Exercise on polynomials
Determine GCD(x4 + x −1, x3 − 2)
Determine GCD(x5 − x3 +x2 −2x +1, x4 +x3 + 2x2 + x +1)
Calculate GCD(x3 + x2 -6x +1, x4 −2x3 -2x -1) with coefficients in ℤ7 (ℤ7[x] means poly whose coeffs are from ℤ7).
Find if they exist h(x) and k(x) such that
h(x) (x2 + 1) + k(x) (x3 −3) = 20
Find the factorization that is described in the Unique Factorization Theorem for the polynomial f(x) = 2x4 + x3 + 3x2 + 2x + 4, over the field ℤ5.
Factorize the polynomial x4 + 4x + 6 over the field ℤ11.
Solutions
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x4 + x − 1 = (x3 − 2) ⋅ x + (3x − 1)
x3 − 2 = (3x − 1) (x2/3 + x/9 + 1/27) − 53/27
3x − 1 = 53/27 (27⋅3/53 ⋅ x - 27/53)
The last non-zero remainder is −53/27, which is the polynomial (-53/27)⋅x0, and we make it monic by dividing by -53/27, 1x0 = 1.
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x5 − x3 + x2 − 2x +1 = (x − 1) ⋅ (x4 +x3 + 2x2 + x +1) + (−2x3 + 2x2 − 2x + 2)
(x4 + x3 + 2x2 + x + 1) = (−x/2 − 1) ⋅ (−2x3 + 2x2 − 2x + 2) + (3x2 + 3)
(-2x3 + 2x2 − 2x + 2) = (−2x/3 + 2/3)⋅(3x2+3)
so the last non-zero remainder is 3x2 + 3, and we turn it to a monic per convention: x2 + 1.
x4 −2x3 in ℤ7[x] is -2x − 1 is x4 + 5x3 + 5x +6. The coefficient 5 is obtained because −2 ≡ 5 (mod 7). We have
x4 + 5x3 + 5x +6 = (x + 4)(x3 + x2 + x +1) + (2x2 + 2)
(x3 + x2 + x +1) = (4x + 3)(2x2 + 2)
so 2x2 + 2 -> x2+1. By the Euclidean division algorithm
x3 − 3 = x(x2 + 1) −x −3
(x2 + 1) = (−x − 3) (−x + 3) + 10
(−x − 3) = 1/10 (−x −3) · 1/10
by substituting −x − 3 = (x3 − 3) − x · (x2 + 1), into (x2 + 1) = (−x − 3) (−x + 3) + 10.
(x2 + 1) =[x3 − 3 − x · (x2 + 1)] (−x+3) + 10
(x2 + 1) = (x3 − 3) (−x + 3) −x (x2 + 1) (−x+3) + 10
(x2 + 1) +x (x2 + 1) (−x + 3) + (x3 − 3) (−x + 3) = 10
(x2 + 1) + [1 + x (−x + 3)] + (x3 − 3) (x − 3) = 10
if we multiply by 2 we get
20 = (x2 + 1) (−2x +6x +2) + (x3 − 3) (2x − 6)
so we obtain h(x) = −2x +6x +2 and k(x) = (2x − 6).
Note. In ℚ, 10 is 10x0, if we take the monic we can write gcd(x2 + 1, x3 −3) =1.
We first determine the zeros of f(x) in Z 5:
f (0) = 4, f(1) = 2, f(2) = 0, f(3) = 1, f(4) = 1.
Thus 2 is the only zero of f(x) in ℤ5, and the Factor Theorem assures us that x − 2 is a factor of f(x). Dividing by x − 2, we get
f(x) = (x − 2)(2x3 + 3x + 3).
The zeros of f(x) are 2 and the zeros of g(x) = 2x3 + 3x + 3. We therefore need to determine the zeros of g(x), and the only possibility is 2, since this is the only zero of f(x) in ℤ5. We find that g(2) = 0, and this indicates that x − 2 is a factor of g(x). Performing the required division, we obtain
2x3 + 3x + 3 = (x − 2)(2x2 + 4x + 1)
and
f(x) = (x − 2)(x − 2)(2x2 + 4x + 1)
We now find that 2x2 + 4x + 1is irreducible over ℤ5, since it has no zeros in ℤ5. To arrive at the desired factorization, we need only factor the leading coefficient of f(x) from the factor 2x2 + 4x + 1:
f(x) = (x − 2)2(2x2 + 4x + 1) = (x − 2)2 [2x2 + 4x + (2)(3)] = 2(x − 2)2 (x2 + 2x + 3) ■
The polynomial p(x) x4 + 4x + 6 has two roots x = ±1 over ℤ11. Then p(x) is divisible by x + 1. By long division we get x3 − x2 + 5x + 6 as quotient then
p(x) = (x + 1) (x3 − x2 + 5x + 6)
Since p(x) is also divisible by x − 1, we divide x3 − x2 + 5x + 6 by x − 1 and obtain.
p(x) = (x + 1) (x − 1) (x2 + 5x)