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# Factorising by taking out common factors

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1. I'm stuck. Help would be appreciated.

This is an example from a book:
Q:Factorise 5(x+2)^2 -3(x+2)
A: (x+2)(5x+7)
I expanded the question and got: 5x^2 +17x+14
I expanded the answer and got 5x^2 +14-3x

I'm assuming they should both be equal so I'm guessing I did something wrong when I expanded the question?
2. 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.
3. (Original post by Chrollo-Lucilfer)
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 +14-3x

I'm assuming they should both be equal so I'm guessing I did something wrong when I expanded the question?
Right 5(x+2)^2 - 3(x+2)
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+10-3= 5x+7 so the answer is (x+2)(5x+7)
4. You don't have to expand the equation first.
Notice how in you have the common factor .
Factored out, you get .
5. (Original post by Chrollo-Lucilfer)
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 +14-3x

I'm assuming they should both be equal so I'm guessing I did something wrong when I expanded the question?
In the question, whats the common term?
Hint: (x+2)^2=(x+2)(x+2)
Posted from TSR Mobile
6. (Original post by Chrollo-Lucilfer)
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 +14-3x

I'm assuming they should both be equal so I'm guessing I did something wrong when I expanded the question?
You have expanded the question part properly, but the expansion you did for the answer is wrong.You should get 5x^2 +10x (2 multiplied by the 5x) +7x (7 multiplied by a single X) +14

Hope this helps
7. I'm glad everyone agrees with me.
8. 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) + (10-3)(x+2)

5x(x+2) + 7(x+2)

(5x + 7)(x + 2)

EDIT: never mind, the above answers are so much better
9. (Original post by B_9710)
I'm glad everyone agrees with me.
We have yet to match your typing speed.
10. (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 bit-by-bit 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.
11. (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.
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?
12. (Original post by Chrollo-Lucilfer)
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?
Spoiler:
Show

13. (Original post by MartyO)
Spoiler:
Show

Thanks, that makes sense.

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.
14. (Original post by Chrollo-Lucilfer)
...
(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:
15. (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:
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
16. (Original post by Chrollo-Lucilfer)
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
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
17. (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
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)
18. (Original post by Chrollo-Lucilfer)
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)
15 is not a factor of 35, you must've confused it with 45 hence the 3 is incorrect.
19. (Original post by RDKGames)
15 is not a factor of 35, you must've confused it with 45 hence the 3 is incorrect.
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?
20. (Original post by Chrollo-Lucilfer)
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?
That's because the equation is wrong. It's not linear. I think you meant otherwise as the x's cancel out.

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