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Math Help - polynomial

  1. #1
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    polynomial

    Find a possible formula for the polynomial with these properties: f is the third degree with f(-3)=0, f(1)=0, f(4)=0, f(0)=4.
    f(x)=?

    *I don't even know where to start. Please help, step by step if possible.
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  2. #2
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    f is the third degree with f(-3)=0, f(1)=0, f(4)=0, f(0)=4.
    f is the third degree, in general case we have

    f(x)=ax^3+bx^2+cx+d

    f(-3)=0
    f(-3)=a(-3)^3+b(-3)^2+c(-3)+d
    f(-3)=-27a+9b-3c+d, then
    -27a+9b-3c+d=0 (I)

    f(1)=0
    f(1)=a*1^3+b*1^2+c*1+d
    f(1)=a+b+c+d, then
    a+b+c+d=0 (II)

    f(4)=0
    f(4)=a*4^3+b*4^2+c*4+d
    f(4)=64a+16b+4c+d, then
    64a+16b+4c+d=0 (III)

    f(0)=4
    f(0)=a*0^3+b*0^2+c*0+d
    f(0)=d, then
    d=4 (IV)

    Form (I), (II), (III), (IV) we need to solve the system
    -27a+9b-3c+d=0
    a+b+c+d=0
    64a+16b+4c+d=0
    d=4
    and find a, b, c
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  3. #3
    Member realintegerz's Avatar
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    The function f, which is actually called f(x) has 3 roots

    so when it says f(-3) = 0

    that means at x = -3, there is no y value, so that is one point

    the same goes for 1 and 4, those are also "zeroes" which are places where the function, f, touches the x-axis

    and as for f(0) = 4, another point is at (0,4)



    Now, from these zeroes, -3, 1, and 4, we can get the equation which is

    (x+3)(x-1)(x-4)

    And so when you multiply it out you get an x^3, which means it is a third degree polynomial, always remember that the highest exponent tells you how many roots/"zeroes" there are and what degree the equation is

    I can't explain why it is, but if anyone here can, please feel free to do so...
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  4. #4
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    After solving the system you have:
    a=1/3
    b=-2/3
    c=-11/3
    Then you have the next polynomial
    f(x)=1/3 x^3 - 2/3 x^2 - 11/3 x + 4
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  5. #5
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    Quote Originally Posted by realintegerz View Post
    The function f, which is actually called f(x) has 3 roots

    so when it says f(-3) = 0

    that means at x = -3, there is no y value, so that is one point

    the same goes for 1 and 4, those are also "zeroes" which are places where the function, f, touches the x-axis

    and as for f(0) = 4, another point is at (0,4)



    Now, from these zeroes, -3, 1, and 4, we can get the equation which is

    (x+3)(x-1)(x-4)

    And so when you multiply it out you get an x^3, which means it is a third degree polynomial, always remember that the highest exponent tells you how many roots/"zeroes" there are and what degree the equation is

    I can't explain why it is, but if anyone here can, please feel free to do so...
    I think you have a mistake:
    (x+3)(x-1)(x-4)=x^3-2x^2-11x+12
    In this case you dont have f(0)=4 , but f(0)=12
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  6. #6
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    Math_helper's method is entirely correct, but realintegerz is very very close to a much easier way to do this.

    An alternative form of the cubic is A(x+B)(x+C)(X+D). It isn't quite as general as the one math_helper posted but is perfectly sufficient for this question.
    f(-3) = 0 \implies A(-3+B)(-3+C)(-3+D)=0 \implies one of the factors is 0, using the null factor theorem. Without loss of generality we can say that this factor is -3-B = 0, giving B = 3.
    Repeating this for the remaining zeroes gives C=-1 and D = -4. We then solve for A in the usual manner.

    When you are using this to solve a problem simply write down the factors as realintegerz did, but don't forget the constant A

    f(x) = A(x--3)(x-1)(x-4)
    =A(x+3)(x-1)(x-4)
    Then use the
    remaining peice of information f(0) = 4 to find A
    f(0) =4
    A3(-1)(-4) = 4
    A = 1/3

    So the solution, which is equivalent to that found by math_helper is
    f(x) = (1/3)(x+3)(x-1)(x-4)
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