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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 (n1) n Answer Stage 1 1/2 n (2(2n+1)  (n1)) 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: 2,885 Thanks: 1504 
note the common factors in the two terms ... $\displaystyle {\color{red}{\frac{1}{2}}} \cdot 2{\color{red}{n}} \cdot (2n+1)  {\color{red}{\frac{1}{2}}} \cdot (n1) \cdot {\color{red}{n}}$ $\displaystyle {\color{red}{\frac{1}{2} \cdot n}} \left[2(2n+1)(n1)\right]$ 
November 13th, 2014, 02:04 PM  #3  
Newbie Joined: Oct 2014 From: england Posts: 23 Thanks: 0  Quote:
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  
Newbie Joined: Oct 2014 From: england Posts: 23 Thanks: 0  Quote:
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 $\displaystyle a\times(bc)=(a\times b)(a\times c)$. 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 $\displaystyle c=n1$. Thus, you can separate a, which is $\displaystyle \frac{1}{2}n$ from the others. Good luck.

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


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algebra, factorising, stumped 
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