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    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?
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    (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.
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    (Original post by notnek)
    Are you sure you posted the expression correctly?

    Check your plus signs.
    Amended. Thank you.
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    (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.
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    (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?
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    (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.
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    (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.
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    (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).
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    (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.
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    (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.
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    (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 + 153x^2

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

    Factorise:

    24xb + 153x^2

    Let us know if you can do this and post your answer.
    3x(8b + 51x)?
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    (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.
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    (Original post by Exceptional)
    3x(8b + 51x)?
    That's correct.

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

    24xb + 153x^2

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

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

    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.
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    (Original post by notnek)
    That's correct.

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

    24xb + 153x^2

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

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

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

    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 x and the second uses (a+b).
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    (Original post by notnek)
    \displaystyle 24\boldsymbol{x}b + 153\boldsymbol{x}^2 = 3\boldsymbol{x}\left(8b+51 \boldsymbol{x} \right)

    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 x and the second uses (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)'.
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    (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.
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    (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 + 59b

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