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Math Help - Monic polynomial help

  1. #1
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    Monic polynomial help

    Find the monic polynomial of degree 5 which has 1, 1 + i and 2 i as three of its roots..

    any help on this question id appreciate.
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  2. #2
    MHF Contributor red_dog's Avatar
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    I think the polynomial has real coefficients. Then

    x_1=1, \ x_2=1+i, \ x_3=1-i, \ x_4=2-i, \ x_5=2+i

    P=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)=

    =(x-x_1)[(x^2-(x_2+x_3)x+x_2x_3]\cdot[x^2-(x_4+x_5)x+x_4x_5]=

    =(x-1)(x^2-2x+2)(x^2-4x+5)
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  3. #3
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    Quote Originally Posted by red_dog View Post
    I think the polynomial has real coefficients. Then

    x_1=1, \ x_2=1+i, \ x_3=1-i, \ x_4=2-i, \ x_5=2+i

    P=(x-x_1)(x-x_2)(x-x_3)(x-x_4)(x-x_5)=

    =(x-x_1)[(x^2-(x_2+x_3)x+x_2x_3]\cdot[x^2-(x_4+x_5)x+x_4x_5]=

    =(x-1)(x^2-2x+2)(x^2-4x+5)
    wow looks very complicated didnt understand really.
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  4. #4
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    Quote Originally Posted by jvignacio View Post
    wow looks very complicated didnt understand really.
    Did you replace x_1, x_2, x_3, and x_4 with 1, 1+ i, 1- i, 2- i and 2+ i, respectively?

    First, the reason red-dog said, "I think the polynomial has real coefficients" is that if you allow complex coefficients, there are an infinite number of such polynomials. Assuming complex coefficients, all roots must be in complex conjugate pairs. That is why 1- i and 2+ i must also be roots. Of course, it follows (You've solved polynomial equations by factoring haven't you. Same idea here). Saying "monic" polynomial means the leading coefficient (the coefficient of x^5 here) is 1.

    That is why the polynomial must be the product, (x-1)(x-1-i)(x-1+i)(x-2+i)(x-2-i). That is the polynomial. Whether you are required to multiply that out depends on the particular statement of the question.

    If you do, it's simplest to multiply the complex conjugates first: (x-1-i)(x-1+i) can be done as "sum and difference product": ((x-1)-i)((x-1)+i)= (x-1)^2- i^2= x^2- 2x+ 1+ 1= x^2- 2x+ 2

    (x-2+i)(x-2-i)= ((x-2)+i)((x-2)-i= (x-2)^2- i^2= x^2- 4x+ 4+ 1= x^2- 4x+ 5

    The polynomial you want is (x-1)(x^2- 2x+ 2)(x^2- 4x+ 5)
    Can you multiply that out?
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    Did you replace x_1, x_2, x_3, and x_4 with 1, 1+ i, 1- i, 2- i and 2+ i, respectively?

    First, the reason red-dog said, "I think the polynomial has real coefficients" is that if you allow complex coefficients, there are an infinite number of such polynomials. Assuming complex coefficients, all roots must be in complex conjugate pairs. That is why 1- i and 2+ i must also be roots. Of course, it follows (You've solved polynomial equations by factoring haven't you. Same idea here). Saying "monic" polynomial means the leading coefficient (the coefficient of x^5 here) is 1.

    That is why the polynomial must be the product, (x-1)(x-1-i)(x-1+i)(x-2+i)(x-2-i). That is the polynomial. Whether you are required to multiply that out depends on the particular statement of the question.

    If you do, it's simplest to multiply the complex conjugates first: (x-1-i)(x-1+i) can be done as "sum and difference product": ((x-1)-i)((x-1)+i)= (x-1)^2- i^2= x^2- 2x+ 1+ 1= x^2- 2x+ 2

    (x-2+i)(x-2-i)= ((x-2)+i)((x-2)-i= (x-2)^2- i^2= x^2- 4x+ 4+ 1= x^2- 4x+ 5

    The polynomial you want is (x-1)(x^2- 2x+ 2)(x^2- 4x+ 5)
    Can you multiply that out?
    yeah umm is it x^5 - 7x^4 + 21x^3 -25x^2 + 20x - 10 ?
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