I'm stuck. Help would be appreciated.
This is an example from a book:
I expanded the question and got: 5x^2 +17x+14Q:Factorise 5(x+2)^2 3(x+2)
A: (x+2)(5x+7)
I expanded the answer and got 5x^2 +143x
I'm assuming they should both be equal so I'm guessing I did something wrong when I expanded the question?
Can someone please explain?
Factorising by taking out common factors
Announcements  Posted on  

TSR's new app is coming! Sign up here to try it first >>  17102016 

 Follow
 1
 02082016 17:48

 Follow
 2
 02082016 17:56
Don't expand. You can write it as , by taking out the common factor, (x+2), of both terms. The expression in the square brackets can be simplified further.
Post rating:1 
 Follow
 3
 02082016 17:56
(Original post by ChrolloLucilfer)
I'm stuck. Help would be appreciated.
This is an example from a book:
I expanded the question and got: 5x^2 +17x+14
I expanded the answer and got 5x^2 +143x
I'm assuming they should both be equal so I'm guessing I did something wrong when I expanded the question?
Can someone please explain?
Think of this as two terms [5(x+2)^2] and [3(x+2)]
Since both the terms have (x+2) you take that out as a common factor
You are left with (x+2)[5(x+2)3]
Then you solve the bracket
[5(x+2)3] = 5x+103= 5x+7 so the answer is (x+2)(5x+7)Last edited by Potatoo; 02082016 at 17:58.Post rating:1 
 Follow
 4
 02082016 17:57

 Follow
 5
 02082016 17:57
(Original post by ChrolloLucilfer)
I'm stuck. Help would be appreciated.
This is an example from a book:
I expanded the question and got: 5x^2 +17x+14
I expanded the answer and got 5x^2 +143x
I'm assuming they should both be equal so I'm guessing I did something wrong when I expanded the question?
Can someone please explain?
Hint: (x+2)^2=(x+2)(x+2)
Posted from TSR Mobile 
 Follow
 6
 02082016 17:57
(Original post by ChrolloLucilfer)
I'm stuck. Help would be appreciated.
This is an example from a book:
I expanded the question and got: 5x^2 +17x+14
I expanded the answer and got 5x^2 +143x
I'm assuming they should both be equal so I'm guessing I did something wrong when I expanded the question?
Can someone please explain?
Hope this helps 
 Follow
 7
 02082016 18:00
I'm glad everyone agrees with me.

 Follow
 8
 02082016 18:03
I'm really bad at explaining things so sorry if this series of steps makes absolutely no sense:
5(x+2)^2  3(x+2)
5(x+2)(x+2)  3(x+2)
(5x + 10)(x + 2)  3(x + 2)
5x(x+2) + 10(x+2)  3(x+2)
5x(x+2) + (103)(x+2)
5x(x+2) + 7(x+2)
(5x + 7)(x + 2)
EDIT: never mind, the above answers are so much better 
 Follow
 9
 02082016 18:05
(Original post by B_9710)
I'm glad everyone agrees with me. 
 Follow
 10
 02082016 18:18
(Original post by surina16)
[full solution...]
I see that you've been posting in the maths forum a bit now; you should be aware that we only give hints and guidance here, we help the OP reach the answer and understanding bitbybit instead of dumping a full solution on them which, in most cases, is quite unhelpful in terms of actual understanding and detrimental in the long run. So, can you please switch to hints and tips instead of full blown solutions from now on? Thanks. 
 Follow
 11
 02082016 19:44
(Original post by B_9710)
Don't expand. You can write it as , by taking out the common factor, (x+2), of both terms. The expression in the square brackets can be simplified further.
So how would you expand the original question: 5(x+2)^2 3(x+2) to get the same answer as (x+2)(5x+7) expanded? 
 Follow
 12
 02082016 19:52
(Original post by ChrolloLucilfer)
Thanks. I understand how to factorise it, but the reason I expanded was because when i did it to the answer I didn't get the same answer as the question. Does that make sense?
So how would you expand the original question: 5(x+2)^2 3(x+2) to get the same answer as (x+2)(5x+7) expanded?Post rating:1 
 Follow
 13
 03082016 08:27
I also had some problems with factorising the following expressions. Could you explain where I am going wrong:
Q: 1.
2.
* > isn't being used as greater than, just showing what I do next.
I did this for 1 but I'm not sure if it's right:
Common factor: P so
2:
I know this is wrong because I've seen the answer, but I don't understand what I did wrong.
3. Q:
This is a question I answered correctly, but I want to know if my method for doing it is right.
A:
My concern here is the +1 in the 1st square bracket. Is it correct to use +1 or could I have gone without it? So if it was 7 in the question instead, it would have been a 1 in the square bracket right?
Sorry for how lengthy this is. I just want to make sure I have a good understanding of it. 
 Follow
 14
 03082016 09:43
(Original post by ChrolloLucilfer)
...
(2) You're right but you can also factor out the 2 for completeness.
(3) Why did you use the +1 and where did the y come from? Just do:
If it was:
Last edited by RDKGames; 03082016 at 09:54.Post rating:2 
 Follow
 15
 03082016 10:19
(Original post by RDKGames)
(1) Just use \rightarrow to show progression anyway, you're correct
(2) You're right but you can also factor out the 2 for completeness.
(3) Why did you use the +1 and where did the y come from? Just do:
If it was:
(3) The y was just a typo.the +1 from this:
Answering: I see 7 and 14 and see 7 as common factor. So then I divide 7 & 14 by the 7 so I get this:
I put the 1 in front of the plus so I know I'm multiplying by +1. I realise now that it's unnecessary and only need to put an operator in if it's a negative number. E.g. if it was 7. It would be 
 Follow
 16
 03082016 10:26
(Original post by ChrolloLucilfer)
Thanks. I'll use the right arrow next time, didn't know there was an action like that.Can you show the working for (2) so I get the idea.
(3) The y was just a typo.the +1 from this:
Answering: I see 7 and 14 and see 7 as common factor. So then I divide 7 & 14 by the 7 so I get this:
I put the 1 in front of the plus so I know I'm multiplying by +1. I realise now that it's unnecessary and only need to put an operator in if it's a negative number. E.g. if it was 7. It would be
The +1 is not necessary as you said but if it helps you with the signs within the overall expression, then go ahead and use it. And you are correct for (3) though at the very end there, you can notice that you can factor out the 1 because 
 Follow
 17
 03082016 15:06
(Original post by RDKGames)
Working for (2)? No need. Simply look at your right bracket and you can notice that you can factor out a 2 from it which makes the whole thing
The +1 is not necessary as you said but if it helps you with the signs within the overall expression, then go ahead and use it. And you are correct for (3) though at the very end there, you can notice that you can factor out the 1 because
I have another factorising question. Q:
I answered: Which is supposedly wrong, even though I get the same expression as the question when I expand. So what am I doing wrong here? the answer is (from the answer sheet) 
 Follow
 18
 03082016 15:15
(Original post by ChrolloLucilfer)
Thank you
I have another factorising question. Q:
I answered: Which is supposedly wrong, even though I get the same expression as the question when I expand. So what am I doing wrong here? the answer is (from the answer sheet) 
 Follow
 19
 08082016 18:54
(Original post by RDKGames)
15 is not a factor of 35, you must've confused it with 45 hence the 3 is incorrect.
Could you help me with linear equation I've been having?
I used the lcm between 7&5 (35) then multiplied all the terms. Got rid of the 7 in first term, then was left with tried to multiply all terms by 2x, but then I end up unable to find x. Could you explain please? 
 Follow
 20
 08082016 18:58
(Original post by ChrolloLucilfer)
Thank you.
Could you help me with linear equation I've been having?
I used the lcm between 7&5 (35) then multiplied all the terms. Got rid of the 7 in first term, then was left with tried to multiply all terms by 2x, but then I end up unable to find x. Could you explain please?
Write a reply…
Reply
Submit reply
Register
Thanks for posting! You just need to create an account in order to submit the post Already a member? Sign in
Oops, something wasn't right
please check the following:
Sign in
Not got an account? Sign up now
Updated: August 12, 2016
Share this discussion:
Tweet
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.