Factoring & Polynomial Division

[Possibilities] [Fundamental Theorem of Algebra]

Possibilities

(Theorems & Lemmas)

If p(x) and q(x) are nonzero elements in F[x], then deg (p(x)q(x)) = deg p(x) + deg q(x)

Proof

Let m = deg (p(x)) and n = deg (q(x)). Let p(x) = a_0 + (a_1)x + ... + (a_m)x^m where a_m 0 and q(x) = b_0 + (b_1)x + ... + (b_n)x^n where b_n 0 then p(x)q(x) = c_0 + (c_1)x + ... + (c_(m+n))x^(m+n) so the highest power of p(x)q(x) is x^(m+n) where its coefficient is (a_m)(b_n) which is not zero. So a_m 0 and b_n 0. Therefore the highest power of p(x) is x^m and q(x) is x^n so the deg p(x) + deg q(x) = m + n. Therefore deg (p(x)q(x)) = deg p(x) + deg q(x).

If p(x) and q(x) are in F[x] and p(x) + q(x) 0, then deg(p(x) + q(x)) max(deg p(x), deg q(x)).

Proof

Let p(x) and q(x) be in F[x] where p(x) + q(x) 0. Let p(x) = a_0 + (a_1)x + ... + (a_m)x^m where a_m 0 and q(x) = b_0 + (b_1)x + ... + (b_n)x^n where b_n 0 and where mn. The deg p(x) = m and deg q(x) = n. so max(deg p(x), deg q(x)) = n if m n. p(x) + q(x) = c_0 + (c_1)x + ... + (c_n)x^n. So deg (p(x) + q(x)) = n if m<n or m=n and b_n -a_m and deg (p(x) + q(x)) < n if m = n and b_n = -a_n. Thus deg (p(x) + q(x)) max(deg p(x), deg q(x)) = n.

Division Algorithm: Given the polynomial f(x) and g(x) in F[x] where g(x) 0, f(x) = q(x)g(x) + r(x) where q(x) and r(x) are in F[x] and r(x) = 0 or deg r(x) < deg g(x)

Proof

Assume that deg f(x) deg g(x) where f(x) = a_0 + (a_1)x + ... + (a_m)x^m, a_m 0, and g(x) = b_0 + (b_1)x + ... + (b_n)x^n, b_n 0 such that mn. Look at (a_m/b_n)x^(m-n)g(x) = (a_m/b_n)x^(m-n)(b_0 + (b_1)x + ... + (b_n)x^n) = ((a_m)(b_0)/(b_n))x^(m-n) + ... + (a_m)x^m. So ((a_m)/(b_n))x^(m-n)g(x) has the same degree and leading coefficient as f(x), so f(x) - ((a_m)/(b_n))x^(m-n)g(x) = h(x) where deg h(x) < deg f(x).

Assume h(x) = q_1(x)g(x) + r(x) where q_1(x) and r(x) in F[x] and r(x) = 0 or deg r(x) < deg g(x). Looking back at what h(x) is, h(x) = f(x) - ((a_m)/(b_n))x^(m-n)g(x) = q_1(x)g(x) + r(x) (by our assumotion) so f(x) = ((a_m)/(b_n))x^(m-n) + q_1(x))g(x) + r(x). If q(x) = (a_m/b_n)x^(m-n) + q_1(x) then f(x) = q(x)g(x) + r(x) where q(x) and r(x) are in F[x] and deg r(x) < deg g(x).

In Algebra, we discuss two ways to divide polynomials, synthetic division and long division. The reason these methods works is because of the division algorithm. Looking specifically at the connections between long division and the division algorithm, the remainder is r(x), the dividend is f(x), the divisor is g(x) and the quotient is q(x).

If I (0) is an ideal of F[x] then I consists of all the multiples of a fixed polynomial in F[x]. (I = {f(x)g(x)| f(x) F[x]})

Proof

Let I (0). So I contains elements of nonnegative degree. Let g(x) 0 be in I and be of minimal degree. So if 0 t(x) is in I then deg g(x) deg t(x). So by the division algorithm t(x) = q(x)g(x) + r(x) where r(x) = 0 or deg r(x) < deg g(x). Since g(x) is in I, q(x)g(x) is in I and so t(x) will be in I. However, we said that deg g(x) was minimal so deg r(x) deg g(x), thus r(x) = 0. So t(x) = q(x)g(x) + 0 = q(x)g(x). So every element in I will be a multiple of g(x).

Given f(x) 0 and g(x) 0 in F[x], then their greatest common divisor d(x) in F[x] exists and d(x) = a(x)f(x) + b(x)g(x) for some a(x) and b(x) in F[x].

Proof

Let I equal the set of all r(x)f(x) + s(x)g(x) where r(x) and s(x) are in F[x]. Suppose I is an ideal of R. (r_1(x)f(x) + s_1(x)g(x)) + (r_2(x)f(x) + s_2(x)g(x)) = (r_1(x) + r_2(x))f(x) + (s_1(x) + s_2(x))g(x) which is in I. Let t(x) in F[x], t(x)(r(x)f(x) + s(x)g(x)) = (t(x)r(x))f(x) + (t(x)s(x))g(x) is also in I. So I is in fact an ideal of F[x].

Since I 0 is an ideal of F[x] there exists a unique monic polynomial d(x) that generates I and since f(x) and g(x) are in I they must be mulitples of d(x). So d(x)|f(x) and d(x)|g(x). Because I consists of all the elements r(x)f(x) + s(x)g(x) and d(x) in I, d(x) = a(x)f(x) + b(x)g(x) for soem a(x) and b(x) in F[a]. Therefore if h(x)|f(x) and h(x)|g(x) then h(x)|d(x). So d(x) will be the greatest common divisor.

Suppose the greatest common divisor is not unique. Suppose d(x) and d'(x) are both the greatest common divisor of f(x). Then d(x)|f(x)g(x) and d'(x)\f(x)g(x). So d(x)|d'(x) and d'(x)|d(x), therefore d(x) = ad'(x) but by the definition of a greatest common divisor, d(x) must be monic so a =1 and d(x) = d'(x).

If f(x) 0 and g(x) 0 are in F[x] and f(x)|g(x) and g(x)|f(x), then f(x) = ag(x), where a is in F.

Proof

Let f(x)|g(x) and g(x)|f(x), so f(x) = a(x)g(x) and g(x) = b(x)f(x). Given two nonzero polynomials p(x) and g(x) in F[x], deg (a(x)g(x)) = deg a(x) + deg q(x). Therefore deg f(x) deg g(x) deg f(x) and deg f(x) = deg (a(x)g(x)) = deg a(x) + deg g(x). Since deg f(x) = deg g(x), deg a(x) = 0, a(x) = a which is an element of F. Thus f(x) = ag(x) with a in F.

If q(x) and f(x) are relatively prime and if q(x)|f(x)g(x) then q(x)|g(x).

Proof

Let q(x)|f(x)g(x) where q(x) and f(x) are relatively prime. Then a(x)f(x) + b(x)q(x) = 1 for some a(x) and b(x) in F[x]. So, g(x)a(x)f(x) + g(x)b(x)q(x) = g(x) (by multiplying everything by g(x)). Since q(x)|g(x)b(x)q(x) and q(x)|f(x)g(x), q(x)|g(x).

What do all of these proofs have to do with high school mathematics?

These proofs show that the properties of addition, subtraction, multiplication and division hold true for polynomials.

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Fundamental Theorem of Algebra

In high school, teachers focus mainly on polynomials of degree less than or equal to two. The question becomes, why? There are polynomials of much greater degree. In fact, there is an infinite number of them. Why don't we care?

It would sound somewhat reasonable at first to think it is because polynomials of larger degree are more complicated and would be confusing for high school students. But this is not the true reason. All polynomials of larger degree can be broken down into quadratic and linear polynomials over the reals.

Look at a polynomial over the complex numbers. Gauss came up with The Fundamental Thoerem of Algebra which says that the only finite extension of the field of complex numbers is the complex numbers itself.

From this we can get that a polynomial, f(x) of degree n will have exactly n zeros. If f(x) is a polynomial over the complex numbers where f(z) = 0, then its complex conjugate z` will also equal to zero where z = a + bi and z` (complex conjugate) is a - bi s.t. a and b are real numbers. Since these are both roots, (x - z)(x - z`) = x^2 - (a+bi)x - (a - bi)x + (a + bi)(a - bi) = x^2 -2ax + (a^2 + b^2) where all of the coefficients are real. So under the complex numbers f(x) breaks down into polynomials of degree 1, f(x) = (x - z)(x - z`)(x - y)(x - z`)(x - w)(x - w`)...(x - c)(x - c`).

However, under the real numbers, x^2 -2ax + (a^2 + b^2) cannot be factored into linear polynomials, since its zeros are not real, so f(x) will consist of linear and quardatic polynomials. After all this we realize that polynomials of degree greater than two are just polynomials that can be factored into polynomials of smaller degree we know why we don't need to focus our studies on polynomials with a large degree.

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