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    Okay, there off the C3 Specimen paper (link below)

    http://www.edexcel.org.uk/VirtualCon...r_MkScheme.pdf

    Q4a. I have got to the stage of: (x³+2x²-3)/ (x+3)(x-1) but cant see what they've done to get to the next step.

    Q5c. i've put y=0 to get e^0 = (2x+3)^3

    Hence: ³√1 = ³√(2x+3)

    so: ±1 = 2x+3

    Why have they taken the +ve root of 1 and not the -ve?

    Thanks
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    Q4a. You divide x³+2x²-3 by (x-1) by algebraic division to give (x²+3x+3)(x-1)/(x+3)(x-1)... The (x-1)'s then cancel et voila!
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    Q5c. I did it this way:

    y=0, so 3ln(2x+3) = 0 then divide both sides by 3 to give ln(2x+3) = 0
    then 2x+3 = e^0 so 2x+3 = 1 so x = -1 therefore p = -1
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    Why have they taken the +ve root of 1 and not the -ve?
    you don't get any negative roots from cube rooting i.e the cube root of 1 is 1
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    (Original post by !Laxy!)
    Q5c. i've put y=0 to get e^0 = (2x+3)^3

    Hence: ³√1 = ³√(2x+3)

    so: ±1 = 2x+3

    Why have they taken the +ve root of 1 and not the -ve?

    Thanks
    after you put y = 0, you got

    e^0 = (2x+3)^3
    or
    1 = (2x+3)^3

    .: (2x+3) must be +ve, since if it was -ve, then so also would be its cube!

    So, express (2x+3) as the (+ve) cube root of 1.

    Another reason is that you can't get a -ve (cube) root of 1!!
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    (Original post by violet)
    Q4a. You divide x³+2x²-3 by (x-1) by algebraic division to give (x²+3x+3)(x-1)/(x+3)(x-1)... The (x-1)'s then cancel et voila!

    Why do you divide by (x-1)???
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    Because otherwise you wouldn't know what the other factor of x³+2x²-3 was, so you wouldnt know what was left once you'd cancelled the (x-1)s.
 
 
 
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Updated: June 12, 2005
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