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How to Simplify this Expression?

Hi,

So to factorise this:

24(a+b) b + 153 (a+b)^2

What would be the first step?

Find the HCF for 24 and 153?
(edited 7 years ago)
Reply 1
Avatar for Notnek
Notnek
21
Original post by Exceptional
Hi,

So to factorise this:

24(a+b) b + 153 + (a+b)^2

What would be the first step?

Find the HCF for 24 and 153?

Are you sure you posted the expression correctly?

Check your plus signs.
Original post by notnek
Are you sure you posted the expression correctly?

Check your plus signs.


Amended. Thank you.
Original post by Exceptional
Hi,

So to factorise this:

24(a+b) b + 153 (a+b)^2

What would be the first step?

Find the HCF for 24 and 153?


Well the first step would be to take out any common factors you see that the two terms share.
Original post by RDKGames
Well the first step would be to take out any common factors you see that the two terms share.


3 for 24 and 153. And the a's and b's. Would I need to expand the brackets first before I can factorise the whole thing?
Original post by Exceptional
3 for 24 and 153. And the a's and b's. Would I need to expand the brackets first before I can factorise the whole thing?


Expanding is the opposite of factorising, so no you do not need to expand the brackets. If you treat the bracket (a+b) as whatever number you want, you can see that you can easily factor it out of the whole expression without expansion.
Original post by RDKGames
Expanding is the opposite of factorising, so no you do not need to expand the brackets. If you treat the bracket (a+b) as whatever number you want, you can see that you can easily factor it out of the whole expression without expansion.


Hmm, I've been told a few contradictory methods to solve it. Someone has said that it's necessary to expand and simplify it before factorising it.
Original post by Exceptional
Hmm, I've been told a few contradictory methods to solve it. Someone has said that it's necessary to expand and simplify it before factorising it.


Most likely the reason they say that is that they don't fully appreciate that (a+b) is a common factor.
Hence write it as:
24b(a+b) + 153(a+b)(a+b)
= (a+b)(24b + 153(a+b))
= 3(a+b)(8b + 51(a+b))
= 3(a+b)(51a + 59b).
Original post by HapaxOromenon3
Most likely the reason they say that is that they don't fully appreciate that (a+b) is a common factor.
Hence write it as:
24b(a+b) + 153(a+b)(a+b)
= (a+b)(24b + 153(a+b))
= 3(a+b)(8b + 51(a+b))
= 3(a+b)(51a + 59b).


Thank you!

But, I just don't understand the thought process behind it.

How did you know to write '24b(a+b)' from '24(a+b)b'?

Where did '(a+b)' go in 24b(a+b) + 153(a+b)(a+b)' when you got to the next step, '3(a+b)(8b+51(a+b))'?

Lastly, where did you get 59b from?

Basically, I need an explanation because I don't understand how to do this when faced with new problems.
Original post by Exceptional
Thank you!

But, I just don't understand the thought process behind it.

How did you know to write '24b(a+b)' from '24(a+b)b'?

Where did '(a+b)' go in 24b(a+b) + 153(a+b)(a+b)' when you got to the next step, '3(a+b)(8b+51(a+b))'?

Lastly, where did you get 59b from?

Basically, I need an explanation because I don't understand how to do this when faced with new problems.


24*b*(a+b) = 24*(a+b)*b as multiplication is commutative (can be done in any order)
Notice that 3(a+b)(8b+51(a+b)) = 3(a+b)(8b) + 3(a+b)(51(a+b)) = 3(8b)(a+b) + 3(51)(a+b)(a+b) etc.
The key idea is that (a+b)^2 = (a+b)(a+b).
8b + 51(a+b) = 8b + 51a + 51b = 51a + 59b.
All of this is basic algebra.
Reply 10
Avatar for Notnek
Notnek
21
Original post by Exceptional
Hi,

So to factorise this:

24(a+b) b + 153 (a+b)^2

What would be the first step?

Find the HCF for 24 and 153?

I think you should try something a bit simpler.

Factorise:

24xb+153x224xb + 153x^2

Let us know if you can do this and post your answer.
Original post by notnek
I think you should try something a bit simpler.

Factorise:

24xb+153x224xb + 153x^2

Let us know if you can do this and post your answer.


3x(8b + 51x)?
Original post by HapaxOromenon3
24*b*(a+b) = 24*(a+b)*b as multiplication is commutative (can be done in any order)
Notice that 3(a+b)(8b+51(a+b)) = 3(a+b)(8b) + 3(a+b)(51(a+b)) = 3(8b)(a+b) + 3(51)(a+b)(a+b) etc.
The key idea is that (a+b)^2 = (a+b)(a+b).
8b + 51(a+b) = 8b + 51a + 51b = 51a + 59b.
All of this is basic algebra.


Perhaps, but I haven't done it for a few years and don't remember the steps. I've just been given various questions and asked to solve them without an explanation.
Reply 13
Avatar for Notnek
Notnek
21
Original post by Exceptional
3x(8b + 51x)?

That's correct.

Now compare the question you just did with the original question:

24xb+153x224xb + 153x^2

24(a+b)b+153(a+b)224(a+b) b + 153 (a+b)^2

Notice that the difference is that xx is now (a+b)(a+b). But you can still take out (a+b)(a+b) as a factor just like you took out xx.

Try your original question again with this in mind and see how you go.

A lot of people have trouble with kind of question - you're not the only one.
Original post by notnek
That's correct.

Now compare the question you just did with the original question:

24xb+153x224xb + 153x^2

24(a+b)b+153(a+b)224(a+b) b + 153 (a+b)^2

Notice that the difference is that xx is now (a+b)(a+b). But you can still take out (a+b)(a+b) as a factor just like you took out xx.

Try your original question again with this in mind and see how you go.

A lot of people have trouble with kind of question - you're not the only one.


It's making more sense, but then why isn't the answer 3(a+b)(8b + 51a)? That way you'd get 24 and 153 like in the original question if you tried to expand.
Reply 15
Avatar for Notnek
Notnek
21
Original post by Exceptional
It's making more sense, but then why isn't the answer 3(a+b)(8b + 51a)? That way you'd get 24 and 153 like in the original question if you tried to expand.

24xb+153x2=3x(8b+51x)\displaystyle 24\boldsymbol{x}b + 153\boldsymbol{x}^2 = 3\boldsymbol{x}\left(8b+51 \boldsymbol{x} \right)

24(a+b)b+153(a+b)2=3(a+b)(8b+51(a+b))24\boldsymbol{(a+b)}b+153 \boldsymbol{(a+b)}^2 = 3\boldsymbol{(a+b)} \left(8b + 51\boldsymbol{(a+b)} \right)

Both have been factorised in exactly the same way. The only difference is that the first one uses xx and the second uses (a+b)(a+b).
Original post by notnek
24xb+153x2=3x(8b+51x)\displaystyle 24\boldsymbol{x}b + 153\boldsymbol{x}^2 = 3\boldsymbol{x}\left(8b+51 \boldsymbol{x} \right)

24(a+b)b+153(a+b)2=3(a+b)(8b+51(a+b))24\boldsymbol{(a+b)}b+153 \boldsymbol{(a+b)}^2 = 3\boldsymbol{(a+b)} \left(8b + 51\boldsymbol{(a+b)} \right)

Both have been factorised in exactly the same way. The only difference is that the first one uses xx and the second uses (a+b)(a+b).


Okay, I can see that. Thank you.

So, is that an acceptable answer? Because others have said the answer is '3(a+b)(51a + 59b)'.
Original post by Exceptional
Okay, I can see that. Thank you.

So, is that an acceptable answer? Because others have said the answer is '3(a+b)(51a + 59b)'.


Both forms are equivalent. The 51a+59b one has just distributed the 51 between the (a+b) term.
Reply 18
Avatar for Notnek
Notnek
21
Original post by Exceptional
Okay, I can see that. Thank you.

So, is that an acceptable answer? Because others have said the answer is '3(a+b)(51a + 59b)'.

You should simplify it:

8b+51(a+b)=8b+51a+51b=51a+59b8b+51(a+b) = 8b + 51a + 51b = 51a + 59b

That;'s where the 51a+59b51a + 59b comes from.

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