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Math Help - urgent.i have exam tmr.roots & remainder

  1. #1
    Member helloying's Avatar
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    urgent.i have exam tmr.roots & remainder

    the cubic polynomial f(x) is such that the coefficient of x^3 is 1 and the roots of f(x)=0 are 1, k , k^2 . it is given that f(x) has a remainder of 7 when divided by x-2.

    show that k^3 - 2k^2 - 2k - 3 =0.

    thanks teach me how to solve. thank you very much
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  2. #2
    MHF Contributor
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    Quote Originally Posted by helloying View Post
    the cubic polynomial f(x) is such that the coefficient of x^3 is 1 and the roots of f(x)=0 are 1, k , k^2 . it is given that f(x) has a remainder of 7 when divided by x-2.

    show that k^3 - 2k^2 - 2k - 3 =0.

    thanks teach me how to solve. thank you very much

    f(x)=ax^3+bx^2+cx+d .. a=1 so

     <br />
f(x)=x^3+bx^2+cx+d <br />

    f(x)=(x-1)(x-k)(x-k^2) .. expand this part

    =x^3-(k^2+k+1)x^2+(k^3+k^2+k)x-k^3

    Here you can compare

    -k^2-k-1=b --- 1

    k^3+k^2+k=c --- 2

    ----------------------------------------

    Now make use of the factor and remainder theorem

    You know one of the factors is (x-1) so

    f(1)=0

    1+b+c+d=0\Rightarrow b+c+d=-1 --- 3

    Then it will give a remainder of 7 when divided by (x-2)

    f(2)=7

    8+4b+2c+d=7\Rightarrow 4b+2c+d=-1 --- 4

    4-3  3b+c=0\Rightarrow c=-3b

    Now look back . From 2 above , k^3+k^2+k=-3b --6
    (Replace the c with -3b)

    Multiply 1 by 3 ... -3(-k^2-k-1)=-3(b) -- 5

    6-5 i will leave this for you .
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  3. #3
    Senior Member
    Joined
    Jul 2009
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    Hi helloying
    The factors are x = 1, x = k, and x = k^2, so :

    (x-1)(x-k)(x-k^2) = 0

    f(x) = (x-1)(x-k)(x-k^2)

    Subs. x = 2
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