Polynomial

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Uchiha Sasuke
25/05/2018 at 08:52
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Suppose that the polynomial \(f\left(x\right)=x^5-x^4-4x^3+2x^2+4x+1\) has 5 solutions \(x_1;x_2;x_3;x_4;x_5\).The other polynomial \(K\left(x\right)=x^2-4\). Find the value of P=K(\(x_1\))×K(\(x_2\))×K(\(x_3\))×K(\(x_4\))×K(\(x_5\))

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Uchiha Sasuke
25/05/2018 at 08:46
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Suppose that the polynomial \(f\left(x\right)=2x^5-9x^3+2x^2+9x-3\) has 5 solutions \(x_1;x_2;x_3;x_4;x_5\). The other polynomial \(K\left(x\right)=x^2-4\). Find the value of \(P=K\left(X_1\right)\times K\left(X_2\right)\times K\left(X_3\right)\times K\left(X_4\right)\times K\left(X_5\right)\)

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Uchiha Sasuke
25/05/2018 at 08:39
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Let P(x) be the polynomial given by \(P\left(x\right)=\left(2+x+2x^3\right)^{15}\). Suppose that \(P\left(x\right)=a_0+a_1x+a_2x^2+...+a_{45}x^{45}\).The value of \(S=a_1-a_2+a_3-a_4+...-a_{44}+a_{45}\).
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    Uchiha Sasuke 26/05/2018 at 01:19

    Thank you very much yeu

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    Dao Trong Luan Coodinator 25/05/2018 at 12:30

    P(x) = a0 + a1x + a2x2 + ... + a45x45

    => P(1) = a0 + a1 + a2 + ... + a45 = (2 + 1 + 2)15 = 515

    => P(-1) = a0 - a1 + a2 - ... - a45 = (2 - 1 - 2)15 = (-1)15

    ==> P(1) - P(-1) = 2a1 + 2a3 + ... + 2a45 = 515 + 1

            P(1) + P(-1) = 2a0 + 2a2 + ... + 2a44 = 515 - 1

    => (a1 + a3 + ... + a45) - (a0 + a2 + ... + a44) = \(\dfrac{5^{15}+1-5^{15}+1}{2}=1\)

    => S = -a0 + a1 - a2 + .... - a44 + a45 = 1

    P/s: Not S in the topic banh


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Uchiha Sasuke
26/01/2018 at 07:48
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Given a third degree polynomial P (x). Find the coefficient of x3 of P (x) such that P(0) = 10; P(1) = 12; P(2) = 4; P(3) = 1 

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tranthuydung
09/03/2017 at 18:52
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Let P(x) be a polynomial of degree 2015.

Suppose P(n)=\(\dfrac{n}{n+1}\) for all n = 0, 1 , 2 ,..., 2015.

The value of P(2016) is ...
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    John 11/03/2017 at 09:07

    Let \(Q\left(x\right)=\left(x+1\right)P\left(x\right)-x\),  (*)

    since \(P\left(n\right)=\dfrac{n}{n+1}\) we infer:

     \(Q\left(n\right)=\left(n+1\right)P\left(n\right)-n=0\) for \(n=0,1,..,2015\).

    Because P is a polynomial of dgree 2016, Q is a polynomial of dgree 2016. Q has 2016 solutions (0, 1, .. , 2015) so Q can be expressed in a form:

      \(Q\left(x\right)=a\left(x-0\right)\left(x-1\right)...\left(x-2015\right)\)   (**)

    In other hand, in (*) we set x = -1 then  \(Q\left(-1\right)=1\). And replace to (**) we have:

      \(1=Q\left(-1\right)=a\left(-1\right)\left(-2\right)...\left(-2016\right)\)

     \(\Rightarrow a=\dfrac{1}{2016!}\)

    Finally \(Q\left(x\right)=\dfrac{1}{2016!}\left(x-0\right)\left(x-1\right)...\left(x-2015\right)\)

    \(\Rightarrow Q\left(2016\right)=\dfrac{1}{2016!}2016.2015...1=1\)

    from (*) \(Q\left(2016\right)=\left(2016+1\right)P\left(2016\right)-2016\)

    \(\Rightarrow\left(2016+1\right)P\left(2016\right)-2016=1\)

    \(\Rightarrow P\left(2016\right)=\dfrac{2017}{2017}=1\)

    tranthuydung selected this answer.
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    FA KAKALOTS 28/01/2018 at 22:05

    Let Q(x)=(x+1)P(x)−x

    ,  (*)

    since P(n)=nn+1

     we infer:

     Q(n)=(n+1)P(n)−n=0

     for n=0,1,..,2015

    .

    Because P is a polynomial of dgree 2016, Q is a polynomial of dgree 2016. Q has 2016 solutions (0, 1, .. , 2015) so Q can be expressed in a form:

      Q(x)=a(x−0)(x−1)...(x−2015)

       (**)

    In other hand, in (*) we set x = -1 then  Q(−1)=1

    . And replace to (**) we have:

      1=Q(−1)=a(−1)(−2)...(−2016)

     ⇒a=12016!

    Finally Q(x)=12016!(x−0)(x−1)...(x−2015)

    ⇒Q(2016)=12016!2016.2015...1=1

    from (*) Q(2016)=(2016+1)P(2016)−2016

    ⇒(2016+1)P(2016)−2016=1

    ⇒P(2016)=20172017=1