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Math Help - Find a polynomial p(t) of degree 6 which...

  1. #1
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    Find a polynomial p(t) of degree 6 which...

    Find a polynomial p(t) of degree 6 which has a zero of multiplicity 2 at t = 1 and a zero of multiplicity 3 at
    t = 2, and also satisfying: p(0) = 2 and p`(0) = 1. What is the other root of p(t)?

    Attempt at solution:

    zero of multiplicity 2 at t =1 implies (t-1)^2 is a factor or p(1) = 0 and p`(1) = 0
    zero of multiplicity 3 at t = 2 implies (t-2)^3 is a factor or p(2) = 0, p`(2) = 0, p``(2) = 0

    so now I seem to have 7 pieces of data...

    p(0) = 2,
    p`(0) = 1,
    p(1) = 0,
    p`(1) = 0,
    p(2) = 0,
    p`(2) = 0,
    p``(2) = 0

    so I thought I would try to put these in a linear system and solve them to get the coefficients.

    p(t) = t0 + t1*t + t2*t^2 + t3*t^3 + t4*t^4 + t5*t^5
    p`(t) = t1 + 2t*2t + 3t3*t^2 + 4t*t^3 + 5t*t^4
    p``(t) = 2t2 + 6t3*t + 12t4*t^2 + 20t5*t^3

    p(0) = t0 = 2
    p`(0) = t1 = 1
    p(1) = t0 + t1*t + t2*t^2 + t3*t^3 + t4*t^4 + t5*t^5 = 0
    p`(1) = t1 + 2t*2t + 3t3*t^2 + 4t*t^3 + 5t*t^4 = 0
    p(2) = t0 + 2t1 + 4t2 + 8t3 + 16t4 + 32t5 = 0
    p`(2) = t1 + 4t2 + 12t3 + 32t4 + 80t5 = 0
    p``(2) = 2t2 + 12t3 + 48t4 + 160t5 = 0

    so I put these in the calculator and rref them but that doesn't give me any useful information.

    I am not sure what to do at this point...

    Thank you
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  2. #2
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    Re: Find a polynomial p(t) of degree 6 which...

    Why do you not use the information you are given?
    zero of multiplicity 2 at t =1 implies (t-1)^2 is a factor
    zero of multiplicity 3 at t = 2 implies (t-2)^3 is a factor
    And since p has degree six, any other factor must be linear:
    p(t)= (t- 1)^2(t- 2)^3(at+ b)

    Now put the other information into that.
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  3. #3
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    Re: Find a polynomial p(t) of degree 6 which...

    Okay, so that makes sense, the factor thing, but now it seems we have two unknowns...How would I get the polynomial from its factors, given two unknowns?

    Unless taking the derivative is the key?
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  4. #4
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    Re: Find a polynomial p(t) of degree 6 which...

    we have, as Plato so graciously hinted at:

    p(x) = (x - 1)2(x - 2)3(ax + b)

    therefore:

    p(0) = (-1)2(-2)3(b) = -8b

    since p(0) = 2, this becomes:

    2 = -8b

    b = -1/4

    now let's compute p'(x):

    p'(x) = [(x - 1)2(x - 2)3]'(ax + b) + (x - 1)2(x - 2)3(a)

    = [(x - 1)2(3(x - 2)2) + 2(x - 1)(x - 2)3](ax + b) + (x - 1)2(x - 2)3(a)

    = [(x - 1)(3) + (2)(x - 2)](x - 1)(x - 2)2(ax + b) + (x - 1)2(x - 2)3(a)

    = [3x - 3 + 2x - 4](x - 1)(x - 2)2(ax + b) + (x - 1)2(x - 2)3(a)

    = (5x - 7)(x - 1)(x - 2)2(ax + b) + (x - 1)2(x - 2)3(a)

    = [(5x - 7)(ax + b) + a(x - 1)(x - 2)](x - 1)(x - 2)2

    = [5ax2 + (5b - 7a)x - 7b + ax2 - 3ax + 2a](x - 1)(x - 2)2

    = (6ax2 + (5b - 10a)x + (2a - 7b))(x - 1)(x - 2)2

    therefore:

    p'(0) = (2a - 7b)(-1)(-2)2 = (2a - 7b)(-4) = 28b - 8a

    now we know that b = -1/4, so p'(0) = -7 - 8a.

    since p'(0) = 1, we have:

    1 = -7 - 8a
    8 = -8a
    a = -1

    so p(x) = -(x - 1)2(x - 2)2(x + 1/4). <---the third root is -1/4. expanding this becomes:

    p(x) = -x6 + (31/4)x5 - 23x4 + (127/4)x3 - (37/2)x2 + x + 2

    as a check, we find:

    p(0) = 2, and:

    p'(x) = -6x5 + (155/4)x4 - 92x3 + (381/4)x2 - 37x + 1, so that:

    p'(0) = 1.

    (i can't vouch for my algebra, i often make mistakes, so i recommend checking my work)
    Last edited by Deveno; October 15th 2012 at 02:06 AM.
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