Higher_Mathematics_Part_2
.pdfTheorem 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
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Pk (x) |
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Qm (x) |
Qm (x) |
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Or, equivalently, |
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Pn(x) = Qm(x) Pn–m(x) + Pk(x). |
Qm(x) is the divisor, Pn–m(x) is the quotient, and Pk(x) is the remainder. |
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Here |
Pn−m (x) is the integer part of the polynomial of ( n − m )th degree), |
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Pk (x) |
is a proper rational ratio. |
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Qm (x) |
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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,
x2 + 6x − 4 |
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x(x − 1)(x + 2) |
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x + 2 |
To determine the constants A, B, and C, we first combine the terms on the right side:
x2 + 6x − 4 |
A(x − 1)(x + 2)+ Bx(x + 2)+ C x(x − 1) |
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x(x − 1)(x + 2) |
x(x − 1)(x + 2) |
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
x2 + 6x − 4 |
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x(x − 1)(x + 2) |
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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:
Pn (x) |
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(x − a)m |
x − a |
(x − a)2 |
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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
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Mx + N |
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x 2 + px + q |
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If the denominator of |
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contains linear and quadratic factors, some of |
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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:
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P (x) |
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Bβ |
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Qm (x) |
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M μ x + Nμ |
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L x + S |
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+ … + |
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x2 + lx + s |
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Typical problems |
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Т.2 |
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1. Rewrite the functions in the factored form: |
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а) P (x) = 3x2 |
+ x − 14 ; |
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b) P (x) = x3 + 2x2 − x − 2 ; |
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c) P (x) = x4 |
− x3 − 6x2 + 14x − 12 . |
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Solution. а) as is generally known, ax2 + bx + c = a(x – x1)(x – x2). The roots |
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of the quadratic equation 3x2 + x − 14 = 0 are |
x = 2, x |
= − |
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P (x) = 3(x − 2)(x + |
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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
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x 4 − x3 − 6x 2 + 14x − 12 |
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x −2 |
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x 4 − 2x3 |
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x3 + x 2 − 4x + 6 |
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x3 −6x |
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x3 − 2x 2 |
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− 4x 2 + 14x |
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− 4x 2 + 8x |
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6x− 12 |
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6x − 12 |
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Hence, |
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0 |
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reminder |
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P (x) = (x − 2)(x3 + x 2 − 4x + 6) . |
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A number |
x = −3 is |
the root of |
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polynomial |
Q3 (x) = x3 + x 2 − 4x + 6 , |
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Q3 (−3) = −27 + 9 + 12 + 6 = 0 . Using |
the long-division of |
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Q3 (x) to х+3, we will have |
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x 4 − x3 − 6x 2 + 14x − 12 = (x − 2)(x + 3)(x 2 − 2x + 2) . |
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2. Express |
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x4 + 2x2 − 3x + 2 |
in the mixed-number form – polynomial plus |
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x2 |
+ 4 |
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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
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x4 + 2x2 − 3x + 2 |
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(x2 |
+ 4) − 2(x2 + 4) − (3x − 10) |
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3x −10 |
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x2 + 4 |
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x2 + 4 |
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3. Transform the given fraction into linear factors |
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3x 2 − 21x |
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x3 − 3x 2 − 6x + 8 |
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Solution. There are roots –2; 1 and 4 of |
Q3 (x) = x3 − 3x 2 − 6x + 8 . Then |
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Q3 (x) = (x − 1)(x + 2)(x − 4) . So, |
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3x 2 − 21x |
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3x 2 − 21x |
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x3 |
− 3x 2 − 6x + 8 |
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(x + 2)(x − 1)(x − 4) |
x + 2 |
x − 1 |
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hence |
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3x2 − 21x |
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A(x − 1)(x − 4) + B(x + 2)(x − 4) + C(x + 2)(x − 1) |
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(x + 2)(x − 1)(x − |
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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,
3x 2 − 21x |
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(x + 2)(x − 1)(x − 4) |
x + 2 |
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x − 4 |
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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:
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x 4 − 3x 2 − 3x − 2 |
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x(x3 |
− x2 |
− 2x) + x3 − x2 |
− 3x − 2 |
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x3 − x 2 − 2x |
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x3 − x2 − 2x |
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= |
x(x3 |
− x2 − 2x) + (x3 |
− x2 − 2x) − x − 2 |
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x + 2 |
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x3 − x2 |
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Step 2. If there are n such distinct linear factors, there will be n such partial
fractions. In our case |
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x + 2 |
and we have 3 distinct linear |
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x3 − x2 − 2x |
x(x − 2)(x + 1) |
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factors:
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x + 2 |
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A(x − 2)(x + 1) + Bx(x + 1) |
+ Cx(x − 2) |
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x(x − 2)(x + 1) |
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2 |
x + 1 |
x(x − 2)(x + 1) |
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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
x = 2 , then 4 = 6B and B = 2 / 3 . The last, |
if, |
x = −1, |
then 1 = 3C and |
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C = 1/ 3 . Thereby |
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x + 2 |
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x(x − 2)(x + 1) |
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3(x − 2) |
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Namely, |
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x 4 − 3x 2 − 3x − 2 |
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− x 2 − 2x |
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3(x −2) |
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5. Indicate the form of the partial-fraction representation of |
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(x − 1)(x + 1)2 |
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,
x 2 + 2 |
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x − 1 |
x + 1 |
(x + 1)2 |
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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
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setting x = –1; |
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setting x = 0. |
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and B = |
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Thus,
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x 2 + 2 |
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(x − 1)(x + 1)2 |
4(x − 1) |
4(x + 1) |
2(x + 1)2 |
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6. Indicate the form of the partial-fraction representation of |
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(x + 1)(x3 |
+ 1) |
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Solution. As is generally known (x + 1)(x3 + 1) = (x + 1)(x + 1)(x 2 − x + 1) = |
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= (x + 1)2 (x 2 − x + 1) . Since x + |
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x 2 − x + 1 is irreducible, we seek constants A, B, C, and D such that |
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x 2 |
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(x + 1)2 |
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Then |
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x 2 = A(x 2 − x + 1) + B(x 2 − x + 1)(x + 1) + (Cx + D)(x + 1)2 |
or |
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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,
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x2 : A + 2C + D = 1, |
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x : – A + C + 2D = 0, |
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x0 : A + B + D = 0. |
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Thereby |
A = |
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B = − |
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, D = 0 . |
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3 |
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Hence, |
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Т.2 |
Exercises for class and homework |
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Rewrite the functions in the factored form |
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1. |
P (x) = x2 |
− 2x − 8 . |
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2. P (x) = 2x2 − x − 10 . |
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3. |
P (x) = x3 |
+ x2 + x + 1. |
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4. P (x) = x5 |
− 16x . |
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5 |
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5. |
P (x) = x4 |
− 3x3 + x2 + 3x − 2 . |
6. P (x) = x4 |
+ 3x3 + 6x2 + 5x + 3 . |
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Express the given fractions in the mixed-number form |
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7. |
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x3 + 1 |
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8. |
x3 |
− x + 1 |
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x3 |
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x2 + 5x + 6 |
x2 |
+ x + 1 |
x + |
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Express the given expressions as the sum of partial fractions
10. |
x2 |
− 6x − 7 |
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x4 |
− 4x3 + 8x2 − 4 |
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12. |
8x3 |
+ 2x − 2 |
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(x2 + 4x − 5)(x |
− 3) |
(x + 1)(x2 − 2x)2 |
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− 1 |
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Answers
1. |
(x +2)(x −4) . |
2. (x + 2)(2x − 5) . |
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3. (x +1)(x2 +1) . |
4. x(x +2)(x −2)(x2+4) . |
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5. |
(x + 1)(x − 1)2(x − 2) . |
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6. |
(x2 + x + 1)(x2 + 2x + 3) . |
7. |
x − 5 + |
26 |
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x + 3 |
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8. |
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x − 1 + |
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x2 − 2x + 4 − |
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10. |
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x2 + x + 1 |
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12. |
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(x − 2)2 |
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x + 1 |
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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
Is degree of |
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Divide Qm(x) into Pn(x): |
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Pn(x) less |
no |
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Pn (x) |
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= Pn - m(x) + |
Pk (x) |
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than degree |
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, degree |
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Qm (x) |
Qm (x) |
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of Qm(x) ? |
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of Pk(x) is less than degree of Qm(x) |
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yes |
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for |
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Pk (x) |
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For each (ax + b)n write |
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Qm (x) |
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Express Qm(x) as product |
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of powers of first-degree |
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∑i 1 |
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(l) |
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(ax + b)i |
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and irreducible second- |
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degree polynomials |
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For each (ax + bx + c)m write |
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c j x + d j |
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∑ |
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(2) |
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Find constants ki, cj, dj such that the |
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+ bx |
+ c) |
j |
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j=1 (ax |
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sum of (l) and (ll) is |
Pn (x) |
Pk (x) |
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Qm (x) |
or |
Qm (x) |
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98 |
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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.
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Т.2 |
Individual test problems |
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2.1. Rewrite the functions in the factored form |
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2.1.1. |
P (x) = x3 |
+ x2 −2x −8 . |
2.1.2. |
P (x) = x3 |
−x2 −2x +8 . |
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2.1.3. P (x) = x3 |
−3x2 −4x +12 . |
2.1.4. |
P (x) = x3 |
+4x2 +5x +2 . |
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2.1.5. P (x) = x3 |
−7x2 +16x −12 . |
2.1.6. P (x) = x3 |
+5x2 +8x +4 . |
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2.1.7. P (x) = x3 |
+3x2 +6x −10 . |
2.1.8. P (x) = 2x3 |
−7x2 +8x −3 . |
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2.1.9. P (x) = 3x3 |
+4x2 |
−2x −5 . |
2.1.10. P (x) = 2x3 |
+3x2 −6x −7 . |
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2.1.11. P (x) = x4 |
+3x2 |
−4 . |
2.1.12. P (x) = x4 |
+2x2 −8 . |
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2.1.13. P (x) = x4 |
−2x3 |
−x2 +2x . |
2.1.14. P (x) = 2x4 |
+7x2 −9 . |
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2.1.15. P (x) = x3 |
+6x2 |
+13x +10 . |
2.1.16. P (x) = 5x4 |
−3x2 −8 . |
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2.1.17. P (x) = x4 |
−6x3 +11x2 −6x . |
2.1.18. P (x) = 4x4 |
−4x2 +1 . |
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2.1.19. P (x) = 4x3 + x2 |
−3x −2 . |
2.1.20. P (x) =9x4 |
+6x2 +1 . |
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2.1.21. P (x) = x3 |
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2.1.22. P (x) = x3 −x2 −4x −6 . |
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2.1.23. P (x) = x3 |
+2x2 |
+5x +24 . |
2.1.24. P (x) = 2x3 |
+ x2 −3x +6 . |
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2.1.25. P (x) =−x3 −7x2 +6x +2 . |
2.1.26. P (x) = 2x4 |
+3x2 −5 . |
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2.1.27. P (x) = 3x3 + x2 |
−9x −10 . |
2.1.28. P (x) = 4x4 |
−20x2 +25 . |
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2.1.29. P (x) =−x3 −x2 |
+11x +3 . |
2.1.30. P (x) = 9x4 |
+ 12x2 + 4 . |
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Тopic 3. Integration of rational functions by partial fractions
Integrating of partial fractions. Integrating of rational functions.
Literature: [1, section 6, ch. 6.4], [2, section 2, ch. 2.1], [3, section 7, § 1], [4, section 7, § 22], [6, section 8], [7, section 10, §§7—9], [9, § 31].
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Т.3 |
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Main concepts |
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Occur the following partial fractions: |
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ІІ. |
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ІІІ. |
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Mx + N |
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ІV. |
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where a, p, q, A, M , N are |
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numbers, n = 2, |
3, …, |
D = p 2 − 4q < |
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namely |
x 2 + px + q is irreducible. |
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Let’s consider the integrals of the partial fractions: |
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І. ∫ |
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A |
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dx = A∫ |
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d (x − a) |
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ІІ. ∫ |
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A |
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dx = A∫(x − a)−n d(x − a) = |
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+ C ( n ≠ 1 ). |
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(x − a) |
n |
(1− n)(x − a) |
n−1 |
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ІІІ. |
J1 |
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∫ |
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dx |
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( a ≠ 0 ). |
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ax2 + bx + c |
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Suppose a quadratic factor ax2 + bx + c 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.
First perform algebraic manipulations, as is shown:
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x + |
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ax 2 + bx + c = |
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= a |
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a |
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4a |
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x |
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b |
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= a |
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± m2 |
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2a |
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where |
± m |
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b2 |
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c |
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= − |
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− 4ac |
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D |
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. There is a sign of |
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being |
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4a 2 |
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4a 2 |
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opposite of a sign of a discriminate D . Then |
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d(x + |
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dx |
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1 |
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dx |
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1 |
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) |
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1 |
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dz |
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J1 = |
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2a |
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, |
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∫ ax2 + bx + c |
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∫ |
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(x + |
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2 |
± m |
2 |
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a ∫ |
(x + |
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2 |
± m |
2 |
a |
∫ |
z2 |
± m2 |
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) |
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) |
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b |
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2a |
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where |
z = x + |
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There are three cases: |
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1 |
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z |
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1) |
if D < 0 , then J |
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arctg |
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+ C ; |
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1 |
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am |
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