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    http://www.ocr.org.uk/Images/62241-q...hematics-3.pdf
    For 8iii of the paper OCR FP3 paper above, in this examiners report: http://www.ocr.org.uk/Images/61726-e...ports-june.pdf
    It says on page 26 that 'it was very worrying that 1 was oftenretained on the right hand side and statements' but I think this gives the correct answer so what's wrong?
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    What have you done?
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    (Original post by B_9710)
    What have you done?
    Sorry heres the mark scheme http://www.ocr.org.uk/Images/60318-m...ics-3-june.pdf
    So basically for 8ii as can be seen in the mark scheme you begin with a factorised expression. Why can't you make that expression equal to ne like in the examiners report for 8iii( i would copy and paste but it becomes messed up). I say this becuase you end up with 2cos(squared theta)-1=1 form which you get that cos(theta)=1 or-1 which is the answer in the mark scheme
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    What you have is  \cos 4\theta \cos 2\theta = 1 .
     \displaystyle \Rightarrow 16\cos ^6\theta -24\cos^4 \theta +10\cos^2 \theta -1=1 .So you get
     \displaystyle 16c^6-24c^4+10c^2-2=0 , where  c=\cos \theta .
    Is this what you did?
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    What the examiner is trying to get at is that candidates tried to factorise without first ensuring that the RHS was zero. For the most part, when solving equations, factorising is only useful if the RHS is zero.

    If candidates at FP3 do not realise that the RHS has to be zero when factorising is your solution strategy, this is worrying.

    At GCSE/C1 you might often see ( foo )( bar )=1 >>> either foo = 1 or bar = 1 but if it happens at FP3 you wonder how?
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    (Original post by nerak99)
    What the examiner is trying to get at is that candidates tried to factorise without first ensuring that the RHS was zero. For the most part, when solving equations, factorising is only useful if the RHS is zero.

    If candidates at FP3 do not realise that the RHS has to be zero when factorising is your solution strategy, this is worrying.

    At GCSE/C1 you might often see ( foo )( bar )=1 >>> either foo = 1 or bar = 1 but if it happens at FP3 you wonder how?
    thanks for the reply. why does the right have to equal zero where factorising because 0 is just a number and you are trying where the graph crosses the line y=0 but if u can rearrange it in the another way the curve changes (but the equation is the same overall) you can find the solutions when y=c which is equivalent to finding when y=0 when you have the original equation? I hope it this clear
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    I think that the following is a gap in your understanding.

    A multiplication sum with the result zero can only happen if one or more of the factors is zero. You use this fact to find the roots. Remember that factorise means 'make into a multiply' not 'make into brackets', although in polynomial factorisation this factorisation results in the form of brackets.

    If a multiply happens to result in (say) 1, well you can nothing about the factors. They could be any one of an infinite set of multiplies that happend to result in 1 e.g. 0.000001 X 1000000 or 0.5 X 2 or -i X i ...
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    (Original post by nerak99)
    I think that the following is a gap in your understanding.

    A multiplication sum with the result zero can only happen if one or more of the factors is zero. You use this fact to find the roots. Remember that factorise means 'make into a multiply' not 'make into brackets', although in polynomial factorisation this factorisation results in the form of brackets.

    If a multiply happens to result in (say) 1, well you can nothing about the factors. They could be any one of an infinite set of multiplies that happend to result in 1 e.g. 0.000001 X 1000000 or 0.5 X 2 or -i X i ...
    thank you i get it know.
 
 
 
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