# Factorising algebra!!! Stumped.Watch

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#1
Hi
Ive 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

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 has the 1/2 and the n at the end gone?

Thanks for any help
p.s Im ok with satge 2 & 3 but the final stage stumps me, so ill write that later.
0
5 years ago
#2
(Original post by nicevans1)
Hi
Ive 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

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 has the 1/2 and the n at the end gone?

Thanks for any help
p.s Im ok with satge 2 & 3 but the final stage stumps me, so ill write that later.
Hi!

So with this question, I believe the n has come from the red side.
See, it is a balanced equation. So notice the n on the red side in the question? In the first stage it has dissapeared from the red side and onto the blue side.

With balanced equations, if you add something, you add it to both sides. In the question, there is a minus sign between the two sections, therefore the red section is negative. So, as I'm sure you know, a single number in algebra is equal to 1? Therefore on the red section, the n is equal to -1.

If you add n to both sides, the n in the red section dissapears as -1 +1= 0. However it adds to the blue section as there were no n's there before (not including whats in the brackets).

I'm afraid I'm not sure about the '2', but I hope I've helped solve half your issue. :-)
0
5 years ago
#3
(Original post by nicevans1)
Hi
Ive 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

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 has the 1/2 and the n at the end gone?

Thanks for any help
p.s Im ok with satge 2 & 3 but the final stage stumps me, so ill write that later.
they "pulled out" 1/2 n from red and blue

This leaves from the first term the 2 and the (2n+1)

minus

from the second term just the (n-1)

Any good?
0
5 years ago
#4
(Original post by mediageek123)
Hi!

So with this question, I believe the n has come from the red side.
See, it is a balanced equation. So notice the n on the red side in the question? In the first stage it has dissapeared from the red side and onto the blue side.

With balanced equations, if you add something, you add it to both sides. In the question, there is a minus sign between the two sections, therefore the red section is negative. So, as I'm sure you know, a single number in algebra is equal to 1? Therefore on the red section, the n is equal to -1.

If you add n to both sides, the n in the red section dissapears as -1 +1= 0. However it adds to the blue section as there were no n's there before (not including whats in the brackets).

I'm afraid I'm not sure about the '2', but I hope I've helped solve half your issue. :-)
I know you are only trying to help but this is not an equation 0
5 years ago
#5
(Original post by TeeEm)
I know you are only trying to help but this is not an equation It isn't? Oh I'm so sorry! Well, you can see why I only got a 'B' in maths gcse! Haha.
0
5 years ago
#6
(Original post by mediageek123)
It isn't? Oh I'm so sorry! Well, you can see why I only got a 'B' in maths gcse! Haha.
no worries

It is the good intention which matters.
0
#7
(Original post by TeeEm)
they "pulled out" 1/2 n from red and blue

This leaves from the first term the 2 and the (2n+1)

minus

from the second term just the (n-1)

Any good?
But the 1/2 & n are still there in the first term?
0
5 years ago
#8
(Original post by nicevans1)
But the 1/2 & n are still there in the first term?
definitely they are not looking at your stage 1

(Are you a GCSE student or AS?)
0
5 years ago
#9
(Original post by nicevans1)
But the 1/2 & n are still there in the first term?
In your first expression, 1/2 and n are in both terms. In the second expression, you have 1/2*n*(term 1 + term 2), and neither 1/2 nor n are in either of those terms.
0
#10
AS level!!!

I'v multiplied the whole expression (both terms) by 2 then later I divide by 2 and that's how the half will still be there.

Then I realised the lone n at the end of the first expression is still there but it is repositioned at the start of the second expression.

Now can you explain why the 2n at the start of the first expression became a lone 2 in the second expression?

P.s I have problems seeing what may be simple to others.... Lol
0
5 years ago
#11
(Original post by nicevans1)
AS level!!!

I'v multiplied the whole expression (both terms) by 2 then later I divide by 2 and that's how the half will still be there.

Then I realised the lone n at the end of the first expression is still there but it is repositioned at the start of the second expression.

Now can you explain why the 2n at the start of the first expression became a lone 2 in the second expression?

P.s I have problems seeing what may be simple to others.... Lol
I'm not sure what you have done but you cannot just multiply it by 2. It isn't an equation.
0
5 years ago
#12
(Original post by nicevans1)
AS level!!!

I'v multiplied the whole expression (both terms) by 2 then later I divide by 2 and that's how the half will still be there.
This is really GCSE stuff - what grade did you get in GCSE?

You have this:
(1/2)(2n)(2n + 1) - (1/2)(n-1)n

The factors in the first term are 1/2, 2, n and 2n+1. If you pull out 1/2 and n you are left with factors of 2 and 2n + 1.

The factors in the second term are 1/2, n-1 and n. If you pull out 1/2 and n you are left with a factor of n - 1.

So, the whole expression is (putting the common factors 1/2 and n at the front of each term):

(1/2)n[2(2n+1)] - (1/2)n[n-1]

Now take the (1/2)n factor outside a larger bracket to get:

(1/2)n[2(2n+1) - (n-1)]
0
5 years ago
#13
(Original post by nicevans1)
Hi
Ive 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

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 has the 1/2 and the n at the end gone?

Thanks for any help
p.s Im ok with satge 2 & 3 but the final stage stumps me, so ill write that later.
TBH I think your colour coding shows the issue

Consider

1/2 (2n) (2n+1) - 1/2 (n-1) n

becomes

1/2 (2n) (2n+1) - 1/2 (n-1) n

Giving

1/2n(2(2n+1) - (n-1))
0
#14
Tut sorry all. Yes I get it now. Thanks for your help.

P.s Im doing a distance learning a level and last year I did a distance learning igcse and its difficult when you don't have someone to speak face to face.

I didn't do anything like the above in my igcse. Yes I did expressions advanced algebra differentiation you name it but that swapping around numbers inside and outside the brackets is proper simple pre gcse stuff.

Thanks again.

I'll post the second stage in another thread but another day. Pmsl
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