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 November 12th, 2014, 01:36 PM #1 Newbie   Joined: Oct 2014 From: england Posts: 23 Thanks: 0 Factorising algebra!!! Stumped. Hi I've been given this question and told to factorise it and I have been shown the answer in stages, but I'm stuck on the first stage. Question: 1/2 (2n) (2n+1) - 1/2 (n-1) n Answer Stage 1 1/2 n (2(2n+1) - (n-1)) Stage 1, (blue section) where has the first single n come from? Is it because 1/2 (2n) = n? If so where has the 2 come from inside the first bracket? Then red section, where have the 1/2 and the n at the end gone? Thanks for any help p.s. I'm ok with stages 2 & 3, but the final stage stumps me, so I'll write that later. Last edited by skipjack; November 12th, 2014 at 10:58 PM.
 November 12th, 2014, 03:34 PM #2 Math Team     Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 note the common factors in the two terms ... $\displaystyle {\color{red}{\frac{1}{2}}} \cdot 2{\color{red}{n}} \cdot (2n+1) - {\color{red}{\frac{1}{2}}} \cdot (n-1) \cdot {\color{red}{n}}$ $\displaystyle {\color{red}{\frac{1}{2} \cdot n}} \left[2(2n+1)-(n-1)\right]$ Thanks from topsquark
November 13th, 2014, 02:04 PM   #3
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Joined: Oct 2014
From: england

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Quote:
 Originally Posted by skeeter note the common factors in the two terms ... $\displaystyle {\color{red}{\frac{1}{2}}} \cdot 2{\color{red}{n}} \cdot (2n+1) - {\color{red}{\frac{1}{2}}} \cdot (n-1) \cdot {\color{red}{n}}$ $\displaystyle {\color{red}{\frac{1}{2} \cdot n}} \left[2(2n+1)-(n-1)\right]$

I see the common factors 1/2 and n... but why have they been taken away from the second term? And swapped round in the first term?

Sorry

Last edited by nicevans1; November 13th, 2014 at 02:13 PM.

November 13th, 2014, 02:11 PM   #4
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Joined: Oct 2014
From: england

Posts: 23
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Quote:
 Originally Posted by skeeter note the common factors in the two terms ... $\displaystyle {\color{red}{\frac{1}{2}}} \cdot 2{\color{red}{n}} \cdot (2n+1) - {\color{red}{\frac{1}{2}}} \cdot (n-1) \cdot {\color{red}{n}}$ $\displaystyle {\color{red}{\frac{1}{2} \cdot n}} \left[2(2n+1)-(n-1)\right]$

Yes, I see the c factors, but I still don't understand why they have been taken away from the second term and why the 2n has swapped round in the first term?

Sorry

Last edited by skipjack; November 16th, 2014 at 10:53 AM.

 November 14th, 2014, 12:20 AM #5 Senior Member     Joined: Nov 2010 From: Indonesia Posts: 2,001 Thanks: 132 Math Focus: Trigonometry Do you know about distributive property? This property says that $\displaystyle a\times(b-c)=(a\times b)-(a\times c)$. Compare it to your problem and you may understand. If it's applied to your problem, $\displaystyle a=\frac{1}{2}n$, $\displaystyle b=2(2n+1)$, and $\displaystyle c=n-1$. Thus, you can separate a, which is $\displaystyle \frac{1}{2}n$ from the others. Good luck. Thanks from topsquark and nicevans1
 November 16th, 2014, 09:28 AM #6 Newbie   Joined: Oct 2014 From: england Posts: 23 Thanks: 0 Thank you for your help and time. The info on the distributive properties clinched the deal. Lol

 Tags algebra, factorising, stumped

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