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    How do you do Q2 on JAN 11 CORE 1 OCR?

    http://www.ocr.org.uk/Images/65379-q...hematics-1.pdf

    Please explain in detail, I've look at previous explanations and I don't get it!
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    Well I do not know which explanations you have not understood. However since it is an identity (true for all x).
    Substitute in x=0 to get 5p=2q and also some other value of x to get another equation.

    If you factorise both sides fully you find a common factor of (x+2) on both sides which cancels to give (x-p)(2x+5)=(x-2)(2x+q) which makes life easier.
    Once you sub in x=2, the RHS is zero and the LHS is fairly easy.

    Next use x=-5/2 which makes the LHS zero and the RHS fairly easy

    The key point is knowing what an identity is. You can put in any value of x you like and choosing a sensible value will give you a relatively straightforward thing to do.

    See repeat Q answer for comment about mark scheme
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    Here is a detailed look at a simple version of what the mark scheme wants.
    First factorise both sides to give
    (x-p)(2x+5)(x+2)=(x-2)(x+2)(2x+q)
    We can divide out the (x+2) to get
    (x-p)(2x+5)=(x-2)(2x+q) (if you don't get this divide out of (x+2) life is just harder but it will work out.
    and multiply both sides out to get
    2x^2+5x-2px-5p=2x^2+qx-4x-2q\\\;\Rightarrow\;2x^2+(5-2p)x-5p=2x^2+(q-4)x-2q

    Compare coefficients of x^1 to get 5-2p=q-4\;\Rightarrow\;9-2p=q
    Compare coefficients of x^0 to get 5p=2q
    Hence we can solve the pair of SEs to get p=2 and q=5

    If you miss dividing out the (x+2) then you get a cubic on each side but it still works out.
 
 
 
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