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Math Help - function is defined by folliowing rule question.

  1. #1
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    function is defined by folliowing rule question.

    a function is defiend by the following rule:

    f (x) is
    0 if x lesser than/equal to -3
    -1 if -3<x<0
    x if x is greater then or equal to 0



    find i) F (-3) + f (-2) + f (2)
    ii) f (a ^2)

    another question:
    4 ^ (x+1)/ ( 4 ^(x + 2) – 4 ^ (x) )

    and another.

    x-1 is a factor of x ^ 3 - (2 X ^ 2) + px + 6
    find the value of P

    and another
    fully factorise the polynomial P(X) = X ^4 - 7x^3 + 12x^2 + 4x -16
    given that x = 2 is a double root of P(x)
    Last edited by thcbender; December 8th 2009 at 09:44 PM.
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  2. #2
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    Quote Originally Posted by thcbender View Post
    a function is defiend by the following rule:

    f (x) is
    0 if x lesser than/equal to -3
    -1 if -3<x<0
    x if x is greater then or equal to 0



    find i) F (-3) + f (-2) + f (2)
    ii) f (a ^2)
    (i) f(-3)+f(-2)+f(2)=0+(-1)+2=1
    (ii) Since a^{2}\geq0, so f(a^{2})=a^{2}


    Quote Originally Posted by thcbender View Post
    another question:
    4 ^ (x+1)/ ( 4 ^(x + 2) 4 ^ (x) )
    \frac{4^{x+1}}{4^{x+2}-4^{x}}=\frac{4(4^{x})}{4^{2}4^{x}-4^{x}}
    =\frac{4(4^{x})}{4^{x}(4^{2}-1)}=\frac{4}{4^{2}-1}=...

    Quote Originally Posted by thcbender View Post

    x-1 is a factor of x ^ 3 - (2 X ^ 2) + px + 6
    find the value of P
    Since (x-1) is a factor, by factor theorem, f(1)=0, where f(x)=x^{3}-2x^{2}+px+6
    So 1^{3}-2(1)^{2}+p(1)+6=0...solve it you will get your p

    Quote Originally Posted by thcbender View Post
    fully factorise the polynomial P(X) = X ^4 - 7x^3 + 12x^2 + 4x -16
    given that x = 2 is a double root of P(x)
    Since x=2 is a double root, so (x-2)^{2} is a factor, get the other quadratic factor by either long division, comparison or synthetic division. You will get f(x)=(x-2)^{2}(x^{2}-3x-4)=(x-2)^{2}(x-4)(x+1)
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