MA132 - Polynomials

Product of two polynomials formula

(Σaₖxᵏ from k=0 to m)(Σbₖxᵏ from k=0 n) = Σ(Σaᵢbₖ₋ᵢxᵏ from i=0 to k) from k=0 to m+n alternatively, Σ(Σaᵢbⱼxᵏ for i+j=k) from k=0 to m+n

Prove remainder theorem

- P=(x-a)Q+R, R=0 or degR<deg(x-a)=1 - so R must be constant in either case - sub x=a to get P(a)=0Q(a)+R so R=P(a)

How to prove x²+1 is irreducible in ℝ[x]?

- assume it is reducible in ℝ[x] so x²+1=Q₁Q₂ where Q₁,Q₂∈ℝ[x] - because it's reducible degQ₁, degQ₂ < deg(x²+1) = 2 - degQ₁Q₂ = degQ₁+degQ₂=2 - so degQ₁=degQ₂=1 meaning Q₁ and Q₂ are both linear factors - compare cooefficients and arrive at a contradiction

Lemma about common factors of P₁, P₂ and R in division algorithm

- basically Euclid's algorithm for polynomials - if P₁=QP₂+R then the set of common factors of P₁ and P₂ is equal to set of common factors of P₂ and R - suppose F is a common factor of P₁ and P₂ then P₁=FQ₁ and P₂=FQ₂ - then from P₁=QP₂+R we have R=P₁-QP₂ = FQ₁-QFQ₂ = F(Q₁-QQ₂) so F|R and therefore F is a common factor of R and P₂ so ℱ(P₁, P₂)⊆ℱ(P₂, R) - conversely, suppose F is a common factor of R and P₂ then easy to show that F|P₁ so F is a common factor of P₁ and P₂ so ℱ(P₂, R)⊆ℱ(P₁, P₂) - therefore ℱ(P₁, P₂)=ℱ(P₂, R) so the hcf of P₁ is equal to hcf of P₂ and R QED

Prove if H₁ and H₂ are hcfs of P₁ and P₂ then H₁ and H₂ are equivalent

- by division algorithm we know P₁=Q₁P₂+R₁, R₁=0 or degR₁<degP₂ - if R₁=0 then we're done, else continue - P₂=Q₂R₁+R₂, R₂=0 or degR₂<degR₁ - again, if R₂=0 then we're done, else continue ... - R_(N-3)=(Q_N-1)R_(N-2)+R_(N-1) - R_(N-2)=(Q_N)R_(N-1) - so degP₂>degR₁>degR₂>...degR_(N-1) - the degrees of the remainders are decreasing at each step and the degrees are natural numbers so the process must terminate after at most degP₂ steps - if degR_(N-1)=0 for some N then R_(N-1) is a non-zero constant and R_N=0 so you end with exact division - so ℱ(P₁, P₂)=...=ℱ(R_(N-1), 0) = ℱ(R_(N-1)) - RHS has unique element of highest degree R_(N-1) so it must be unique element of highest degree on LHS i.e. R_(N-1) is hcf of P₁ and P₂ - so can write ℱ(P₁, P₂)=ℱ(H) where H=hcf(P₁, P₂) - this means H is divisible by every other common factor of P₁ and P₂ QED

Prove that hcf of two non-zero polynomials exists

- if Q is a common factor of P₁ and P₂ then it's degree is bounded above by min(degP₁, degP₂) - so set of degrees of common factors of P₁ and P₂ is bounded above so it has a greatest element

Prove α is a multiple root of P iff P(α)=P'(α)=0

- if α is a multiple root of P then P=(x-α)ᵏR where k≥2 and R∈ℂ[x] - differentiate to give P'=(x-α)ᵏ⁻¹[R'(x-α)+kR] - since k-1>0, P'(α)=0 - conversely if P(α)=P'(α)=0 then P(x)=(x-α)Q for some Q∈ℂ[x] because of remainder theorem - differentiating gives P'=Q+(x-α)Q' - subbing x=α gives Q(α)=0 so (x-α)|Q by corollary of remainder theorem - so (x-α)²|P QED

Define minimal polynomial of the algebraic number α

- if α is a root of P and Q then it is a root of hcf(P, Q) - if you let P be the polynomial of least degree with α as a root then every other polynomial with α as a root is a multiple of P - this P (scaled to be monic) is called the minimal polynomial of the algebraic number α i.e. P is a minimal polynomial of α if P(α)=0 and if Q(α)=0 then degQ>=degP

Prove uniqueness of minimal polynomial of a number α

- let P₁, P₂∈ℚ[x] be minimal polynomials of α - then (x-α)|P₁, (x-α)|P₂ - consider H=hcf(P₁,P₂) so (x-α)|H - H∈ℚ[x] if Euclidean algorithm is applied - H(α)=0, H∈ℚ[x] and degH<=degP₁, degP₂ by definition of hcf of polynomials - so degH=degP₁ and degH=degP₂ because degH can't be less than the degrees of the minimal forms - so P₁=λ₁H and P₂=λ₂H so P₁=λP₂ - P₁, P₂ are monic so λ=1 and P₁=P₂ QED

Prove that a polynomial with real coefficients of degree n can have at most n roots

- let degP=0 then P(x)=a₀≠0 so P has no roots so true for base case - assume polynomials of degree n-1 have at most n-1 roots - take P∈ℝ[x] with degree n - case 1: P has no roots and we're done because 0<=n - case 2: P has a root x₀ so P(x₀)=0 - by corollary 1, (x-x₀)|P ⇒ P=(x-x₀)Q for some Q∈ℝ[x] - degQ=depP-deg(x-x₀)=n-1 - roots(P) = roots(Q)∪{x₀} - by our assumption, |roots(Q)|<=n-1 so |roots(P)|<=n QED

Show that sqrt(77) is irrational

- sqrt(77) is a root of x²-77 - assume sqrt(77)=m/n where hcf(m,n)=1 - using previous proposition n|1 (n divides the leading coefficient) - so sqrt(77)=m/n=m has to be an integer - but 8<sqrt(77)<9 so sqrt(77) isn't an integer hence it must be irrational

Prove P has no multiple roots if hcf(P, P')=1

- we'll prove the contrapositive i.e. if P has a multiple root than hcf(P,P')≠1 - by previous part of prop, P has a multiple root α ⇔ P(α)=P'(α)=0 ⇔ (x-α)|P and (x-α)|P' ⇒ deg(hcf(P, P'))>0 so hcf(P, P') can't be 1 QED

Define transcendental number

A complex number that is not algebraic

Define algebraic number

A number if algebraic if it is a root of a non-zero polynomial P∈ℚ[x]

Another corollary from remainder theorem

A polynomial P∈ℝ[x] of degree n can have at most n roots in ℝ

Define leading coefficient

Coefficient of leading term

Define degree of a polynomial

Highest power of x appearing in it with non-zero coefficient

Relationship between degree of 2 polynomials and degree of the sum

If P₁ and P₂ are two non-zero polynomials and P₁+P₂≠0 then deg(P₁+P₂)≤max(deg(P₁), deg(P₂))

Relationship between degree of 2 polynomials and degree of the product

If P₁ and P₂ are two non-zero polynomials then deg(P₁P₂)=deg(P₁)+deg(P₂)

Proposition for division with remainder for polynomials with real coefficients

If P₁,P₂∈ℝ[x] and P₂≠0 then ∃Q,R∈ℝ[x] st P₁=QP₂+R with either R=0 or deg(R)<deg(P₂)

Corollary from fundamental theorem of algebra + how to prove

If P∈ℂ[x] has leading coefficient aₙ then P factorises completely into linear factors in ℂ[x] P(x)=aₙ(x-α₁)...(x-αₙ) - prove by induction, base case is linear polynomial - assume a polynomial of order k can be factorised in such way - then consider a polynomial P of order k+1 - P has a root x₀ in ℂ by FTAlg so by corollary from remainder theorem P=(x-x₀)Q - order of Q is k+1-1=k so Q can be factorised in such way - rest is easy

Fundamental theorem of algebra

If P∈ℂ[x], degP≥1 then P has a root in ℂ

Corollary factor theorem from remainder theorem

If P∈ℝ[x] and a∈ℝ and P(a)=0 then (x-a)|P

Remainder theorem

If P∈ℝ[x] and a∈ℝ then on division of P by (x-a) the remainder is P(a)

Define highest common factor of two polynomials with real coefficients

Let P₁,P₂∈ℝ[x] then H∈ℝ[x] is called a highest common factor of P₁ and P₂ if - H|P₁, H|P₂ - if Q|P₁, Q|P₂ then degQ≤degH

Proposition about multiple roots of P in relation to P and it's derivative P' and hcf(P, P')

P(x)∈ℂ[x] - α is a multiple root of P iff P(α)=P'(α)=0 - if hcf(P, P')=1 then P has no multiple roots

Monic

Polynomial with leading coefficient 1

Way of saying corollary from FTAlg in terms of roots

Polynomials over ℂ have as many roots as their degree, counting multiplicities

Prove division with remainder for polynomials with real coefficients

Prop: If P₁,P₂∈ℝ[x] and P₂≠0 then ∃Q,R∈ℝ[x] st P₁=QP₂+R with either R=0 or deg(R)<deg(P₂) Proof: - let ℛ={R∈ℝ[x] : R=P₁-QP₂ for some Q∈ℝ[x]} Case 0∈ℛ then P₂|P₁ exactly and we're done Case 0∉ℛ then consider 𝒟={d∈ℕ : d=deg(R) for R∈ℛ} - i.e. 𝒟=deg(ℛ) - 𝒟 is not empty since P₁∈ℛ so degP₁∈𝒟 - by WOP ∃d₀∈𝒟 - the least element of 𝒟 ⇒ ∃R₀∈ℛ st deg(R₀)=d₀ and WTS d₀<deg(P₂) - by construction ∃Q₀∈ℝ[x] st R₀=P₁-Q₀P₂ - assume d₀≥deg(P₂) - leading term of Q₀ is some constant multiple of x^d₀, say cx^d₀ - let m=deg(P₂) and leading term of P₂ be bxᵐ - then the polynomial R₀-(c/b)x^(d₀-m)P₂ = P₁-(Q₀+(c/b)x^(d₀-m))P₂ is also in ℛ and has degree less than d₀ - this contradicts the fact that d₀ is the smallest possible degree of a polynomial in ℛ - therefore d₀<deg(P₂) QED

Proposition + proof of uniqueness of Q and R in polynomial division

Prop: If P₁=Q₁P₂+R₁ = Q₂P₂+R₂ with R₁=0 or deg(R₁)<P₂ and R₂=0 or deg(R₂)<P₂ then Q₁=Q₂ and R₁=R₂ Proof: - rearranging gives P₂(Q₁-Q₂)=R₂-R₁ - if Q₁≠Q₂ then LHS is a polynomial of degree at least deg(P₂) - RHS is a polynomial of degree ≤ max(deg(R₁), deg(R₂)) < deg(P₂) - so LHS can't equal RHS unless both are 0 i.e. R₁=R₂ and Q₁=Q₂ QED

Proposition to do with the root of a polynomial being rational, the constant term and the leading coefficient + proof

Prop: Let P(x)=aₖxᵏ+...+a₁x+a₀ and let α be a root of P then - if α∈ℤ then α|a₀ - if α=m/n∈ℚ where hcf(m,n)=1 then m|a₀ and n|aₖ Proof: α is a root so P(α)=0 - if α∈ℤ then a₀=-a₁α-...-aₖαᵏ is a multiple of α so α|a₀ - if α=m/n then sub and multiply out to get rid of denominator to get aₖmᵏ+aₖ₋₁mᵏ⁻¹n+...+a₁mnᵏ⁻¹+a₀nᵏ=0 - using previous part, m|a₀nᵏ but hcf(m,n)=1 so m|a₀ - similarly, n|aₖmᵏ so n|aₖ QED

Define simple root and multiple root

Simple root is a root of multiplicity 1 Multiple root is a root of multiplicity greater than 1

Define leading term of a polynomial

Term of the highest degree

Define irreducible polynomial

The polynomial P∈ℝ[x] is irreducible in ℝ[x] if degP>0 and ∀Q₁,Q₂∈ℝ[x] st P=Q₁Q₂, either Q₁ or Q₂ is a constant Else, it is reducible

Define equivalent polynomials

Two polynomials Q and P are equivalent if one is a non-zero constant multiple of the other

Degree of the polynomial 0

Undefined

Leading term of the polynomial 0

Undefined

Are all polynomials of degree 1 irreducible?

Yes

If α is a root of P and Q is it a root of hcf(P, Q)?

Yes

Is hcf of two polynomials unique?

Yes, but only up to multiplication by a non-zero constant

Sum of two polynomials formula

Σaₖxᵏ + Σbₖxᵏ = Σ(aₖ+bₖ)xᵏ

Define root of multiplicity k

α is said to be a root of multiplicity k if x-α appears k times in the factorisation of P(x), so that (x-α)ᵏ|P(x) but (x-α)ᵏ⁺¹∤P(x)

Set of transcendental numbers

ℂ\𝔸

What field does polynomial division with remainder proposition not hold for?

ℤ

Define set of algebraic number in terms of union of roots

𝔸=union of roots(P) over all P∈ℤ[x]\{0}