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Maths Edexcel C1 "Simplifying sums and differences of fractional powers" watch

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    I'm trying to understand the example in the photo which is from page 8 of "Modular Maths for Edexcel 2nd Edition Core Maths 1 & 2" - Sykes.


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    I get Example 1.6 but 1.7 baffles me.

    From the first line of the solution starting "The expression is ...", If I take out the trailing

    (x + 7)^1/2

    I can't see how
    3(x + 7)^1/2 - 2/3(x + 7)

    can become

    {3 - 2/3(x - 7)} (x+7)^1/2
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    Well, there is a common factor and also note that (x+7)^3/2=(x+7)times (x+7)^1/2 and so (x+7)^1/2 can also be a common factor.

    Edit, there is also typo which does not help where (x-7) is written instead of (x+7) in line 3
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    (Original post by SoCentral2)
    I'm trying to understand the example in the photo which is from page 8 of "Modular Maths for Edexcel 2nd Edition Core Maths 1 & 2" - Sykes.


    Name:  IMG_0962.jpg
Views: 88
Size:  392.7 KB

    I get Example 1.6 but 1.7 baffles me.

    From the first line of the solution starting "The expression is ...", If I take out the trailing

    (x + 7)^1/2

    I can't see how
    3(x + 7)^1/2 - 2/3(x + 7)

    can become

    {3 - 2/3(x - 7)} (x+7)^1/2
    What you've written in your post isn't complete.

    You get to 3(x + 7)^1/2 - 2/3(x + 7)(x+7)^1/2, then the two expressions have a common factor.

    If you can't see that, try substituting in a=(x+7) and rewrite the expression above and see if you can factorise it.
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    As an additional point. If I was taking out that (x+7)^1/2 as a common factor I would bring it out to the left (for clarity on line 3). Even if later on I moved it to the right.
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    (Original post by nerak99)
    Well, there is a common factor and also note that (x+7)^3/2=(x+7)times (x+7)^1/2 and so (x+7)^1/2 can also be a common factor.

    Edit, there is also typo which does not help where (x-7) is written instead of (x+7) in line 3
    Lovely! Yes, a mate came over and put me right on the typo.
 
 
 
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