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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!

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