2
y = 1

(2x1) + 2(2x1)(1+1)2(1) = 18

only A. = 18

(2x1)(2x1 + 1)^2 = 2(3)^2 = 18
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