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#1
12u^2 + 6u -6 This is the question.

My textbook says the answer is (3u+3) (4u-2)

When checking online answers I found the answer to be: 6(u+1) (2u-1). I completely understand how that answer came about because you take 6 out as a common factor.

I'm not sure if the textbook answer is correct or if it's my answer or the online answer.

Thank you
Last edited by Advanced-08234; 1 month ago
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1 month ago
#2
Both answers are correct - not your one, I mean the text books one and the online one.
You can check which one is correct by expanding the brackets, assuming you know how to do that.

I'm assuming you got your answer by diving every term by 6, then factorising. Well, you can't do that. You need to factor out the 6 to get 6(2u^2 +u -1) and then factorise inside the brackets to get 6(2u-1)(u+1) same as the online answer
Last edited by username5737602; 1 month ago
1
1 month ago
#3
12u^2 + 6u -6 This is the question.

My textbook says the answer is (3u+3) (4u-2)

When checking online answers I found the answer to be: 6(u+1) (2u-1). I completely understand how that answer came about because you take 6 out as a common factor.

I'm not sure if the textbook answer is correct or if it's my answer or the online answer.

Thank you
Your answer cannot be correct - if you multiply it out then you don't get 12u^2 for example.

The other 2 factorizations are equivalent because the online version takes out factors 3 and 2 from the respective brackets in the textbook answer.
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1 month ago
#4
Aye, (u+1) (2u-1) is incorrect as that would be 2u^2 + u - 1
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1 month ago
#5
Online and textbook both correct, but the online answer is in the simplest form
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#6
(Original post by tej3141)
Both answers are correct - not your one, I mean the text books one and the online one.
You can check which one is correct by expanding the brackets, assuming you know how to do that.

I'm assuming you got your answer by diving every term by 6, then factorising. Well, you can't do that. You need to factor out the 6 to get 6(2u^2 +u -1) and then factorise inside the brackets to get 6(2u-1)(u+1) same as the online answer
I have a different method for factorising by grouping, but what I've noticed is that this questions allows you to take out a common factor but all the other questions I've done don't. May I ask you get the textbook answer, please?

My normal method is the following:

6x^2 - 13x + 6 This is the question

AC = 6 x 1 = 6

B = -13

Find two numbers that multiply to give 6 but add to give -13

The following numbers work: -9, -4

Place 6x in each bracket

(6x - 9) (6x -4)

Cancel down common factors.

(2x - 3) (3x -2) <----------------This is the final answer

The textbook says that this answer is correct.

I suspect this method doesn't work for expressions that require you/ is possible to take out a common factor. I haven't done any more questions than this so far so I could be wrong.

May I ask how it's possible to get the textbook answer to my original question without taking out a common factor?

Thank you for your help. 🙂
Last edited by Advanced-08234; 1 month ago
0
1 month ago
#7
I have a different method for factorising by grouping, but what I've noticed is that this questions allows you to take out a common factor but all the other questions I've done don't. May I ask you get the textbook answer, please?

My normal method is the following:

6x^2 - 13x + 6 This is the question

AC = 6 x 1 = 6

B = -13

Find two numbers that multiply to give 6 but add to give -13

The following numbers work: -9, -4

Place 6x in each bracket

(6x - 9) (6x -4)

Cancel down common factors.

(2x - 3) (3x -2) <----------------This is the final answer

The textbook says that this answer is correct.

I suspect this method doesn't work for expressions that require you/ is possible to take out a common factor. I haven't done any more questions than this so far so I could be wrong.

May I ask how it's possible to get the textbook answer to my original question without taking out a common factor?

Thank you for your help. 🙂
The method that use is not really good. First of all AC =6×6=36 not 6 as you said but you corrected this in your later working out so it doesn't matter ig. The method is fine until you do the (6x-9)(6x-4) part - that's not correct because if you expand the brackets you get 36 as the x^2 coefficient when it is supposed to be 6.
And I could show you how to manipulate it to get the textbook answer, but theres really no point as you would not be penalised in the exam is you wrote the fully factorised form 6(2u-1)(u+1)
Also, is this for gcse maths?
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#8
(Original post by tej3141)
The method that use is not really good. First of all AC =6×6=36 not 6 as you said but you corrected this in your later working out so it doesn't matter ig. The method is fine until you do the (6x-9)(6x-4) part - that's not correct because if you expand the brackets you get 36 as the x^2 coefficient when it is supposed to be 6.
And I could show you how to manipulate it to get the textbook answer, but theres really no point as you would not be penalised in the exam is you wrote the fully factorised form 6(2u-1)(u+1)
Also, is this for gcse maths?
Yeah, it's for GCSE. It's just a method I learnt that always gave me the right answer. I guess I'll just learn this the textbook way because I don't want to end up in an exam situation and get it wrong.
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#9
(Original post by tej3141)
The method that use is not really good. First of all AC =6×6=36 not 6 as you said but you corrected this in your later working out so it doesn't matter ig. The method is fine until you do the (6x-9)(6x-4) part - that's not correct because if you expand the brackets you get 36 as the x^2 coefficient when it is supposed to be 6.
And I could show you how to manipulate it to get the textbook answer, but theres really no point as you would not be penalised in the exam is you wrote the fully factorised form 6(2u-1)(u+1)
Also, is this for gcse maths?
I figured out why the AC method (my method as I've been referring to) didn't work

12u^2 + 6u -6 This is the question.

AC = 12 x (-6) = -72

B = 6

Two numbers that multiply to give -72 but add to give 6:

Those two numbers are 12 and -6

(12u + 12) (12u - 6)

The textbook answer takes 3 out as a common factor: (12u + 12) (12u - 6) ----------> (3u+3) (4u-2)

The online answer takes 6 out as a common factor at the end: (12u + 12) (12u - 6) ------------> 6(u+1) (2u-1).

What I did was take 12 out as a common factor of this bracket: (12u + 12)

Then I took out 6 as a common factor of this bracket (12u - 6)

I guess I can only use one common factor to simplify both brackets rather than one common factor for EACH bracket?

If that's the case then may I ask why?

Whenever I've used the AC method you only ever have to cancel down one bracket.

Thank you
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#10
(Original post by davros)
Your answer cannot be correct - if you multiply it out then you don't get 12u^2 for example.

The other 2 factorizations are equivalent because the online version takes out factors 3 and 2 from the respective brackets in the textbook answer.
I don't understand why I can't factorise to this:

(12u + 12) (12u - 6)

And then take out 12 as a common factor from the left bracket and then 6 from the right bracket leaving you with: (u + 1) ( 2u - 1)

I guess when you expand them you get (2u^2 + u - 1)

Multiply it out by 6 would give the original form.

I supposed I'm just supposed to expand it at the end and check if it's the correct answer.

That's the first time I've ever come across such as issue. Usually, you take out the highest common factor for each bracket (I've only ever had to do it for one bracket until this question), but I guess sometimes taking out the Highest common factor doesn't work and so you're supposed to use a smaller common factor.

It's very strange to come across this question, but I have noticed that this is the only question so far where you can take out a common factor right at the start in the quadratics original/ expanded form.

Thank you
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1 month ago
#11
I don't understand why I can't factorise to this:

(12u + 12) (12u - 6)
Because that's completely wrong, isn't it?!

If you multiply that out you're going to get something starting with 144u^2, not 12u^2 + 6u -6 which is what you are supposed to be aiming at according to your original post
1
1 month ago
#12
I figured out why the AC method (my method as I've been referring to) didn't work

12u^2 + 6u -6 This is the question.

AC = 12 x (-6) = -72

B = 6

Two numbers that multiply to give -72 but add to give 6:

Those two numbers are 12 and -6

(12u + 12) (12u - 6)

The textbook answer takes 3 out as a common factor: (12u + 12) (12u - 6) ----------> (3u+3) (4u-2)

The online answer takes 6 out as a common factor at the end: (12u + 12) (12u - 6) ------------> 6(u+1) (2u-1).

What I did was take 12 out as a common factor of this bracket: (12u + 12)

Then I took out 6 as a common factor of this bracket (12u - 6)

I guess I can only use one common factor to simplify both brackets rather than one common factor for EACH bracket?

If that's the case then may I ask why?

Whenever I've used the AC method you only ever have to cancel down one bracket.

Thank you
I think you need to get a new method of factorising as this one is wrong and may cause you to lose marks - even if you do manage to get the correct answer In the end. I recommend looking on YouTube for some videos on how to factorise
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