The leading coefficient of a polynomial is the coefficient in front of the highest power that x is raised to provided that that coefficient does not equal zero. If that coefficient did equal zero, then the polynomial could actually be rewritten using 1 less power of x, changing the degree of the polynomial.
Note: In the polynomial ax^n + b^(n-1) + c^(n-2) +
………d, the leading coefficient = a, provided that a does not equal zero. If a does not equal zero, then the degree of
this polynomial is n.
Recall, the highest power of x in a polynomial is referred to as the degree of the polynomial provided that its coefficient is not zero. In other words, the leading coefficient of a polynomial cannot equal zero. For example, the polynomial f(x) = x^2 + 2x – 3 has a degree of 2 since the highest power that x is raised to in f(x) is 2 and the coefficient in front of x^2 is 1 and not zero. Also, the leading coefficient of f(x) is 1 since 1 is the coefficient in front of the highest power that x is raised to in f(x).
Application Problem (Created by Ben Mueller)
If all of the coefficients in the polynomial H(x), given below, are elements in Z4, state the degree and the leading coefficient of the polynomial H(x).
H(x) = [12]x^3 – [16]x^2 + [4]x + [8]
Solution (Provided by Ben Mueller):
I will admit that this problem is a bit tricky, but in Z4 this polynomial actually represents the zero polynomial.
At first glance, one might think that [12] is the leading coefficient, but [12] = [0] in Z4 and remember that the degree of a polynomial must have a non-zero leading coefficient.
Similarly, [16] = [0] in Z4, [4] = [0] in Z4, and [8] = [0] in Z4.
To answer the question, the polynomial H(x) represents the zero polynomial in Z4, has a leading coefficient of zero, and also, a degree of zero.
In general, F[x] denotes the set of all polynomials that have coefficients in field F. Recall that elements in a field have no zero divisors. For example, if x and y are both elements in the field F, xy cannot equal zero unless x = 0 or y = 0. From this, we can make the statement that Deg (g(x)f(x)) = Deg g(x) + Deg f(x).
Prove that if f(x)
and g(x) are non-zero elements of F[x] such that f(x) divides g(x) and g(x)
divides f(x), then f(x) = ag(x) for some non-zero a (constant).
Proof (Written by Ben Mueller):
If f(x) divides g(x), then g(x) = f(x)p(x) where p(x) is an element of F[x]. Also, we have that g(x) divides f(x), which means that f(x) = g(x)h(x) where h(x) is an element of F[x].
Now, to recap g(x) = f(x)p(x) and f(x) = g(x)h(x). From this, f(x) = g(x)h(x) can be rewritten as f(x) = f(x)p(x)h(x) simply by substituting f(x)p(x) in for g(x). Because the coefficients of our polynomials are contained in a field (fields cannot have zero divisors), f(x) = f(x)p(x)h(x) implies that the degree of f(x) must equal the degree of f(x)p(x)h(x). In other words, the polynomial produced by f(x)p(x)h(x) can not have a degree greater than the degree f(x) already has (Deg (f(x)p(x)h(x) = Deg f(x) + Deg p(x)+ Deg h(x)). Now, since we cannot talk about negative degrees, the only way that the statement f(x)p(x)h(x) = Deg f(x) + Deg p(x)+ Deg h(x) holds true is if the degree of p(x) is zero and the degree of h(x) is zero. Thus, h(x) = a, a being a constant and p(x) = b, b being a constant. Then f(x) = ag(x) and the proof is complete./(Proof complete)
Application Problem (Created by Ben Mueller)
Provide a polynomial p(x) such that the following is true:
Solution (By Ben Mueller):
2(x – 2)(x – 1)(x – i) would certainly give us a polynomial such that all of the above conditions hold true. Multiplying this polynomial out produces 2(x^2 – 3x + 2)(x – i) =
2(x^3 – ix^2 – 3x^2 + 3ix + 2x – 2i) = 2x^3 – (2i + 6)x^2 + (6i + 4)x – 4i.
p(x) = 2x^3 – (2i + 6)x^2 + (6i + 4)x – 4i
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