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November 12th, 2014, 02:36 PM   #1
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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 11:58 PM.
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November 12th, 2014, 04:34 PM   #2
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note the common factors in the two terms ...

$\displaystyle {\color{red}{\frac{1}{2}}} \cdot 2{\color{red}{n}} \cdot (2n+1) - {\color{red}{\frac{1}{2}}} \cdot (n-1) \cdot {\color{red}{n}}$

$\displaystyle {\color{red}{\frac{1}{2} \cdot n}} \left[2(2n+1)-(n-1)\right]$
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November 13th, 2014, 03:04 PM   #3
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Quote:
Originally Posted by skeeter View Post
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 03:13 PM.
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November 13th, 2014, 03:11 PM   #4
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Quote:
Originally Posted by skeeter View Post
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 11:53 AM.
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November 14th, 2014, 01:20 AM   #5
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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.
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November 16th, 2014, 10:28 AM   #6
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Thank you for your help and time. The info on the distributive properties clinched the deal. Lol
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