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Math Help - Multiplying out brackets containing multiplication

  1. #1
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    Post Multiplying out brackets containing multiplication

    Apologises in advance if i'm being a bit thick, or this has been answered before, my searches on this forum and google were unsuccessful.
    My problem is that I'm trying to find the rule for expanding brackets which contain multiplied variables. I've found hundreds of examples with addition and subtraction, but no multiplication...which I thought was a bit odd.

    Examplea*b)(x*y)

    How would this be expanded to remove the brackets? I have a feeling theres a few conditions such as like bases. My specific problem involves terms that are constants raised to a variable power (plus or minus a constant).

    Many, many thanks in advance!
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  2. #2
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    Quote Originally Posted by NitrousUK View Post
    Apologises in advance if i'm being a bit thick, or this has been answered before, my searches on this forum and google were unsuccessful.
    My problem is that I'm trying to find the rule for expanding brackets which contain multiplied variables. I've found hundreds of examples with addition and subtraction, but no multiplication...which I thought was a bit odd.

    Examplea*b)(x*y)

    How would this be expanded to remove the brackets? I have a feeling theres a few conditions such as like bases. My specific problem involves terms that are constants raised to a variable power (plus or minus a constant).

    Many, many thanks in advance!
    Hi NitrousUK,

    If I assume your little purple frown guy is hiding one of your parentheses, I am reading:

    (ab)(xy)

    All the bases are different here, so the product is simply:

    abxy

    These factors can be put in any order since multiplication is commutative.
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  3. #3
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    Thanks for the quick response!
    Sorry I wasnt clear with my example. It should of read (a^x a^y)(a^w a^z)
    The problem im dealing with involves the same bases but raised to a variable power.

    Thanks!
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  4. #4
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    Quote Originally Posted by NitrousUK View Post
    Thanks for the quick response!
    Sorry I wasnt clear with my example. It should of read \textcolor{red}{(a^x a^y)(a^w a^z)= (a^{x+y})(a^{w+z}) = a^{x+y+w+z}}
    The problem im dealing with involves the same bases but raised to a variable power.

    Thanks!
    ...
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  5. #5
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    Thanks!
    However I see I'm missing out crucial info in my example. I'll use the same example that i'm stuck on, it's more complicated than the original question but multiplication of different bases inside two brackets was where i stumbled: a^{n-1}b^{n+1}+b^{n-1}a^{n+1}-2(a^nb^n)=a^{n-1}b^{n-1}(b^2+a^2-2*a*b)

    I can't seem to see the steps/rules involved in getting from the first equation to the second.
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  6. #6
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    Quote Originally Posted by NitrousUK View Post
    Thanks!
    However I see I'm missing out crucial info in my example. I'll use the same example that i'm stuck on: a^{n-1}b^{n+1}+b^{n-1}a^{n+1}-2(a^nb^n)=a^{n-1}b^{n-1}(b^2+a^2-2*a*b)

    I can't seem to see the steps/rules involved in getting from the first equation to the second.
    exponent review ...

    note that a^{n-1} \cdot a^2 = a^{(n-1)+2} = a^{n+1}<br />

    also ...

    note that a^{n-1} \cdot a = a^{(n-1)+1} = a^n


    left side expression is a^{n-1}b^{n+1}+b^{n-1}a^{n+1}-2(a^nb^n)

    now note that each of the three terms on the left side have the following factors in common ...

    a^{n-1} and b^{n-1} ...

    factor these two out ...

    a^{n-1}b^{n-1}[b^2+a^2-2(ab)]



    next time start with the original problem.
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  7. #7
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    Thanks!! That's a massive help. My course book doesn't seem to cover the technique of factoring like that, so that's helped me crack this problem. I find the hardest thing in my math learning is finding the info for something I know the question/problem to, Google tends to turn up random hits for most words I try.
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