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    • RE: O-Level Additional Math

      KSP2013777:
      Anyone can help me?


      ] http://i58.tinypic.com/2ex8o4n.jpg\">

      Thanks

      posted in Secondary Schools - Academic Support
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      Panda168
    • RE: O-Level Additional Math

      KSP2013777:
      Anyone can help me?


      ] http://i58.tinypic.com/2ex8o4n.jpg\">

      Thanks
      a) Let the polynomial be f(x)

      f(x)=(x-2)Q(x) where Q(x) is the quotient when divided by(x-2)
      f(x)=(x+10)P(x)+12 where P(x) is the quotient when divided by (x+10)

      Therefore
      (x-2)Q(x)=(x+10)P(x)+12 -----(1)
      since both have the same polynomial.

      Sub x=2 into (1)
      0=12xP(x)+12

      Therefore P(x)=-1 and f(x)=2-x

      Hence the remainder when f(x) is divided by (x-2)(x+10) is
      R(x)=(2-x)/[(x-2)(x+10)]
      =-1/(x+10)

      posted in Secondary Schools - Academic Support
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      Panda168
    • RE: Lower Secondary Mathematics

      http://i58.tinypic.com/2j9w75.jpg\">

      posted in Secondary Schools - Academic Support
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      Panda168
    • RE: Lower Secondary Mathematics

      http://i59.tinypic.com/6z2649.jpg\"> http://i59.tinypic.com/6z2649.jpg\">

      posted in Secondary Schools - Academic Support
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      Panda168
    • RE: Lower Secondary Mathematics

      Answer

      posted in Secondary Schools - Academic Support
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