please answer with calculation details, thank you! Find the greatest common divi

Question

please answer with calculation details, thank you!

Find the greatest common divisor of each of the following pairs p(x) and q(x) of polynomials. If d(x) = gcd(p(x). q(x)), find two polynomials a(x) and b(x) such that a(x)p(x) + b(x)q(x) = d(x). p(x) = 7x3 + 6x2 - 8x + 4 and q(x) = x3 + x - 2. where p{x), q(x) Q[x] p(x) = x3 + x2 - x + I and q(x) - x3 + x - 1, where p(x),q(x) Z2[x] p(x) = x3 + x2 - Ax + 4 and q(x) = x3 + 3x - 2, where p(x), q(x) Z5[x] p(x) = x3 - 2x + 4 and q(x) = 4x3 + x + 3, where p(x),q(x) Q[x]

Explanation / Answer

(a)
I'll write the division sucessively, this is the polynom division algorithm

7x^3+6x^2-8x+4 | x^3+x-2
7x^3-7x+14 | 7
6x^2-15x+18


So : 7x^3+6x^2-8x+4 = 7(x^3+x-2) + 6x^2-15x+18 => 6x^2-15x+18 = p(x) - 7q(x)

We continue iteratively :

x^3+x-2 | 6x^2-15x+18
x^3-15/6x^2+18/6x | 1/6x+5/12
5/2x^2-2x-2 |
5/2x^2-25/4x+15/2 |
17/4x-19/2 |


So : x^3+x-2 = (6x^2-15x+18)*(1/6x+5/12)+(17/4x-19/2)
=> q(x) = (p(x)-7q(x))*(1/6x+5/12)+(17/4x-19/2)
=> (17/4x-19/2) = q(x)-(p(x)-7q(x))(1/6x+5/12)
=> (17/4x-19/2) = -(1/6x+5/12)p(x)+(7/6x+47/12)q(x)

And again :
6x^2-15x+18 | 17/4x-19/2
6x^2-228/17x | 24/17x-108/289
-27/17x+18 |
-27/17x+1026/289 |
4176/289 |

6x^2-15x+18 = (17/4x-19/2)(24/17x-108/289) + 4176/289

The algorithm stops and d(x) = gcd(p(x),q(x))=1

Using all the previous equality we get :

=> 1 =289/4176*((p(x) - 7q(x))-(-(1/6x+5/12)p(x)+(7/6x+47/12)q(x) )((24/17x-108/289)))
=> 1 =289/4176*(p(x)*(4/17x^2+152/289x+244/289)+q(x)(-28/17x^2-1472/289x-1600/289))
=> 1 = (17/1044x^2+19/522x+61/1044)p(x) + (-119/1044x^2-92/261x-100/261)q(x)

so a(x) = 17/1044x^2+19/522x+61/1044 and b(x) = -119/1044x^2-92/261x-100/261
(I checked with a calculator and a(x)p(x)+b(x)q(x) = 1 is correct with the previous polynom...yes they are ugly !)


(b)

x^3+x^2-x+1 | x^3+x-1
x^3+x-1 | 1
x^2-2x+2(mod2) |
x^2

so x^3+x^2-x+1 = (x^3+x-1)*1 + x^2 => x^2 = p(x)-q(x)


x^3+x-1 | x^2
x^3 | x
x-1 |

So x^3+x-1 = x^2*x + (x-1)

(x-1) = q(x)-x*(p(x)-q(x)) = -xp(x)+(1+x)q(x)

x^2 | x-1
x^2-x | x+1
x
x-1
1

So x^2 = (x-1)(x+1)+1

So d(x) = gcd(p(x),q(x)) = 1

Using all the previous equality we get :
=> 1 = (p(x)-q(x))-(x+1)(-xp(x)+(1+x)q(x)) = p(x)(x^2+x+1)+q(x)*(-x^2)

So a(x) = x^2+x+1 and b(x)=-x^2

(c)

x^3+x^2-4x+4 | x^3+3x-2
x^3+3x-2 | 1
x^2-7x+6 (mod 5) |
x^2+3x+1 |

So x^3+x^2-4x+4 = (x^3+3x-2) + (x^2+3x+1) => (x^2+3x+1) = p(x)-q(x)

x^3+3x-2 | x^2+3x+1
x^3+3x^2+x | x-3
-3x^2+2x-2 |
-3x^2-9x-3 |
11x+1 (mod 5)|
x+1

So x^3+3x-2 = (x^2+3x+1)(x-3) + (x+1) => (x+1) = q(x)-(p(x)-q(x))(x-3) = (3-x)p(x)+(x-2)q(x)

x^2+3x+1 | x+1
x^2+x | x+2
2x+1 |
2x+2 |
-1(mod 5)|
4

So (x^2+3x+1) = (x+1)(x+2) + 4

So again gcd(p(x),q(x))=1

Since 4=-1 (mod 5) :

=> -1 = ( (p(x)-q(x)) - (x+2)*( (3-x)p(x)-(x-2)q(x))) = ( p(x)(1+(x-3)(x+2)) + q(x)(-1+(2-x)(2+x)))
=> 1 = p(x)(-x^2+x) + q(x)(x^2-3)

So a(x)=-x^2+x and b(x)=x^2-3

(d)

x^3-2x+4 | 4x^3+x+3
x^3+1/4x+3/4 | 1/4
-9/4x+13/4 |

So x^3-2x+4 = 1/4(4x^3+x+3) + (-9/4x+13/4)
=> (-9/4x+13/4) = p(x)-1/4q(x)

4x^3+x+3 | -9/4x+13/4
4x^3-52/9x^2 | -16/9x^2-208/81x
52/9x^2+x+3 |
52/9x^2-676/81x |
757/81x+3 |

So : 4x^3+x+3 = (-9/4x+13/4)(-16/9x^2-208/81x) + (757/81x+3)
=> 757/81x+3 = q(x)-(p(x)-1/4q(x))(-16/9x^2-208/81x)
=> 757/81x+3 = p(x)(16/9x^2+208/81x) + q(x)(-36/81x^2-52/81x+1)

-9/4x+13/4 | 757/81x+3
-9/4x-2187/3028 | -729/3028
3007/757 |

So -9/4x+13/4 = (757/81x+3)*-729/3028 + 3007/757

Again gcd(p(x),q(x))=1

And :

1 = 757/3007 * ( p(x)-1/4q(x) +729/3028(p(x)(16/9x^2+208/81x) + q(x)(-36/81x^2-52/81x+1)))
1 = p(x)*(324x^2+468x+757)/3007 + q(x)(-81x^2-117x-7))/3007


So a(x) = 1/3007*(324x^2+468x+757) and b(x) = 1/3007 * (-81x^2-117x-7)

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