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    page 13 question 10
    Given that (x-1) and (x+1) are factors of px3+qx2-3x-7
    find the values of p and q
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      If (x-1) is a factor, therefore f(1)=0

      If (x+1) is a factore, therefore f(-1)=0

      Substitute in the values then solve simultaneously.

      EDIT: Wtf? Question 10? Don't you mean Question 1 as that really is quite a simply question with regards to the C2 syllabus.
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      (Original post by reb0xx)
      page 13 question 10
      Given that (x-1) and (x+1) are factors of px3+qx2-3x-7
      find the values of p and q
      Multiply out (x-1)(x+1) and third factor (found by factor theorem)
      Equate coefficients

      Edit: Factor theorem involves trial and error, but generally with C2 is either -3,-2,-1,0,1,2,3
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      (Original post by im so academic)
      If (x-1) is a factor, therefore f(1)=0

      If (x+1) is a factore, therefore f(-1)=0

      Substitute in the values then solve simultaneously.

      EDIT: Wtf? Question 10? Don't you mean Question 1 as that really is quite a simply question with regards to the C2 syllabus.
      Why shouldn't it be Q10?

      Edit: Could be a revision guide
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        (Original post by Freerider101)
        Multiply out (x-1)(x+1) and third factor (found by factor theorem)
        Equate coefficients

        Edit: Factor theorem involves trial and error, but generally with C2 is either -3,-2,-1,0,1,2,3
        This question doesn't require trial and error as we know that (x-1) and (x+1) are factors. Therefore via the factor theorem, +1 and -1 are factors of f(x).
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          (Original post by Freerider101)
          Why shouldn't it be Q10?

          Edit: Could be a revision guide
          I assumed it to be a past paper question.

          Usually the end questions for C2 are solving differential problems (with minima and maxima) or "complicated" co-ordinate geometry problems.
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          (Original post by im so academic)
          This question doesn't require trial and error as we know that (x-1) and (x+1) are factors. Therefore via the factor theorem, +1 and -1 are factors of f(x).
          Yeah but then you have two solve a cubic anyway
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            (Original post by Freerider101)
            Yeah but then you have two solve a cubic anyway
            So it's simultaneous equations. I know this as I've done many questions pretty much in the same way as the OP's question.

            Also, solving a cubic IS C2. :lolwut:
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            (Original post by im so academic)
            So it's simultaneous equations. I know this as I've done many questions pretty much in the same way as the OP's question.

            Also, solving a cubic IS C2. :lolwut:
            Yeah fairplay but thats what the OP was trying to do in the first place and couldn't do.
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              (Original post by Freerider101)
              Yeah fairplay but thats what the OP was trying to do in the first place and couldn't do.
              Really? He just posted the question and nothing else. He has said nothing of what technique he used.
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              (Original post by Freerider101)
              Multiply out (x-1)(x+1) and third factor (found by factor theorem)
              Equate coefficients

              Edit: Factor theorem involves trial and error, but generally with C2 is either -3,-2,-1,0,1,2,3
              I wouldn't do it that way: I'd do it with simultaneous equations as someone has already posted.
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              (Original post by reb0xx)
              page 13 question 10
              Given that (x-1) and (x+1) are factors of px3+qx2-3x-7
              find the values of p and q
              Another option to solve-
              (x-1) and (x+1) are factors, so (x-1)(x+1)=(x^2-1) is a factor.
              Now (ax-b)(x^2-1)=px^3+qx^2-3x-7
              It's easy to solve for a and b, then find p and q by multiplying out(or binomial terms).
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              what are you talking about.....
              its c2 page 13 exercise 1D
              if there is 2 factors which one should i use
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              (Original post by im so academic)
              This question doesn't require trial and error as we know that (x-1) and (x+1) are factors. Therefore via the factor theorem, +1 and -1 are factors of f(x).
              (Original post by Thrug)
              I wouldn't do it that way: I'd do it with simultaneous equations as someone has already posted.
              Ah I forgot to mention. If your calculator has a cubic solver it gives you the values of x. So you can just "happen" to get the third factor using the factor theorem "first time lucky". Its then not a case of trial and error. Then you just multiply out and equate coefficients.
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                (Original post by Freerider101)
                Ah I forgot to mention. If your calculator has a cubic solver it gives you the values of x. So you can just "happen" to get the third factor using the factor theorem "first time lucky". Its then not a case of trial and error. Then you just multiply out and equate coefficients.
                There are method marks however. I do not advise to use a calculator that does the working out for you.

                It is not a case of trial and error, we know the values of the factors of f(x) as it is pretty much given in the question.

                The "multiplying out and equating coefficients" is one possible way, however the easier and the more methodically way of doing it is via simultaneous equations. Even the Edexcel solution does it that way.
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                (Original post by reb0xx)
                what are you talking about.....
                its c2 page 13 exercise 1D
                if there is 2 factors which one should i use
                They are both factors. You have two unknowns, so you need two equations for this.

                And we have no idea what book you are even using, so page 13 exercise 1D is pretty irrelevant. Plus, you've posted the question, so we don't need to refer to it.
               
               
               
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