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Math Help - Help with Factorising of Polynomials in C

  1. #1
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    Question Help with Factorising of Polynomials in C

    Can someone help me with getting the factors for the following problem? I can't seem to get the other set.

    If z-3i is a factor of 2z^4 - 4z^3 + 21z^2 - 36z +27, find the remaining factors.

    ---

    Let P(z) = 2z^4 - 4z^3 + 21z^2 - 36z +27

    If z[1] = z-3i
    then z[2] = z+3i (Conjugate Root Theorem)
    Let z[3], z[4] = third and forth zeros respectively
    then P(z) = (z - z[1])(z - z[2])(z - z[3])(z - z[4])
    = (z - (z-3i))(z - (z+3i))(z - z[3])(z - z[4])
    = (3i)(-3i)(z - z[3])(z - z[4])
    = (-9i^2)(z - z[3])(z - z[4])
    = 9(z - z[3])(z - z[4])

    ---

    I'm pretty sure this is correct so far but I'm not sure how to get the remaining two factors. I'm not sure if you need to divide P(z) by 9 through long division or divide it by z[1]*z[2].

    If someone would be so kind as to help me get the remaining factors which will likely be conjugates.

    Thank you in advance.

    Note: z[1] is meant to be z with a subscript 1, etcetera. Not sure what the notation for that is though.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by joumas View Post
    Can someone help me with getting the factors for the following problem? I can't seem to get the other set.

    If z-3i is a factor of 2z^4 - 4z^3 + 21z^2 - 36z +27, find the remaining factors.

    ---

    Let P(z) = 2z^4 - 4z^3 + 21z^2 - 36z +27

    If z[1] = z-3i
    then z[2] = z+3i (Conjugate Root Theorem)
    Let z[3], z[4] = third and forth zeros respectively
    then P(z) = (z - z[1])(z - z[2])(z - z[3])(z - z[4])
    = (z - (z-3i))(z - (z+3i))(z - z[3])(z - z[4])
    = (3i)(-3i)(z - z[3])(z - z[4])
    = (-9i^2)(z - z[3])(z - z[4])
    = 9(z - z[3])(z - z[4])

    ---

    I'm pretty sure this is correct so far but I'm not sure how to get the remaining two factors. I'm not sure if you need to divide P(z) by 9 through long division or divide it by z[1]*z[2].

    If someone would be so kind as to help me get the remaining factors which will likely be conjugates.

    Thank you in advance.

    Note: z[1] is meant to be z with a subscript 1, etcetera. Not sure what the notation for that is though.
    We know that z - 3i and z + 3i are factors, so (z - 3i)(z + 3i) = z^2 + 9 is a factor. Thus z^2 + 9 divides 2z^4 - 4z^3 + 21z^2 - 36z +27. Doing the division gives:
    2z^2 - 4z + 3

    There are two ways to proceed from here. The first is by use of the quadratic formula (since the remaining polynomial doesn't factor over the reals.) So solving 2z^2 - 4z + 3 = 0 gives z = 1 + i(sqrt{2}/2) and z = 1 - i(sqrt{2}/2). So the remaining factors of the original polynomial are:
    z - 1 - i(sqrt{2}/2)
    z - 1 + i(sqrt{2}/2)
    However, if you look carefully you will note that we are off by a factor of 2 somewhere. Multiplying these two factors together gives z^2 - 2z +3/2, not the required 2z^2 - 4z + 3. A simple way to fix this is to multiply one of our factors by 2, giving the factors of the original polynomial as:
    z - 3i
    z + 3i
    2z - 2 - i*sqrt{2}
    z - 1 + i(sqrt{2}/2)

    The second way is to note that we have two more factors to find: z - z[3] and z - z[4]. So a(z - z[3])(z - z[4]) = az^2 - a(z[3] + z[4])z + a*z[3]*z[4] must be the same as the polynomial 2z^2 - 4z + 3. Thus
    a = 2
    a(z[3] + z[4]) = 4
    a*z[3]*z[4] = 3
    We may solve the system of equations for a, z[3], and z[4] and we get the same set of factors as we did in the first method.

    I recommend the first method over the second as it is slightly simpler.

    -Dan
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  3. #3
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    Thanks for the help. The problem had me stumped for a while due to the 2 outside the brackets so I thought I had done something wrong.
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  4. #4
    Junior Member Dergyll's Avatar
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    My teacher has taught me a way of long division where you divide an equation by another. Its pretty hard to describe but here's a page that tell you how to do it Polynomial Long Division

    Obviously when you get a remainder of zero you know you've found a factor of the original equation.

    Hope this helps
    Derg
    Last edited by Dergyll; February 21st 2007 at 03:23 PM. Reason: additional info
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