Factor the polynomial over complex numbers. ?

The question gives x=2+3i as a zero, how do you find the 3 remaining zeroes. 

And is it correct to assume that x=2-3i is a zero?

p(x)=x^4-12x^3+62x^2-172x+221

4 Answers

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  • 4 weeks ago

    Yes, it's correct to assume that 2 - 3i is a zero.  It's a basic theorem in any first course in complex algebra or analysis that, when the coefficients of a polynomial p are all real then p(z') is a root of p whenever p(z) is.  I'm using z' to mean the complex conjugate of z.  Since (z')' = z, that's an "if and only if" relation. 

    Since 2 + 3i and 2 - 3i are roots, then (x - 2 - 3i)(x - 2 + 3i)  = x^2 - 4x + 13 is a factor of your polynomial.  Divide p(x)/(x^2 - 4x + 13) to get x^2 - 8x + 17 and a zero remainder.  (This is a good check on your work...if you get a nonzero remainder, something got muffed.)

    [The rest of this is more for cosmo, but you might find it interesting.]

    The proof that complex roots to any polynomial occur in conjugate pairs is simple.  Still using ' for complex conjugate, it's pretty easy to show that, for any complex numbers w and z:

        (z + w)' = z' + w'

        (zw)' = (z')(w')

    The above can be extended to k terms or factors (k is a positive integer):

        (kz)' = k(z')

        (z^k)' = (z')^k

    Apply those properties to a polynomial p(z) = a_0 + a_1*z + a_2*z^2 + ... + a_n*z^n and you get:

        p(z)' = a_0' + a_1'*z' + a_2'*(z')^2 ... + a_n'*(z')^n

    If the coefficients a_k of p are real, then a_k' = a_k each time, and the right side simplifies to p(z').

        p(z)' = p(z')             [whenever the coefficients of p are all real]

    So, when p(z) = 0, then p(z') = p(z)' = (0)' = 0.  That is, if p(z) = 0 then p(z') = 0.

    QED

  • If your polynomial has real coefficients and one of the roots is complex, then it will have another complex root and that root will be the conjugate of the other complex root.  So:

    p(x) = x^4 - 12x^3 + 62x^2 - 172x + 221

    All real coefficients

    x = 2 + 3i is a 0

    Therefore, x = 2 - 3i is a 0

    (x - (2 + 3i)) * (x - (2 - 3i)) =>

    x^2 - x * (2 - 3i) - x * (2 + 3i) + (2 - 3i) * (2 + 3i) =>

    x^2 - x * (2 - 3i + 2 + 3i) + 4 - 9i^2 =>

    x^2 - x * 4 + 4 + 9 =>

    x^2 - 4x + 13

    (x^2 - 4x + 13) * (ax^2 + bx + c) = x^4 - 12x^3 + 62x^2 - 172x + 221

    a * 1 = 1

    a = 1

    13 * c = 221

    c = 17

    (x^2 - 4x + 13) * (1x^2 + bx + 17)

    (x^2 - 4x + 13) * (x^2 + bx + 17)

    x^4 + bx^3 + 17x^2 - 4x^3 - 4bx^2 - 68x + 13x^2 + 13bx + 221

    bx^3 - 4x^3 = -12x^3

    b - 4 = -12

    b = -8

    17x^2 - 4bx^2 + 13x^2 = 62x^2

    17 - 4b + 13 = 62

    30 - 4b = 62

    -4b = 32

    b = -8

    -68x + 13bx = -172x

    -68 + 13b = -172

    13b = -104

    b = -8

    b = -8 works all around, so we know we're right

    (x^2 - 4x + 13) * (x^2 + 8x + 17) = x^4 - 12x^3 + 62x^2 - 172x + 221

    We just need to find when x^2 + 8x + 17 = 0 to find the remaining 0s

    x^2 + 8x + 17 = 0

    x^2 + 8x = -17

    x^2 + 2 * 4x = -17

    x^2 + 2 * 4x + 4^2 = -17 + 4^2

    (x + 4)^2 = -17 + 16

    (x + 4)^2 = -1

    x + 4 = +/- i

    x = -4 +/- i

    Our 0s are:

    2 + 3i

    2 - 3i

    -4 - i

    -4 + i

  • cosmo
    Lv 7
    4 weeks ago

    (x^2 - 8x - 17) (x^2 -4x -13)      The hint here is that 221 factors into 17 * 13

    Since it factors to two quadratics with real coefficients, the complex roots will come in conjugate pairs. 

  • 4 weeks ago

    2+3i, 2-3i, 4+1i, 4-1i are the roots.

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