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Unformatted text preview: 538 I Chapter 6 Systems of Equations
El“
" ” PR DJ E 0 T5
1. FINDING ZEROS OF A POLYNOMIAL One zero of c. What is the slope of the line betweenO and P?
P(x) = x3 + 2x2 + Cx — 6 is the sum of the other two . , . _ .
zeros of Pa). Find C and the three zeros of Rx). d. Prove that line segment OP is perpendlcular to line
segment PQ.
2. PROVING .A GEOMETRY THEOREM Consider the triangle e. How can you conclude from the foregoing that w that is shown at the right inscribed in a circle of radius a triangle OPQ is a right triangle? with one side along the diameter of the circle. Prove that the triangle is a right triangle by completing the following steps. a. Let y =’mx,.. 2 O.‘ Show that the graph of y =mx,
m 2 0, intersects the Circle whose equation is 2a ' Zma > 1+mz'1+m2 (x—a)z+y2=a2,a>0,atP b. Show that the slope of the line through P and Q(2a, 0) W‘WE‘WMWSW rm? 9 PARTIAL FRACTION 'DECOMPOSITION An algebraic application of systems of equations isa technique known as partial :
fractions. In Chapter P, we reViewed the problem of adding two rational expres sions. For example,
' 5 . 1 I 6x‘ + 9
+ .
x —.1 x +2 (x — 1)(x + 2) T . '  Now we will take an opposite approach. That is, giVen a rational expression, We:
0 review RATIONAL . . . . . . . . .
EXPRESSIONS see p 56 Will find Simpler rational expressmns whose sum 15 the given expressmn. The
I method by which a more complicated rational expression is written as a sum of rational expressions is called partial fraction decomposition. This technique 15'
based on the following theorem. ' is a rationalexpréssion in which the degree of numerator is lessthan the degree of the denominator, and 7909) and q‘(x) have no comrhon factors)?
then f(x) can be Written as a partial fraction decomposition 'm the form
. , _ a . . .. .. f<x>;a=.ﬁ<,x>,,:+ ax) :w .+ an... a i _, ‘
' where each fi(x) has one70f the forms? ' ‘ ' ‘ ' ‘ . ' Bxisc.
or. 2 .l'Ptitj‘qt’iill l . ‘ Wit t C)?“ 6.4 Partial Fractions 539 The procedure for finding a partial fraction decomposition of a rational expression l } depends on factorization of the denominator of the rational expression. There are  four cases. ' I Case I Nonrepeated Linear Factors p _ — w l The partial fraction decomposition Will contain an expression of the form x I: a for
each nonrepeated linear factor of the denominator. Example:
l
l 3x—1 E hl' ft fthd ' t
“ ° ac Inear 3C 0” 0 e enomlna OI”
‘ x(3x + 4)(x _ 2) occurs only once. ' 4 Partial fraction decomposition;  p ' g l . " » " ‘ "T ' ‘ ' i 3x — 1' A B c
i *— = — + +
3 x(3le4)(x—2) x 3x+4 x—2 p l Case 2 Repeated Linear Factors m l 4 _ The partial fraction decomposition Will contai_n.a.n expression of the form
A1  A2 ' ' Am
' +’——+l....+.—
(x+a). (x+{z)2 (x+a)’"
foreach repeated linear factor of multiplicity m. Example: ' l
4H5 < 2r ( 2x 2)
° x — = x  x — ,
(x _ 2)2(_2x + 1) a repeated linear factor. 4
Partial fraction decomposition: I
, V 4x + 5 , _ A1 A2 + B
.e. 7 I :(x—2)2..(2x+1)"x—2 (x~2)2 2x+1
 Case 3 Nonrepeated Quadratic Factors ‘ I I ll
The partial fraction decomposition will contain an expression of the form ‘
' , ‘ Ax + B .
ax2 + bx + c ' r ll
for each quadratic factor irreducible over the real numbers. Example: ” A m x _ 4 . ____ , ' > i . p  ( 2 + + 1X ° X2 + x + 1 is irreducible over the real numbers. .
x x x — ‘ Partial fraction decomposition: x—él _ Ax+B + C
(x2+x+1)(x—4)_x2+x+l x—4 Case 4 Repated Quadratic Factors .
Thepariial fraction decomposition will contain an expression‘of the form Alx + B1 + Azx 'l" B2 + . + Amx 'l‘ Bm
axz + bx + c .(axz, + bx + c)2 '(axz + bx + c)“ li' ‘ 540 Chapter 6 Systems of Equations for each quadratic factor irreducible over the real numbers. Example: “ 2x
m ~(xz + 4)2 is a repeated quadratic factor. Partial fraction decomposition: 2x =.A1X+B1+A2X+B2+ C ‘ (xi 2x972 + 4)? x2 + 4 (x2 + 4r x  2. 9 QUESTION Which of the four cases of a partial fraction decomposition apply
" ~—~— I ' _ M_ ' v VN»»___ ‘ ‘4 v r $ ._ There are various methods for ﬁnding the constants of a partial fraction de H '
. composition. One such method is basedon a property of polynomials. 1" degree n, then p‘(x),'=_ r(x) if ‘ Find a Partial Fraction Decomposition
Alternative to Example 1 Case I: Nonrepeated Linear Factors
Find a partial fraction decomposition of
X_5 ' F'd tialfr a d 'a’ f x+11
 ' m a ar ac o ecom 03 o .
Xz”l p ' n P ion x2—2x—15 ‘ .
3 Z
9 _
X + 1 X _ 1 Solution
First factor the denominator. ,
xZ—Zx—15=(x+3)(x—5)
The‘f‘a‘ctors are nonrepeated linear factors. Therefore; the partial fraction
decomposition will have the form _ " '
INSTRUCTUH NOTE + 11 A B .'
Another way to approach solving for A and x = + (1)
. (x+3)(x—5)'x+3vx—‘5 " B is to evaluate the left and right sides of _
To solve for A and B, multiply each side of the equation by the least x+i1=A(X—5)+B(X+_3) _
 common multiple of the denominators, (x + 3) (x — 5).
at "convenient" values of X—that is, the V
valueglofxforwhich X+ 3 = 0 or 35+ 11 = A(x _ 5) + BO“ 'l' 3) ‘ 
X  57= 0. x + 11 = (A + B)x + (—5A + SB) {Combine like terms. .9 ANSWER Cases 1 and 4. Alternative to Example 2
'Find a partial fraction decomposition of . 6.4 ‘Partial Fractions Using the Equality of Polynomials Theorem, equate “coefficients of like
powers. The result will be the system of equations 1’: A+ B R lltht "1
n=~MABB 6“ ax‘ * Solving the system of equations for A and B, we have A = —1 and B = 2. Substituting —1 for A and 2.for B into the form of the partial fraction
decomposition (1), we obtain x+11 _ —1 + 2
«out ,7 F _v(x+3)(x—.5) x+3 »x—=5— You should add the two eXpressions to verify the equality. 'TRY‘Ex‘E'RCISE l4; PAGE"544 ' Find the Partial Fraction Decomposition I 217': Case 2: Repeated Linear Factors ' 2 + 2 + 7
Find a partial fraction decomposition of > , x(x — 1)
Solution The denominator has one~ nonrepeated factor and one repeated factor. The
partial fraction decomposition will have the form x2+2x+7_A B C xx—l)2 x+x—1+Z9:)—2
Multiplying each side by the LCDx0: — '1)2, we haye
x2 + 2x +7 =A(x  1)2 + B(x — 1)x + Cx
Expanding the right side and combining like terms gives
x2+2x+7=(A+B)x2+(—2A—B+C)x+A Using the Equality of Polynomials Theorem, equate coefficients of like
powers. This will result in the system of equations 1= A+B
=—2A—B+C
7= A. The solution is A = 7, 3= —6, and C = 10. Thus the partial fraction
decomposition is V x2+2x+7 1+ —6 ' 10
g" x(x—1)2 x +.__—
x—l (x—l)2 {TRY EXER'ISE‘ZZ,’ PAGE 544 . 54l _m 542 Chapter 6 Alternative to Example 3
Find a partial fraction decomposition of
3x2 + 6X — 21 " '
(X — 5)()(2 + 3)‘ 3 B o
x—5+(x2+3) INSTRUCTUR NOTE Example 3 can be solved in a manner that
is similar to the method discussed in the
instructor note accompanying Example 1. Substitute i\/7 for X in ~ 3X+ 16 = A(XZ + 7)
+ (BX + C)(X — 2) and solve tot Band be equating complex (numbers. This may be a good challenge problem for some students. Alternative to Example 4
Find a partialfraction decomposition of
2x3 — X2 — 6X —— 7 (Xz—X—3)Z
2x+1 + x—4”
Xzi—X3 (x?—)(3)2 Systems of Equations _ Find the Partial Fraction Decomposition
Case 3: No‘nrepeated Quadratic Factor 3r+16 Find the partial fractlon decomposmon of I Solution Because (at — 2) is a nonrepeated linear factor and x2 + 7 is an irreducible
quadratic over the real numbers, the partial fraction decomposition Will
have the form 3x + 16 A Bx + C — = +
" (x —2)(x2 l‘7) x — 2 at2 +7
Multiplying each side by the LCD (x h— 2)(x2 + 7) yields
' 3x + 16_= A(x2 + 7) + (Bx + C)(x — 2) Expanding the right side and combining like terms, we have
3x + 16 é (A + B)x2 + (—2B + C)x + (7A — 2C) Using the Equality of Polynomials Theorem, equate coefficients of like
powers. This will result in the system of equations ' Fb= A+ B Thinkof3x__+ioasoxé+3x+ia
‘ 3= —2B+ C
16=7A —‘2C The solution is A = 2, B = —2, and C = ~1. Thus the partial fraction
decomposition is 3x+1e '2 —2x—1 ———= ,+— (x—2)(x2+7) x—2 x2+7 ‘ {Existtentatsptaezs‘ms; ‘ Finda Partial Fraction Decomposition
Case 4: Repeated Quadratic Factors 4x3+5x2+7x—1 Find the partial fraction decomposition of (x2 + x + Dz Solution The quadratic factor (x2 + x + 1) is irreducible over the real numbers and is
a repeated factor. The partial fraction decomposition will be of the form v ‘ 4x3+5x2+7x—1__ Ax+B + Cx+D ' (x2+x+1)2 x2+x+1 (252+x+1)2 VMultiplying each side by the LCD (x2 + x + l)2 and collecting like terms, We.
obtain ‘ iv ' 4x3+5x2+7x—1=(Ax+B)(x2+x+1)+Cxp+D A =Ax3+Ax2+Ax+Bx2+Bx+B+Cx+D =Ax3+(A+B)x2+(A+B+C)x+(B+D) Alternative to Example 5
Find a partial fraction decomposition 'of
2x3 — iiXZ + 8X+9 X2—3X b ii'IRv{Examsesoxtliés15.4515» 6.4 Partial Fractions 543 Equating coefficients of like powers gives the system of equations 4=A
'5=A+B
7=A+B+C
—1= B +D Solving this system, we have A = 4, B = 1, C = 2, and D = —2. Thus the
partial fraction decomposition is 4x3+5x2+7x—1_4x+1 I 2x—2
(362+x+1)2 x2+x+1l(x2+x+1)Z The Partial Fraction Decomposition Theorem requires that the degree of the
numerator be less than the degree of the denominator. If this is not the case, use
long division to ﬁrst write the rational expression as a polynomial plus a remain
der over the denominator. ‘ ’ I Find a Partial Fraction Decomposition
When the Degree of the Numerator
Exceeds the Degree of the Denominator x3 — 4952 19x — 35 ‘ Find the partial fraction decomposition of F(x) = x2 _ 7x Solution Because the degree of the denominator is less than the degree of the
numerator, uselong division ﬁrst to obtain _  » ' 2x — 35
= + + —
P(x) x 3 x2 _ 7x p
2x — 35 ' The partial fraction decomposition of n will have the form n—mgmeng B x2—7x_xx—7)px x—7 Multiplying each side by x(x  7) combining like terms, we have . q
‘ 2x — 35 = (A + B)x + (714)
Equating coefﬁcients of like powers yields a 2= A+B
—35=—7A The solution of this system is A = 5 and B = —3. The partial fraction
decomposition is '_3__ 2_ __ _3 ‘x 4x 19x 35=x+3+g+ : x2—7x x x—7 . i > ' intimates 34,:PAGE 545:3: ...
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This note was uploaded on 04/13/2008 for the course MATH 2412 taught by Professor Matroy during the Spring '08 term at Alamo Colleges.
 Spring '08
 MatRoy
 Fractions

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