Higher_Mathematics_Part_2

Theorem 8. Any polynomial with real coefficients can be given as a product of linear and quadratic factors with real coefficients, namely

Pn (x) = a0 (x − x1 )k1 ... (x − xr )kr (x2 + p1 x + q1 )s1 ... (x2 + pm x + qm )sm .

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Moreover, k1 + k2 + ... + kr + 2(s1 + s2 + ... + sm ) = n , Di = pi2 − 4qi < 0, i = 1, m.

2.2. Rational functions

A rational function is a function that can be expressed as the ratio of two polynomial functions. That is, f(x) is a rational function if there exist polynomial functions Pn(x) and Qm(x) such that

f (x) = Pn (x) . Qm (x)

The domain of f(x) is the set of all x such that the denominator Qm(x) is not zero (otherwise the ratio would be undefined). The zeros of f(x) are those zeros of the numerator Pn(x) that are in the domain of f(x) since a ratio is zero if and only if its numerator is zero.

If the degree of the numerator Pn(x) is less than the degree of the denominator Qm(x), then f(x) is a proper rational function. If, on the other hand, the degree of the numerator Pn(x) is larger than or equal to the degree of denominator Qm(x), then f(x) is an improper rational function.

Just as an improper fraction can be rewritten as the sum of an integer and a proper fraction (for example, 215 = 4 + 15 ), we can rewrite an improper rational

function as the sum of a polynomial function and a proper rational function. Suppose that Pn(x) and Qm(x) are nonconstant polynomial functions such that

the degree n of Pn(x) is greater than or equal to the degree m of Qm(x). There

exist two unique polynomials Pn–m(x) and Pk(x) such that the degree k of Pk(x) is less than the degree of Qm(x) and

Given a rational algebraic expression in which a higher-degree polynomial is divided by a lower-degree polynomial, use long division to transform to mixednumber form – polynomial plus proper fraction.

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Observe that the denominator consists only of distinct linear factors, each factor occurring exactly once. It can be shown that to each such factor x - a there corresponds a partial fraction of the form

A (A is a constant)

x− a

such that the composite fraction is the sum of the partial fractions. If there are n such distinct linear factors, there will be n such partial fractions. For instance,

To determine the constants A, B, and C, we first combine the terms on the right side:

Since the denominators of both sides are equal, we may equate their

numerators:

x2 + 6x – 4 = A(x − 1)(x + 2)+ Bx(x + 2)+ C x(x − 1).

Although condition is not defined for x = 0, x = 1, and x = –2, we want to find values for A, B, and C that will make the last equation true for all values of x. That is, it will be an identity. By successively setting x equal to any three different numbers, we can obtain a system of equations which can be solved for A, B, and C. In particular, the work can be simplified by letting x be the roots of Qm(x) = 0, in our case x = 0, x = 1, and x = – 2. If x = 0 we have

–4 = A (–1) 2 + B 0 + C 0 = –2A, so A = 2.

If x = 1,

3 = A 0 + B 3 + C 0, so B = 1.

If x = – 2,

–12 = A 0 + B 0 + C (–2) (–3), so C = – 2.

Thus

If the denominator of Pn (x) contains only linear factors, some of which are

Qm (x)

repeated, then for each factor ( x - a)n, where n is the maximum number of times x - a occurs as a factor, there will correspond the sum of n partial fractions:

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Suppose a quadratic factor x2 + px + q occurs in Qm(x) and it cannot be expressed as a product of two linear factors with real coefficients. Such a factor is said to be an irreducible quadratic factor over the real numbers. To each distinct irreducible quadratic factor that occurs exactly once in Qm(x), there will correspond a partial fraction of the form

which are repeated

Qm (x) = a0 (x − a)α …(x − b)β (x 2 + px + q)μ …(x 2 + lx + s)ν ,

there will correspond the following sum of m (m = α +…+β + 2(λ +…+ ν) fractions:

b)P3 (x) = x2 (x + 2) − (x + 2) = (x + 2)(x2 − 1) = (x + 2)(x + 1)(x − 1) .

c)use the following rule.

If Pn (x) with integer coefficients has the integer roots, they are among divisors of an

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There are divisors of an : ±1; ± 2; ± 3; ± 6; ± 12 . It is true that P4 (2) = 16 − 8 − 24 + 28 − 12 = 0 ,

So, P4 (x) = (x − 2)Q3 (x) .

A rational algebraic expression such as

x4 − x3 − 6x2 + 14x − 12

x − 2

can be thought of as an improper fraction too, since its numerator is of a higher degree than the denominator is. The same long-division process can be used to transform this expression to a new polynomial. Here, step by step, is how it works

proper fraction.

Solution. Transform the nominator in the following form x4 + 2x2 − 3x + 2 = x2 (x2 + 4) − 2x2 − 3x + 2 =

= x2 (x2 + 4) − 2(x2 + 4) − 3x + 10 . 94

Then

Unknown constants А, В and С are determined from expression: 3x2 − 21x = A(x − 1)(x − 4) + B(x + 2)(x − 4) + C(x + 2)(x − 1) .

By theorem 5, an identity is an equation that is true for all values of the variable, namely

A+B +C = 3 ; х: −5A−2B +C =−21 ; x0 : 4A−8B −2C = 0,

we have A = 3 , B = 2 , C = −2 . Thus,

4. Express x 4 − 3x 2 − 3x − 2 in the mixed-number form. x3 − x 2 − 2x

Solution. Step 1. Since the degree of the numerator is not less than the degree of the denominator, carry out the integer part of a fraction:

Step 2. If there are n such distinct linear factors, there will be n such partial

factors:

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Therefore,

x + 2 = A(x − 2)(x + 1) + Bx(x + 1) + Cx(x − 2) .

Although condition is not defined for x = 0, x = –1, and x = 2, we want to find values for A, B, and C that will make the last equation true for all values of x. That is, it will be an identity. By successively setting x equal to any three different numbers, we can obtain they by letting x be the roots of Qm(x) = 0, in our case x = 0, x = –1, and x = 2. If x = 0 we have 2 = −2A , then A = −1 ; if

Solution. The degree of the numerator is less than the degree of the denominator (which is 3). Since the denominator is already factored (a common

occurrence in practice), the denominator (x − 1)(x + 1)2 has no second-degree

irreducible factors.

Since x + 1 appears twice in the factorization and x – 1 once,

To find the constants A, B and C remove the denominators by multiplying both sides by (x − 1)(x + 1)2 :

x 2 + 2 = A(x + 1)2 + B(x − 1)(x + 1) + C(x − 1) .

Three equations are needed to find the three unknowns A, B and C. To obtain them, replace x by three different numbers in turn. Since x + 1 when x = –1, and x – 1 when x = 1, replace x by –1 and then by 1. To obtain the third equation, use x = 0. Thus

Thus,

x2.= (B + C)x3 + (A + 2C +D)x2 + (–A + C + 2D)x + (A + B + D).

That is, coefficients of x3, x2 and x0 in both parts will be an identity. Therefore

x3 : B + C = 0,

Express the given expressions as the sum of partial fractions

Answers

Summary of this section

This flow chart describes the procedure for representing a rational function as a sum of partial fractions.

How to represent Pn ((x)) by partial fractions: Qm x

Useful facts. The number of unknown constants equals the degree of Qm(x). (Check that you have the right number.)

If b2 – 4ac < 0, ax2 + bx + c is irreducible. If b2 – 4ac ≥ 0, ax2 + bx + c is reducible.