Before solving a quadratic or other higher degree polynomials, a clear distinction should be made between the two terms, equation and function. Let us consider the difference between a quadratic equation and a quadratic function. A quadratic function, say, f(P), has the general format as:
There is an infinite number of solution for the above equation, as an infinite number of ordered pairs (P, f(P)) can be written for the above equation.
But if we equate the above f(P) = 0, the situation fundamentally changes. Since the variable f(P) is now assigned a 0 value, the result is a quadratic equation in the single variable P.
Now that f(P) is restricted to 0 value, only a select number of P values satisfy and qualify as its solution values, namely, those solution values at which the parabola is figure above intersects the horizontal axis - on which f(P) is 0. The solution P values are often referred to as the roots of the quadratic equation f(P) = 0.
In general, given a quadratic equation of the form:
there are two roots, which can be obtained using the formula,
For
, we would get 
, which gives us identical repeated roots. For 
, we get the find the square root of the -ve numbers, which is not possible in real number system. So for those cases, no real-valued root exists.
Higher Degree Polynomials
For higher degree polynomials, one useful method that might work is that of factoring the function. Given a 3rd-degree polynomial equation, factoring results in three terms of the form (x-root) yields 3 roots. Generally, an nth-degree polynomial equation should a total of n roots. Second, and more important in root search is the relationship between the roots and the constant term: since the constant term must be the product of the three roots, each root must be a divisor of the constant term. This relationship can be formalized in the following theorem:
Theorem I Given the polynomial eqaution
where all coefficients are integers, and the coefficient of
is unity, if there exists integer roots, then each of them must be a divisor of 
.
Theorem II Sometimes because the coefficients of the highest degree term is not unity, we cannot apply Theorem I and rather we resort on to a more general theorem:
if there exists a rational root r/s, where r and s are integers without a common divisor except unity, then r is a divisor of, and s is a divisor of 
.
Note: An example is given below to show how this can be applied to a concrete polynomial function to derive its roots.
Theorem III Given a polynomial
if the coefficients
add up to 0, then x=1 is a root of the equation.
Example - Applying Theorem II to solve the quartic equation:
.
With
, the only possible values for the numerator r in r/s are the set of divisors {1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 12, -12}. And, with 
, the only possible values for s are the set of divisors {1, -1, 2, -2}. Taking each element in the r set in turn, and dividing it by each element in the s set, respectively, we find r/s can only assume the values
1, -1, 1/2, -1/2, 2, -2, 3, -3, 3/2, -3/2, 4, -4, 6, -6, 12, -12
Among these candidates for the roots, many fail to satisfy the given equation. Also, since we are solving a quartic equation, we can expect at most 4 of the listed r/s values to qualify as the roots. The four successful candidates turn out to be 1/2, 2. -2, and -3. According to factoring principle, we can thus write the quartic equation as
(x - 1/2) (x-2) (x+2) (x+3) = 0
=> (2x - 1) (x-2) (x+2) (x+3) = 0
There is an infinite number of solution for the above equation, as an infinite number of ordered pairs (P, f(P)) can be written for the above equation.
But if we equate the above f(P) = 0, the situation fundamentally changes. Since the variable f(P) is now assigned a 0 value, the result is a quadratic equation in the single variable P.
Now that f(P) is restricted to 0 value, only a select number of P values satisfy and qualify as its solution values, namely, those solution values at which the parabola is figure above intersects the horizontal axis - on which f(P) is 0. The solution P values are often referred to as the roots of the quadratic equation f(P) = 0.
In general, given a quadratic equation of the form:
there are two roots, which can be obtained using the formula,
For
Higher Degree Polynomials
For higher degree polynomials, one useful method that might work is that of factoring the function. Given a 3rd-degree polynomial equation, factoring results in three terms of the form (x-root) yields 3 roots. Generally, an nth-degree polynomial equation should a total of n roots. Second, and more important in root search is the relationship between the roots and the constant term: since the constant term must be the product of the three roots, each root must be a divisor of the constant term. This relationship can be formalized in the following theorem:
Theorem I Given the polynomial eqaution
where all coefficients are integers, and the coefficient of
Theorem II Sometimes because the coefficients of the highest degree term is not unity, we cannot apply Theorem I and rather we resort on to a more general theorem:
if there exists a rational root r/s, where r and s are integers without a common divisor except unity, then r is a divisor of
Note: An example is given below to show how this can be applied to a concrete polynomial function to derive its roots.
Theorem III Given a polynomial
if the coefficients
Example - Applying Theorem II to solve the quartic equation:
With
1, -1, 1/2, -1/2, 2, -2, 3, -3, 3/2, -3/2, 4, -4, 6, -6, 12, -12
Among these candidates for the roots, many fail to satisfy the given equation. Also, since we are solving a quartic equation, we can expect at most 4 of the listed r/s values to qualify as the roots. The four successful candidates turn out to be 1/2, 2. -2, and -3. According to factoring principle, we can thus write the quartic equation as
(x - 1/2) (x-2) (x+2) (x+3) = 0
=> (2x - 1) (x-2) (x+2) (x+3) = 0
No comments:
Post a Comment