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    I really can't work out this question, could anyone help me?

    f(x) = x^3 + ax^2 + bx + 6

    When f(x) is divided by (x-1) the remainder is 24,
    When f(x) is divided by (x+2) the remainder is also 24.

    Use the Remadiner theorm to fine the numberical values of a and b.



    I just don't understand how to work it out =( Thanks everyone.
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    so f(x) leaves a remainder of 24
    if only it was 24 smaller...
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    (Original post by RichardF)
    I really can't work out this question, could anyone help me?

    f(x) = x^3 + ax^2 + bx + 6

    When f(x) is divided by (x-1) the remainder is 24,
    When f(x) is divided by (x+2) the remainder is also 24.

    Use the Remadiner theorm to fine the numberical values of a and b.



    I just don't understand how to work it out =( Thanks everyone.
    f(1)=24
    f(-2)=24
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    (1)^3 + a(1)^2 + b(1) + 6 = 24 \Rightarrow a+b = 17

    (-2)^3 + a(-2)^2 + b(-2) + 6 = 24 \Rightarrow 4a-2b = 26

    ...... Just finish off to get a and b.
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    Thanks so much all!! i got it now =)
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    (Original post by Quadrifolium)
    (1)^3 + a(1)^2 + b(1) + 6 = 24 \Rightarrow a+b = 17

    (-2)^3 + a(-2)^2 + b(-2) + 6 = 24 \Rightarrow 4a-2b = 26

    ...... Just finish off to get a and b.
    Exactly this.

    Out of Interest what does your sig represent??
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    (Original post by JamesyB)
    Out of Interest what does your sig represent??
    \displaystyle\frac{13}{8}+\sum_{  n=0}^{\infty}\frac{\left(-1\right)^{\left(n+1\right)}(2n+1  )!}{(n+2)!n!4^{(2n+3)}} = \dfrac{1+\sqrt{5}}{2} = \phi.

    It's an infinite series for \phi. See here and here.
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    (Original post by Quadrifolium)
    \displaystyle\frac{13}{8}+\sum_{  n=0}^{\infty}\frac{\left(-1\right)^{\left(n+1\right)}(2n+1  )!}{(n+2)!n!4^{(2n+3)}} = \dfrac{1+\sqrt{5}}{2} = \phi.

    It's an infinite series for \phi. See here and here.
    Very Nice especially the infinity + 1
 
 
 
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