# 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

- steiner1745Lv 71 decade agoFavorite 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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- fireflyLv 61 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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- aeiouLv 71 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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- a_math_guyLv 51 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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- rajLv 71 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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- woodsLv 44 years ago
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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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- gjmb1960Lv 71 decade ago
factor them mod 7 and look what they have in common.

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