Abstract-Algebra-1-Factoring Polynomials over a FIeld, guaranteed and verified 100% Pass

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Factoring Polynomials over a Field

Our goal is to find zeros of a polynomial. Suppose we can factor a
polynomial over a field 𝐹, i.e. 𝑓 (𝑥 ) = 𝑔(𝑥 )ℎ(𝑥). Recall that if 𝜙𝛼 is the
evaluation homomorphism:

𝑓(𝛼 ) = 𝜙𝛼 (𝑥 ) = 𝜙𝛼 (𝑔(𝑥 )ℎ(𝑥 )) = 𝜙𝛼 (𝑔(𝑥 ))𝜙𝛼 (ℎ(𝑥 )) = 𝑔(𝛼 )ℎ(𝛼).


Since 𝐹 is a field it has no 0 divisors, if 0 = 𝑓 (𝛼 ) = 𝑔(𝛼 )ℎ(𝛼 ) then either
𝑔(𝛼 ) = 0 or ℎ(𝛼 ) = 0.
So if we can factor a polynomial 𝑓(𝑥) = 𝑔(𝑥)ℎ(𝑥), then finding zeros of 𝑓 (𝑥 ) is
reduced to finding zeros of 𝑔(𝑥 ) and ℎ(𝑥).




Theorem: Division Algorithm for 𝐹[𝑥]



Let 𝑓 (𝑥 ) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎0

𝑔(𝑥 ) = 𝑏𝑚 𝑥 𝑚 + 𝑏𝑚−1 𝑥 𝑚−1 + ⋯ + 𝑏0 be elements of 𝐹[𝑥],
with 𝑎𝑛 and 𝑏𝑚 both non-zero, and 𝑚 > 0. Then there are unique
polynomials 𝑞(𝑥) and 𝑟(𝑥) in 𝐹[𝑥] such that:



𝑓(𝑥 ) = 𝑞 (𝑥 )𝑔(𝑥 ) + 𝑟(𝑥)


where either 𝑟(𝑥 ) = 0 or the degree of 𝑟(𝑥 ) is less than the degree 𝑚 of
𝑔(𝑥 ).

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Ex. Let 𝑓 (𝑥 ) = 𝑥 4 + 𝑥 3 − 3𝑥 2 + 2𝑥 + 3 and 𝑔(𝑥 ) = 𝑥 2 − 2𝑥 + 2
in ℤ5 [𝑥]. Find 𝑞 (𝑥 ) and 𝑟 (𝑥 ) such that 𝑓 (𝑥 ) = 𝑔(𝑥 )𝑞 (𝑥 ) + 𝑟(𝑥)
and 𝑟(𝑥) is of degree less than 𝑔(𝑥 ) (i.e. less than 2).



𝑥 2 + 3𝑥 + 1
𝑥 2 − 2𝑥 + 2 𝑥 4 + 𝑥 3 − 3𝑥 2 + 2𝑥 + 3
𝑥 4 − 2𝑥 3 + 2𝑥 2
3𝑥 3 + 2𝑥 (−3 − 2 ≡ 0 𝑚𝑜𝑑 5)
3𝑥 3 − 𝑥 2 + 𝑥 (3(2) ≡ 1 𝑚𝑜𝑑 5)
𝑥2 + 𝑥 + 3
𝑥 2 − 2𝑥 + 2
3𝑥 + 1

So 𝑞 (𝑥 ) = 𝑥 2 + 3𝑥 = 1 and 𝑟(𝑥 ) = 3𝑥 + 1.



Corollary: (Factor Theorem) An element 𝛼 ∈ 𝐹 is a zero of 𝑓(𝑥) ∈ 𝐹[𝑥]
if, and only if, 𝑥 − 𝛼 is a factor of 𝑓(𝑥) in 𝐹[𝑥].


Proof: Assume that 𝑓 (𝛼 ) = 0, for 𝛼 ∈ 𝐹.
By the previous theorem we can write:

𝑓 (𝑥 ) = (𝑥 − 𝛼 )𝑞(𝑥) + 𝑟(𝑥 ), where the degree of 𝑟(𝑥 ) is 0.
Thus 𝑟(𝑥) =constant. But 𝑓 (𝛼 ) = 0 implies that:

0 = 𝑓(𝛼) = (𝛼 − 𝛼)𝑞(𝛼) + 𝐶 ⟹ 𝐶 = 0.

Hence 𝑓 (𝑥 ) = (𝑥 − 𝛼 )𝑞(𝑥) and 𝑥 − 𝛼 is a factor of 𝑓 (𝑥 ).