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 June 1st, 2015, 05:28 PM #1 Member   Joined: May 2015 From: U.S.A. Posts: 45 Thanks: 0 Adding two identical two-variable terms Sorry if I got any terminology wrong in the title. Please correct me if need be. So, $\displaystyle (4x+5y)^{2}$ I keep getting $\displaystyle 16x^{2}+25y^{2}+8x10y$ . What seems to be throwing me off are the two 4x5y 's I get from the two 5y times 4x 's. I don't know what to do with them, apparently. The answer is $\displaystyle 16x^{2}+40xy+25y^{2}$. Where does the 40 come from? I can see (8 times 10) divided by 2 is 40, but I'm not sure if the 8 times 10 is relevant to the problem. June 1st, 2015, 05:42 PM #2 Banned Camp   Joined: May 2015 From: Casablanca Posts: 36 Thanks: 7 let $a =4x$, $b=5y$ $(4x+5y)^2$ becomes $(a+b)^2$ At the end replace a and b by their respective values. June 1st, 2015, 05:47 PM   #3
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 Originally Posted by Assayid let $a =4x$, $b=5y$ $(4x+5y)^2$ becomes $(a+b)^2$ At the end replace a and b by their respective values.
Hm. I'm still getting $\displaystyle 8x10y$. Using the $\displaystyle (a+b)^{2}$ idea, I got the $\displaystyle a^{2}+b^{2}+2ab$ equation. So, $\displaystyle 2(4x5y)$: Multiply the 4 by 2 and the 5 by 2, and you get 8 and 10, right? June 1st, 2015, 06:32 PM #4 Math Team   Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 $4x5y = 20xy$. Also, $2(4x5y)$ is NOT the same as $2(4x)\cdot2(5y)$. Instead, $2(4x5y)= (2\cdot4\cdot5)xy = 40xy$ Thanks from Rexan June 1st, 2015, 07:17 PM   #5
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 Originally Posted by Azzajazz $4x5y = 20xy$. Also, $2(4x5y)$ is NOT the same as $2(4x)\cdot2(5y)$. Instead, $2(4x5y)= (2\cdot4\cdot5)xy = 40xy$
Ah! Okay, thanks! That explanation really helps. Do numbers against parentheses only get distributed when there's addition or subtraction inside, not multiplication or division? June 1st, 2015, 08:42 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 In ordinary arithmetic, yes. Thanks from Rexan June 2nd, 2015, 08:31 AM   #7
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 Originally Posted by skipjack In ordinary arithmetic, yes.
All right, thanks, skipjack! Tags adding, identical, terms, twovariable Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post clgordon Algebra 2 June 2nd, 2014 12:33 AM Niko Bellic Calculus 0 July 19th, 2013 08:08 PM soulrain Algebra 2 January 11th, 2013 06:19 PM Messerschmitt Algebra 3 September 1st, 2012 07:35 AM Messerschmitt Calculus 0 December 31st, 1969 04:00 PM

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