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im having problem with polynomial arithmetic?

Compute the GCD of the following pairs of polynomials mod 7.

gcd (x3 + 6x2 + 6x + 5, x3 + x2 + 4x + 1)

Solution:

x + 5

i cant get the solution as above, plz help. tx.

9 Answers

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  • 1 decade ago
    Favorite Answer

    Some of the other posters failed to notice that you

    wanted the answer mod 7.

    Let's divide each of your polynomials by x+5

    but do all the arithmetic mod 7.

    We get

    x^3 + 6x² + 6x + 5 = (x+5)(x² + x + 1)

    x^3 + x² + 4x + 1 = (x+5)(x² -4x +3),

    all mod 7.

    So x + 5 is certainly a common factor.

    Let's show that the quotients have no

    linear factor in common. Then we will be done!

    We have

    x² + x + 1 = (x-2)(x-4)

    x² - 4x + 3 = (x-3)(x-1),

    again, all mod 7.

    So the quotients are relatively prime and the

    GCD of your 2 original polynomials is indeed x + 5.

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  • 1 decade ago

    When you say mod7 do you mean the coefficients are the integers modulus 7? If you do then (using your notation)

    x3 + 6x2 + 6x + 5 = (x + 5)(x2 + x + 1)

    x3 + x2 + 4x + 1 = (x+5)(x2 + 3x + 3)

    When doing the long division you have to rember that

    -4 ≡ 3 (mod 7)

    and

    15 ≡ 1 (mod 7)

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  • 1 decade ago

    First, factor each of the two terms:

    x^3 + 6x^2 + 6x + 5 is not too hard. Look at factors of 1 and 5, the two "end" coefficients, and they are 1 and 5. Also, since all the terms are positive, it is probably "(x+5) and not (x-5).

    x+5 into the polynomial is x^2 + x + x

    so the first polynomial can be rewritten as (x+5)*(x^2+x+1)

    the second part of this does not factor because using the quadratic formula results in a square root -of 1*1-4*1*1 which is negative

    So, let's now see if x+5 goes into the second polynomial. I can already see that this does not work, at least for whole numbers, since the last term, 1, is not divisible by 5. I would say that either the problem is written incorrectly, or the solution is wrong.

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  • aeiou
    Lv 7
    1 decade ago

    gcd (x3 + 6x2 + 6x + 5), (x3 + x2 + 4x + 1) =

    gcd x(x3 + 6x2 + 6x + 5), x(x3 + x2 + 4x + 1) =

    gcd x(x² + 6x + 6) + 5, x(x² + x + 4) + 1 = x + 5 and x + 1 =

    gcd = x + 5

    <><

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  • 1 decade ago

    You should do the Euclidean algorithm mod 7:

    A: x^3+6x^2+6x+5 and B:x^3+x^2+4x+1

    new A=B: x^3+x^2+4x+1 new B=A-B: 5x^2+2x+4

    new A=B:5x^2+2x+4 new B=A-3x*B (mod 7!!!): 2x^2+6x+1

    new A=B=2x^2+6x+1 new B=A+B: x+5

    new A=B: x+5 new B=A-2x*B:3x+1

    Note that 3(x+5) = 3x+15 = 3x=1 mod 7 so yes, (x+5) is g.c.d.

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  • raj
    Lv 7
    1 decade ago

    x^3+6x^2+6x+5

    f(-5)=-125+150-30+5=0

    so x+5 is a factor

    dividing by x+5

    1 6 6 5

    0 -5 -5 -5

    1 1 1 0

    quotient is x^2+x+1

    this is irreducible

    x^3+x^2+4x+1

    f(-5)=-125+25-20+1 is not zero

    so x+5 cannot be the GCD

    repost the sum

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  • woods
    Lv 4
    4 years ago

    i'm with Chris in this one, i've got welded stainless for years. shop the warmth concentrated on the better ingredient of the steel, then path decrease back to the thinner cloth. I even have additionally stumbled on that a bite of distinctive cloth clamped (if that's a pipe it may be curved) below attracts maximum folk of the warmth faraway from the undertaking cloth. a distinctive style of steel like copper or aluminum, it wont merge along with your piece. It additionally enables an area the place you could upload cloth with out having it fall into the pipe. annoying slag skill the two no longer warm sufficient for separation of slag and steel or no longer waiting long sufficient for fabric to relax.

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  • 1 decade ago

    p(x) = 3x + 12x + 6x + 5

    = 21x + 5

    = 3(7)(x) + 5

    q(x) = 3x + 2x + 4x + 1

    = 9x + 1

    = 3(3)(x) + 5

    GCD = 3x + 5

    hmmmmmmm..........

    when you say 'x' in the question, do you mean a variable, 'x', or multiplication ?!

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  • 1 decade ago

    factor them mod 7 and look what they have in common.

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