SECTION 3.4 Complex Numbers
255
(c) Using the cubic formula.
3/-4
^
A'
|3
3':
+VT+27 +
-4_ /£
3^
2 " V 4 + 27
= \7-2 + \/4TT+ \/-2-v/4+~T = v/-2 + \/5+ V-2-v^
From the graphing calculator, we see that P (x) = a;3 + 3x+ 4 has one zero.
Using methods from this section, P (x) has possible rational zeros ±1, ±2, ±4.
1 I1 0 3 4
-1 I 1
114
114
4
0 3
4
-11-4
=> 1 is an upper bound.
1-14
0
=> x = -1 isa zero.
P(x) = x3 + 3.x + 4 = (x + 1) (x2 - x + 4). Using the quadratic formula we have:
-(-i)±v/(-D3-«(i)(-») _ i±^EH which is not a real number. Since it is not easy to see that
3/_2+ ^5 + ^-2 - \/5 = -1, we see that the factoring method was much easier.
Complex Numbers
5 - 7/: real part 5, imaginary part -7.
2. -6 + 4/: real part -6, imaginary part 4.
-2 - 5i
4.
= -| - |i: real part -§, imaginary part —|.
4 + 7?
•
7
= 2 + ^i: real part2, imaginary
part 5.
5. 3: real part 3, imaginary part 0.
6. -^: real part -\, imaginary part 0.
7. —§i: real part 0, imaginary part -|.
8. i\/3: real part 0, imaginary part y/S.
11.
^3+ 7=4 = v/3 +2/: real part n/3, imaginary part 2.
(2-5i) + (3 + 4i) = (2 + 3) + (-5 +4)i= 5-i
13.
(-6 + 60 + (9- i) = (-6 + 9) + (6 - l)i = 3 + 5i
9.
10. 2- v/=5 = 2- i-y/5: real part 2, imaginary part —s/5.
12. (2 +5z) + (4-6i) = (2 + 4) + (5-6)» = 6-i
14.
(3 - 20 + (-5 - }«) = (3 - 5) + (-2 - |) i = -2- }t
15.
3t + (6 - 4i) = 6 + (3/ - 40 = 6-7/
16.
17.
(i-iO + (i + W-(i + i) + H + i)<sl
(7 _ Ii) _ (5 + fi) = (7 - 5) + (-1 - I) i = 2- 2t
18.
(_4 + 0 - (2 - 50 = -4 + i - 2 + 5i = (-4 - 2) + (1 + 5) i = -6 + (n
19.
(_12+ 80 - (7 + 4i) = -12 + 8i - 7 - 4i = (-12 - 7) + (8 - 4) i = -19 + U
20.
6i- (4-i) = 6i-4 + i = (-4) + (6+ l)i = -4 + 7«
21.
22.
(0.1- 1.10 -(1-2 -3.60 = (0.1 - 1.2) + [-1.1 -(-3.6)] i = -1.1 +2.5/
23.
4(-l +2i) = -4 + 8/
27.
2i(| -i) =i-2r = 2 + /
(7 _ j)(4 + 2/) = 28 + 14/ - 4/ - 2/2 = (28 + 2) + (14 - 4) / = 30 + 10/
(5 - 30 (1 + 0 = 5 + 5/ - 3/ - 3/2 = (5 + 3) + (5 - 3)/ = 8 + 2/
(3 - 4/) (5 - 12/) = 15 - 36/ - 20/ + 48/2 = (15 - 48) + (-36 - 20) / = -33 - 56/
28.
(| + 12/) (I +24/) = §+ 16/ + 2/ +288Z2 = (j - 288) + (16 +2)1 = -2fi + 18/
24.
25.
26.
256
29.
30.
31.
CHAPTER 3 Polynomial and Rational Functions
(6+ 50 (2 - 3i) = 12 - 18* + lOi - 15i2 = (12 + 15) + (-18 + 10) i = 27 - Si
(-2 + 0 (3 - 70 = -6 + 14t + 3* - 7i2 = (-6 + 7) + (14 + 3)i = 1 + 17i
1
1
i
i
%
32.
33.
34.
35.
1+ i
i
2 -
i
_!
1
\-i
\-i
1+i
l-i
l-i2
37.
38.
39.
40.
2
2
5-i
5-i 3 - 4i _ 15 - 20i - 3t+ 4i2 _ (15 - 4)+ (-20-3) t _ 11 - 23i — i i 23.
3+ 4t 3 + 4t ' 3 - 4*
9 - 16i2
~
9+ 16
~ 25~ ~ 25 25'
26 + 39i _ 26_+39i 2+ 3i _ 52 + 78i + 78i + 117i2 (52 - 117) + (78 + 78) i __ -65 + 156t
2-3i
2- 3i "2+ 3i
4 - 9i2
~
4+ 9
~
il
= 13(-5 + 12Q
25 _ 25
4 - 3i
= -5 + 12i
4+ 3i _ 100 + 75i = 100 + 75i _ 100 + 75i _ 25 (4 + 3t) _
4 - 3i ' 4 + 3i
10i _
1 - 2i
16 - 9i2
16 + 9 ~
10i
1 + 2i
lOi + 20i2
-20 + lOi
1 - 2i
1 + 2t
1 -4i2
1 + 4
P-*'-' =2^31 =—
2-3i
25
~
5(-4 + 20
2 + 3i
2 + 3i
2 + 3i
4-9i2
4+ 9
13
-3
= —+ -2-i
= rl + -^« = 2-4t
-3 + 5i
-3 + hi
-3i + 5i2
-5 - 3i
-5
-3 . _ t
15i
15*
15i2
-15
-15
-15*~~3 +
1 + i
l-i
1
l-i
1
1+ i
l-i
l-i
(1 + 2Q (3 - Q _ 3 - i + 6i - 2i2
2 + i
2 + i
15 + 5i
3i
= -4 + 2*
2 + 3z
-6 + 4i
~4+
25
2 + 3i
4 + 6i _ 4 + 6i i
4t + 6ta
3i ~ 3i ' i '
3i2
41.
42.
2
2-3i _ 2 - 3i 1+ 2i _ 2+ 4t - 3i - 6i2 _ (2 + 6) + (4- 3)i _ 8 + i
l-2i
1 - 2i ' 1 + 2i
1 - 4i2
~
1+ 4
~ ~~5~~ or f + \i
13
36.
l-i
1+ 1
l+i =
11 +
+ ii
l-i
1+ i
l-i2
l-i2
5 + 5i 2-i
2+i
2-i
l-i , -l-i
~—
2 H
= —i
10- 5i + lOi- 5i = (10 + 5) + (-5 + 10) i
4-i2
5
= ¥ + t* = 3 + i
43.
i3 = i2. i = -1. i = -i
44.
(2i)4 = 24 •i4 = 16 •(1) = 16
45.
i100 = (i<)25 = (l)2= = i
46.
.002 = ^ 4\250
47.
'=25 = 5i
48.
4
.2
v250
= (iru-(-i)
= -i
2"
49.
v/=3v/=12 = i\/3 •2ix/3 = 6i2 = -6
50.
yji\f=2r? =yfl •3\/3i =3i
51.
(3-^/=5)(l +v^I) = (3-iy5)(l +0 =3+3i-iV5-i2^/5=(3 + V5) + (3-^^)
52.
!->/-! _ l-i _ l-i l-i _ 1- i - i +i2
l + ^/=l
1+ i
1+ i
l-i
l-i2
(1 - 1) + (-1 - l)j _ -2i _
1 + 1
53.
2_+yg _ 2_+2iy/2 _ 2+W2 l-iV2 _ 2-2t>/5 +2i>/5-4t2
54.
(x/3 - >/=4) (\/6 - >/=§) = (>/5 - 2i) (x/6 - 2v/2i) = v/l8 - 2y/6i - 2y/6i + 4V2i2
l + x/^
l + iv/2
1+ iv^ l-iv/2
l-2i2
(2 +4) + (-2y/2 +2y^) i 6_
~
= (3\/2 - 4^) + (-2VS - 2n/6) i = -v^ - 4v/6i
1+ 2
= 3= 2
SECTION3.4 Complex Numbers
y^36 _
6i
2_ iy/2 _ 2i-y/2 = iy/2 _ _.^
' x/=2\/=9 ~ iy/2•3i ~ iy/2 iy/2
2i2
„ y^77=49 (y/fi) (7Q _ 7i2 _ 7
56'
V2S
~
57. x2 + 9 = 0
2x/7
4*
" 2
a;2 = -9
58. 9x2 + 4 = 0 «*>
2
=>
9x2 = -4
-1
x = ±3i
«• x2 = -g
=•
x = ±§i
59.x2-4x +5=0 => x= -(-4)±V(-y-lilM = 4J^EM =l^H = 4±2i =2±i
60.x2 + 2x + 2= 0 => x = -(2)±V(2)--^^2.) = ^ O = ^^H = ^±21 = -i ±i
ci
2,
,1
n
_*.
..
-(l)±y/(l)2-4(l)(l) _ -i±^/I=l _ -i±^/=3 _ -i±«>/g _
l • ;^3
61. X +X + 1=0 =* X=
3L2{1)
~~
2
—
2
~~2^ ~ 2ll2
CO 2 o,o n _^ „_ -(-3)±^(-3)2-4(l)(3) _ 3±^/g=12^ _ 3±N/=5 _ 3±v/3t _ 3 . {&
62. x" - 3x + 3 = 0 =$• x =
2(Y)
— 2
— 2 — 2 —2 ^* 2
c, 0 2 o_i_1 n
63. 2x -2x + l=0
64. 2x2 + 3 = 2x
X=
-^
=>
«•
-r- -<-2):iV(-2)a-«(a)<l) _ 2±nA^5 _ 2±^/=4^ _ 2±2i _ i + ii
x=
5L2(2)
4
— 4 — 4 — 2 x 2l
2x2 - 2x + 3 = 0
=>
-(-2)±<y/(-2)2-4(2)(3) _ 2±.>/4^23 _ 2±^/=50 _ 2±2v^5t _ 1 . &i
2(2)
—
4
—
4
—
4
— 2^2*
£
66. 2 + 4 + —
=
0
«»
z2 + 4z + 12
=
0
=•
z
_ -(4)±V(4)2-4(1)(12) _ -4±n/16-48 _ -4±y^52 _ -4±4n/2» _ _0 + 2\/2i
2—
2(1)
—
67. 6x2 + 12x + 7 = 0 =•
-(12)±-v/(12)2-4(6)(7)
^=
3L2(6)
68. 4x2 - 16x + 19 = 0
=
2
2
2
-12±Vl44-l68 _ -12±^/=24 _ -12±2i>/6 _ ^12 . 2ii/5 _
12
—
12
—
12
— 12 ^
12
i + ^6,
±""- 6
=>
_ -(-16)±y/(-16)2-4(4)(19) _ 16±^S6-304 _ 16±y/=45 _ 16±4v^3i _. 16 _j_ 4^ _ 2±i&
X—
co i 2
2(4)
—
.En-.
69. ^x -x + 5 = 0
=4»
70. x2 + |x + l=0
=»
x=
8
8
8
8
8
2
-(-i)±y(-D2-4U)(5) _ 1±yrrio _ , j-x _ j, , 3•
27T)
—
1
~~
v
—
-1^/H)2-4(1)(1) ^^.^^.^ijjgi.^
_ _i + 201
—
4^4
x~~
2(1)
2
2
2
71. LHS = z + w = {a + bi) + (c+ di) = a- bi + c- di= (a + c) + {-b- d)i = {a + c) - {b + d)i.
RHS = ~z + w = {a+ bi) + {c+ di) = {a+ c) + {b + d)i = {a+ c)-{b + d) i.
Since LHS = RHS, this proves the statement.
72. LHS = zw = {a + bi){c + di) = ac + adi + bci + bdi2 = {ac - bd) + {ad + be) i = {ac - bd) - {ad + be) i.
RHS = z •w = a + bi •c + di = {a-bi){c- di) = ac- adi -bci + bdi2 = {ac - bd) - {ad + be) i.
Since LHS = RHS, this proves the statement.
73. LHS ={zf =((a +bi))2 =(o - bi)2 =a2 - 2a6i +62i2 =(a2 - b2) - 2a6i.
RHS = I2 = Ja + bif = a2 + 2abi + b2i2 = (a2 - b2) + 2abi = {a2 - b2) - 2abi.
Since LHS = RHS, this proves the statement.
74. f = a + bi = a- bi = a + bi = z.
75. z + ~z = (a + bi) + {a + bi) = a + bi+ a-bi = 2a,which isa real number.
76. z - z = (a + bi) - {a+ bi) = a + bi - (a - bi) = a + bi - a + bi = 2bi, which is a pure imaginary number.
257
258
CHAPTER 3 Polynomial and Rational Functions
77. z z = (a + bi) • (a + bi) = {a + bi) • (a - bi) = a2 - &2t2 = a2 + &2, which isa real number.
78. Suppose z = z. Then we have (a + bi) = {a + bi)
=>
a + bi = a - bi
=*>
0 = -26/
=>
b= 0, so 2 is real.
Now if 2 is real, then 2 = a + 0Z(where a is real). Since z = a- Oi, we have z = z.
79. Using the quadratic formula, the solutions to the equation are x= ~J± V*2a ~Aac. since both solutions are imaginary,
we have b2 - 4ac <0 *> 4ac - 62 >0, so the solutions are x= —±
^-^—/, where y/4ac - b2 is areal
2a
2a
number. Thus thesolutions arecomplex conjugates ofeach other.
80. / = i, i5 = i4 . i = f, »»=«•-• = <; /2 = -l, j6 = j4 . {2 = _h .10 = -8 . .2 = _1;
i3 = -/, i» = (* .,* = _i} jll = j8 .,-3 = _i; .--I = ^ -8 = -4 . .4 = ^ -12 = -8 . -4 = LBecause .4 = ^we hav£
/" = /'', where r is the remainder when nis divided by 4, that is, n = 4•k+ r, where kis an integer and 0 < r < 4. Since
4446 = 4-1111+2, we must have i4446 = i2 = -1.
81. (-l+n/3)3 (-1 +/v/3)3 ={-l +iVs) {-l+ix/3) (-l+iy/3) =(-1 +/V/3) (1 - 2iy/3 +3i2)
= (-l + /v/3)[l-2/v/3 + 3(-l)] = (-l +?:v/3)(-2-2/v/3)
= 2 + 2/\/3 - 2/^ -2i2- 3 = 2 + 6 = 8
(-1 - /v/3)3 =(-1 - jV3) (-1 - is/3) (-1 - is/3) =(-1 - iV§) (1 +2/v/3 +3i2)
= (-1 - ix/3) (-2 + 2iy/3) = 2 - 2iy/3 + 2iyft -2$ -3 = 2+ 6 = 8
Thus 8 has at least three cube roots (one real and two complex). Two fourth roots of 1G are ±2. If we calculate
(2/)', we get 24 •/'' = 16 •(-1)2 = 16, so 2/ is a fourth root of 16. Also, -2/ is a fourth root of 16, because
(-2/)" = (-l)4(2/)4 = 1.16 = 16.
\
3.5
Complex Zeros and the Fundamental
Theorem of Algebra
1. (a) x4 + 4.x-2 = 0 ^
.r2 (a-2 + 4) = 0. So x = 0or a;2 + 4 = 0. Ifx2 + 4 = 0then x2 = -4 &x = ±2i.
Therefore, the solutions are x = 0 and ±2/.
(b) To get the complete factorization, we factor the remaining quadratic factor P(x) = x2 (x +4) = x2 (x - 2/) (x +2i).
2. (a) x5 +9x3 = 0 <* x3 (x2 +9) = 0. So x = 0or x2 +9= 0. Ifx2 +9= 0then x = ±3/. Therefore, the zeros
ofParex = 0, ±3i.
(b) Since -3/ and 3/ are the zeros from x2 +9 = 0, x+3/ and x- 3/ are the factors ofx2 + 9. Thus the complete
factorization is P (x) = x3 (x2 + 9) = x3 (x + 3/)(x - 3/).
3. (a) x3 - 2x2 + 2x = 0 «. x (x2 - 2x + 2) = 0. So x = 0or x2 - 2x + 2 = 0. Ifx2 - 2x + 2 = 0then
r - -(-2)±V^(-2)3-4(D(2) _ 2±V^4
* _
2
2±2i
.,.-
,
.
.
n H, .
- —2— = ~2~ = 1 ± t. Therefore, the solutions are x = 0,1 ± i.
(b) Since 1- i and 1+ / are zeros, x - (1 - i) = x - 1+ / and x - (1 + •/) = x - 1- i are the factors ofx2 - 2x + 2.
Thus the complete factorization is P (x) = x (x2 - 2x + 2) = x (x - 1+ i) (x -l-i).
4. (a) x3 +x2 +x = 0 x (x2 + x + 1) = 0 . So x = 0 or x2 + x +1=0. Ifx2 + x + 1 = 0 then
_ ~(1)±\/(1)2-4(1)(1) _ -I±yT3
•x ~
27T)
=
I..VJ-
,
,
c„
-
n
/o
2— = ~2±l2- Therefore, the zeros ofP are x = 0, -J ±i%p.
(b) The zeros of x2 + x +1 = 0are -\ - i& and -\ +i&, so factoring we get x2 + x +
1=[x - (-§ - Pf)] [x - (-1 +/^)] =(x- +1+i^3) (x +§- i&y Thus the complete factorization
is P(x) =x(x2 +x +1) =x(W§+p£) (* +§- i^y
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
5. (a) x4 + 2x2 + 1= 0 «»
259
(x2 + l)2 = 0 «*• x2 + 1= 0 <*=> x2 = -1 &x = ±i.Therefore the zeros of
P are x = ±i.
(b) Since -i and i are zeros, x + i and x - i are the factors of x2 + 1. Thus the complete factorization is
P(x) = (x2 + l)2 = [(x + i)(x - i)]2 = (x + i)2 (x - i)2.
6. (a) x4-x2-2 = 0 *>
(z2-2) (x2 + l) = 0. Sox2 - 2 = Oorx2 + 1 = 0. Ifx2 - 2 = 0 then x2 = 2 <&
x = ±\/2. And ifx2 + 1=0 then x2 = -1
«•
x = ±i. Therefore, thezeros of P arex = ±\/2, ±i.
(b) To get the complete factorization, we factor the quadratic factors to get
P{x) = (x2 - 2) (x2 + 1) = (x - v/2) (x+ x/2) (x- i) (x+ i).
7. (a) x4 - 16 = 0 & 0 = (x2 - 4) (x2 + 4) = (x - 2) (x+ 2) (x2 + 4). So x = ±2 orx2 + 4 = 0. Ifx2 +4 = 0
then x2 = -4
=$>
x = ±2i. Therefore thezeros of P are x = ±2, ±2i.
(b) Since -i and i are zeros, x + i and x - i are the factors of x2 + 1. Thus the complete factorization is
P (x) = (x - 2)(x+ 2) (x2 + 4) = (x - 2)(x+ 2)(x - 2i) (x + 2i).
8. (a) x4 + 6x2 + 9 = 0 <&
(x2 + 3)2 = 0 & x2 = -3. So x = ±iy/3 are the only zeros of P (each of
multiplicity 2).
(b) Toget the complete factorization, we factorthe quadratic factorto get
P{x) = (x2+3)2 = [(x-ix/3) (x + iV3)]2 = (x-iV3)2 {x +iy/3)2.
9. (a) x3 + 8 = 0 «•
(x+ 2) (x2 - 2x + 4) = 0. So x = -2 or x2 - 2x + 4 = 0. Ifx2 - 2x + 4 = 0 then
x= -(-2)±>/(-2)2-4(i)(4) = 2±^EH = 2±^5 = 1±iy/^ Therefore, the zeros ofPare x= -2,1 ±iy/3.
(b) Since 1- iy/3 and 1+ iy/3 are the zeros from the x2 - 2x + 4 = 0, x - (l - iy/3) and x - (l + iy/3) are the
factors ofx2 - 2x + 4. Thus the complete factorization is
P{x) = (x + 2) (x2 - 2x + 4) = (x + 2) [x - (1 - iy/3)] [x - (l + iV3)]
= (x+ 2) (x - 1+ iy/3) (x - 1- iy/3)]
10. (a) x3 - 8 = 0 &
(x- 2) (x2 + 2x + 4) = 0. So x = 2or x2 + 2x + 4 = 0. Ifx2 + 2x + 4 = 0 then
x= -(2)±V(y-4(D(42 = -2±y^n = ^2^5 = _1±i^ Therefore, the zeros ofPare x= 2, -1 ±iy/3.
(b) Since -1 - iy/3 and -1 + iy/3 are the zeros from x2 + 2x + 4 = 0, x - (-1 - iy/3) and x - (-1 + iy/3) are the
factors ofx2 —2x + 4. Thus thecomplete factorization is
P(x) = (x-2)(x2 + 2x + 4) = (x- 2) [x - (-1 - iy/3)] [x - (-1 +iy/3)]
= (X - 2) (x + 1+ iy/3) (x + 1 - iy/3)
260
CHAPTER 3 Polynomial and Rational Functions
11. (a) x6-1=0 <* 0= (x3 - 1) (x3+ l) =(x-l)(x2+x +l)(x +l)(x2-x+l). Clearly^ =±1 are
solutions. Ifx2 +x +1=0>thenx=^±^lil2m =^O =_I ±^ sox =_i ±-^ ^ jf
x2-x + l = 0,thenx=ii^i^
2 a: t 2 , 2 n: z 2 .
(b) The zeros of x2 + x + 1 = 0are -\ - i^ and -± + i^, so x2 + x + 1 factors as
[*-(-i-^)][*-(-§+^)] =fc+i+^fc+i-^). Similarly.since
the zeros of x2 - x + 1 = 0are \ - i& and ±+ i^, so x2 - x + 1 factors as
[* ~Q-^)] [x - (* +iJ?)] =(* ~£+*f)(* "i"^). Thus the complete
factorization is
P{x) = (x-l)(x2 + x+ l)(x + l)(x2-x + l)
= (x-l)(x +l)(x+I+if)(x+I-i^)(x-I+i^)(x-I-i^)
12. (a)x6-7x3-8 =0 * 0=(x3-8) (x3 +l)=(x-2)(x2 +2x +4)(x +l)(x2-x +l). Clearly,
x = -1 and x = 2are solutions. Ifx2 + 2x + 4 = 0, then x = ~2:tV4~4(1)(4i
= -a±^/=n
_ _2
2 .2
2^^5
2
2
*=-l±v^Ifx2-x+l =0,menx=±±i/^^
of P are x = -1,2, -1 ± y/3i, ± ± i&.
(b) From Exercise 10, x2 + 2x + 4 = (x + 1+iy/3) (x + 1- iy/3) and from Exercise 11,
x2-x +l=(x-i+ i^j (x-\- i^pj. Thus the complete factorization is
P{x) = (x-2)(x2 + 2x + 4)(x + l)(x2-x + l)
= (x-2)(x +l)(x +l+ix/3)(x +l-iV3)(x-i+i^)(x-i-i^)
13. P(x) = x2 + 25 = (x - 5i) (x +5i). The zeros ofPare 5i and -5i,both multiplicity 1.
14. P(x) = 4x2 +9= (2x - 3i) (2x +3i). The zeros ofPare fi and - §i, both multiplicity 1.
15. Q(x) = x2 +2x +2. Using the quadratic formula x = "(2)±^2)1'-4(1-H2> = -*±V=i = -2±2i = _1±L So
Q(x) = (x + 1- i)(x + 1+ i). The zeros ofQare -1 - i (multiplicity 1) and -1 + i (multiplicity 1).
16. Q(x) =x2-8x +17=(x2-8x +16)+l =(x-4)2 +l = [(x-4)-i][(x-4)+i] = (x-4-i)(x-4 +i).
The zeros ofQ are 4 + i and 4 - i, both multiplicity 1.
17. P(x) = x3 +4x = x(x2 +4) = x(x - 2%) (x +2i). The zeros ofP are 0, 2i, and -2i(all multiplicity 1).
18. P (x) = x3 + x2 + x = x (x2 + x + 1) . Using the quadratic formula, we have
x=-(1)dV(ff-4(D(i) =l±^l =I ±i^. The zeros ofPare 0, \ +i^, and \ - i&, all ofmultiplicity 1. And
P(x) =x(x-i-if)(x-I+if).
19. Q(x) = x4-1 = (x2-l)(x2 + l) =(x-l)(x +l) (x2 + l)=(x -l)(x + l)(x-i) (x +i).The zeros ofQare
1, -1, i, and -i (all of multiplicity 1).
20. Q(x)=x4-625=(x2-25)(x2+25) = (x-5)(x +5)(x2 +25) = (x-5)(x+5)(x-5i)(x +5i).The
zeros of Q are 5, -5, 5i, and5i, all multiplicity 1.
SECTION 3.5 Complex Zeros and theFundamental Theorem ofAlgebra
261
3,
21. P (x) = 16x4 - 81 = (4x2 - 9) (4x2 + 9) = (2x - 3) (2x + 3) (2x - 3i) (2x + 3i). The zeros of P are f, -§,3 fi,
and -§i (all ofmultiplicity 1).
22. P (x) = x3 - 64 = (x - 4) (x2 + 4x+ 16). Using the quadratic formula, we have
_ -4±^/i6-4(i)(i6)
2
-4±/=45
= -4±4V5i
= _2 ± 2y/3i.
The zeros of P are 4, -2 + 2V^i, and -2 - 2^i, all of
2
2
v
j, —
multiplicity 1. And P (x) = (x- 4) (x+ 2- 2y/3i) (x+ 2+ 2^).
23. P(x)=x3+x2 + 9x + 9 = x2(x + l) + 9(x+ l) = (x + l)(x2+9) = (x + 1) (x - 3i) (x+ 3i). The zeros ofP
are -1,3i, and -3i (all of multiplicity 1).
24. P (x) = x6 - 729 = (x3 - 27) (x3 + 27) = (x - 3) (x2 +3x +9) (x + 3) (x2 - 3x + 9). Using the
.
- ^
,
2 . o
, n
u
-3±V9-4(1)(9)
,
. -
,
2
, n
U
3±yfa-4(l)(0) _ 3±v/=27 _ 3 . n/=2T _ 3 , 3^3 •
quadratic formula on x2 + 3x + 9 we have x =
o
x-$
quadratic formula on xz - 3x + 9 we have x = —*—5
=
-3±x/=27 _
J
3, y/=Tf I ]t.:na thp
- _2 ±
2^- usmS tne
- —2— ~ 2 ± "2^ _ 2± 2 *•
The zeros of P are 3, -3, -§ + ^i, -§ - ^i, § + ^i, and § - ^i, all multiplicity 1. And
P(x) =(x-3)(x +3)(x+|-3#i)(x+| +3#i)(x-l-3^i)(x-l +3#i).
25. Q(x) = x4 + 2x2 + 1= (x2 + l)2 = (x - i)2 (x + i)2. The zeros ofQare i and -i (both ofmultiplicity 2).
26. Q(x) =x4 +10x2 +25 =(x2 +5)2 = [(x - iy/b) (x +iy/5)]2 =(x - is/hf (x +iy/5)2. The zeros ofQare i^5
and —iy/5, both of multiplicity 2.
27. P (x) = x4 + 3x2 - 4 = (x2 - 1) (x2 + 4) = (x - 1) (x + 1) (x - 2i) (x + 2i). The zeros ofP are 1, -1,2i, and -2i
(all of multiplicity 1).
28. P(x) = x5 +7x3 = x3 (x2 +7) = x3 (x - iy/l) (x +iy/f). The zeros of P are 0(multiplicity 3), iy/7 and -iy/f,
both of multiplicity 1.
29. P(x) = x5 +6x3 +9x =x(x4 +6x2 +9) = x(x2 +3)2 = x(x - y/3i)2 (x +y/3if''. The zeros of Pare 0
(multiplicity 1), ^3i (multiplicity 2), and -y/3i (multiplicity 2).
30. P(x) = x6 +16x3 +64 = (x3 +8)2 = (x +2)2 (x2 - 2x +4)2. Using the quadratic formula, on x2 - 2x + 4we
have x= -(-2)±V(-2)2-4(i)(4) = 2±^rT5 = 2±^1 = i ±iN/3. The zeros of P are -2,1+ iy/3, and 1- i^3, all
multiplicity 2.
31. Since 1 + i and 1 - i are conjugates, the factorization of the polynomial must be
P (x) = a(x - [1 + i]) (x - [1 - i]) = a (x2 - 2x + 2). Ifwe let a = 1, we get P (x) = x2 - 2x + 2.
32. Since 1 + iy/2 and 1 - iy/2 are conjugates, the factorization of the polynomial must be
P(x) =c(x-[l +iV2])(x- [l-i\/2]) =c(x2-2x +3).Ifweletc=l,wegetP(x)=x2-2x +3.
33. Since 2i and -2i are conjugates, the factorization of the polynomial must be
Q(x) = b{x - 3) (x - 2i) (x + 2i] = 6(x - 3) (x2 +4) = b(x3 - 3x2 + 4x - 12). Ifwe let b= 1, we get
Q(x) = x3 - 3x2 + 4x - 12.
34. Since i is a zero, by the Conjugate Roots Theorem, -i is also a zero. So the factorization ofthe polynomial must be
Q(x) = 6(x + 0) (x - i)(x + i) = bx {x2 + 1) = b(x3 + x). Ifwe let b= 1, we get Q(x) = x3 + x.
35. Since i isa zero, by the Conjugate Roots Theorem, -i is also a zero. So the factorization ofthe polynomial must be
P (x) = a(x - 2) (x - i)(x + i) = a (x3 - 2x2 + x- 2). If we let a = 1, we get P (x) = x3 - 2x2 + x- 2.
262
CHAPTER 3 Polynomial and Rational Functions
36. Since 1+ i is azero, by the Conjugate Roots Theorem, 1- i is also azero. So the factorization ofthe polynomial must be
Q(x) = a(x + 3) (x - [1 +«]) (x - [1 - i]) = a(x + 3) (x2 - 2x + 2) = a(x3 + x2 - 4x + 6). Ifwe let a = 1, we
get Q (x) = x3 + x2 - 4x + 6.
37. Since the zeros are 1 - 2i and 1 (with multiplicity 2), by the Conjugate Roots Theorem, the other zero is 1 + 2i. So a
factorization is
R{x) = c(x - [1 - 2z]) (x- [1 + 2i]) (x- l)2 = c([x - 1] + 2i) ([x - 1] - 2i) (x- l)2
= c([x-l]2-[2i]2)(x2-2x +l)=c(x2-2x+l+4)(x2-2x+l)=c(x2-2x+5)(x2-2x +l)
= c(x4 - 2x3 +x2 - 2x3 +4x2 - 2x +5x2 - lOx +5) =c(x4 - 4x3 +10x2 - 12x +5)
Ifwe letc = 1 weget R (x) = x4 - 4x3 + 10x2 - 12x+ 5.
38. Since S (x) has zeros 2iand 3i, by the Conjugate Roots Theorem, the other zeros ofS (x) are -2i and -3i. So a
factorization of S (x) is
5(x) = C(x-2i)(x +2i)(x-3i)(x +3i) = C(x2-4i2)(x2-9z2)=C(x2+4)(x2 +9)=C(x4 +13x2+36)
Ifwe letC = 1, wegetS (x) = x4 + 13x2 + 36.
39. Since the zeros are i and 1 + i, by the Conjugate Roots Theorem, the other zeros are -i and 1 - i. Soa factorization is
T{x) = C(x-i)(x + i)(x-[l + i])(x-[l-i])
= C(x2-i2)([x-l]-i)([x-l] +i) =C(x2 +l)(x2-2x +l-i2)=C(x2 +l)(x2-2x +2)
= C(x4-2x3 +2x2+x2-2x +2)=C(x4-2x3+3x2-2x +2)=Cx4-2Cx3 +3Cx2-2Cx + 2C
Since the constant coefficient is 12, it follows that 2C = 12 <* C = 6, and so
T(x) = 6(x4 - 2x3 + 3x2 - 2x + 2) = 6x4 - 12x3 + 18x2 - 12x + 12.
40. Since U(x) has zeros ~, -1 (with multiplicity two), and -i, by the Conjugate Roots Theorem, the other zero isi. So a
factorization of U (x) is
C/(x) =c(x-i)(x +l)2(x +i)(x-i) = ic(2x-l)(x2 +2x +l)(x2 +l) = iC(2x5 +3x4 +2x3 +2x2-l)
Since the leading coefficient is 4, we have 4 = \c (2) = c.
Thus we have
U{x) = \ (4) (2x5 + 3x4 + 2x3 + 2x2 - l) = 4x5 + 6x4 + 4x3 + 4x2 - 2.
41. P (x) = x3 + 2x2 + 4x + 8= x2 (x + 2) + 4(x + 2) = (x + 2) (x2 + 4) = (x + 2) (x - 2i) (x + 2i). Thus the zeros
are —2 and ±2i.
42. P (x) = x3 - 7x2 + 17x - 15. We start by trying the possible rational factors ofthe polynomial:
1 | 1 -7 17 -15
1
1
-6
11
-6
11
-4
3 | 1 -7
3
17 -15
-12
15
1-450
=> x = 3 is a zero.
So P(x) = (x- 3)(x2 - 4x+ 5). Using the quadratic formula on the second factor, we have
x= 4±Vi6-4(i)(5) = 4±£3 = 4±2i =2:fci> Thus the zeros are 3,2 ±i.
SECTION 3.5 Complex Zeros and theFundamental Theorem ofAlgebra
263
\3. P (x) = x3 - 2x2 + 2x - 1. By inspection, P (1) = 1- 2+ 2- 1= 0, and hence x = 1is azero.
1 | 1 -2
2-1
1
-1
1
1-110
Thus P(x) = (x - 1) (x2 - x + 1). Sox = 1or x2 - x + 1= 0.
Using the quadratic formula, we have x= ^V1"4*1^ = i±i^. Hence, the zeros are 1and ±±§^.
44. P (x) = x3 + 7x2 + 18x + 18 has possible rational zeros ±1, ±2, ±3, ±6, ±9, ±18. Since all ofthe coefficients are
positive, thereare no positive realzeros.
—111 7 18 18
—211 7 18 18
-311 7 18 18
-1 -6 -12
-2 -10 -16
-3 -12 -18
1
6 12
6
15
8
2
14
6
0 => x = -3 is a zero.
So P (x) = (x - 3) (x2 + 4x + 6). Using the quadratic formula on the second factor, we have
_ -4±yi6-4(i)(6) _ -4±y^8
—
o
-4±ay5» = _2 ± J2i. Thus thezeros are -3, -2 ± y/2i.
2
2
T
45. P (x) = x3 - 3x2 + 3x - 2.
2 I1
-3
3-2
2
-2
1-110
Thus P (x) = (x - 2) (x2 - x+ 1). So x = 2or x2 - x + 1= 0
Using the quadratic formula we have x= ^V1-4^1* - i±%£. Hence, the zeros are 2, and i^.
46. P (x) = x3 - x - 6 has possible zeros ±1, ±2, ±3.
1 I 1 0 -1
XI
-6
2 I 1 0 -1
0-6
-6
2
4
6
12
3
0
=*• x = 2 is a zero.
P(x) =(x - 2) (x2 +2x +3) .Nowx2 +2x +3has zeros x=zH^pM ==2±gb& =_X ±^2. Thus the
zeros are2, -1 ± iy/2.
47. P (x) = 2x3 + 7x2 + 12x + 9has possible rational zeros ±1, ±3, ±9, ±±, ±|, ±|. Since all coefficients are positive,
there are no positive real zeros.
-1 I 2
7
12
9
-2
-5
-7
-2 I 2
2572
7
12
9
-4
-6
-12
236-3
There is a zero between -1 and —2.
12
-3
-6
4
6
-9
0 => x = - § isa zero.
P(x) = (x+f)(2x2+4x+6) = 2(x+|)(x2 + 2x + 3). Now x2 + 2x + 3 has zeros
x = -2*^/4-4(3X1) = -2±2y^2 = _x ± ^
Hence, the zeros are - § and -1 ± iy/2.
264
CHAPTER 3 Polynomial andRational Functions
48. Using synthetic division, we see that (x- 3) is afactor ofthe polynomial:
1 I 2 -8
9-9
2-6
2-6
3 I 2 -8
3
9-9
6-6
3-6
9
2-230
=*x = 3 is a zero.
So P (x) = 2x3 - 8x2 +9x - 9= (x - 3) (2x2 - 2x +3). Using the quadratic formula, we find the other two solutions
~
2±v/4-4(3)(2)
2T2I
~
2±y^20
4
x-VS.-
.
Qlj_V5\
= 2± ~2~l- T""5 the zer°s are 3, 5 ± -^—z.
49. P (x) = x + x + 7x + 9x - 18. Since P (x) has one change in sign, we are guaranteed apositive zero, and since
P (-x) = x4 - x3 + 7x2 - 9x - 18, there are 1or 3negative zeros.
1I1 1 7
1
9-18
1
2
9
18
2
9
18
0
Therefore, P (x) = (x - 1) (x3 + 2x2 + 9x + 18). Continuing with the quotient, we try negative zeros.
-1 I1 2 9 18
1
-2 I1 2 9 18
-1
-1
-8
-2
0
-18
1
8
10
10
9
0
P (x) = (x - 1) (x + 2) (x2 + 9) = (x - 1) (x + 2) (x - 3t) (x + 3i). Therefore,the zeros are 1, -2, and ±3i.
50. P (x) = x4 - 2x3 - 2x2 - 2x - 3 has possible zeros ±1,±3.
1 I 1 -2 -2 -2 -3
3 I 1 -2 -2 -2 -3
1-1-3-5
1
~1
~3
~5
-8
3
3
3
3
11110
=> x = 3 is a zero.
P (x) = (x - 3) (x3 + x2 + x + 1). Ifwe factor the second factor by grouping, we get
x3 + x2 + x + 1 = x2(x +l) + l(x+l) = (x +l)(x2 +l). So we have
P (x) = (x - 3) (x + 1) (x2 + 1) = (x- 3) (x + 1) (x - i)(x + i). Thus the zeros are 3, -1, i,and -i.
51. We see a pattern and use ittofactor by grouping. This gives
P(x) = x5 - x4 +7x3 - 7x2 +12x - 12 =x4 (x - 1) +7x2 (x - 1) +12 (x - 1) =(x - 1) (x4 +7x2 +12)
= (x-l)(x2 +3)(x2+4) =(x-l)(x-zV3) (x +iv/3)(x-2i)(x +2z)
Therefore,the zeros are 1, ±iy/3, and ±2i.
52. P(x)=x5+x3 +8x2+8 = x3(x2 +l)+8(x2 +l) = (x2 +l)(x3+8) = (x2 +l)(x +2) (x2 - 2x +4)
(factoring asum ofcubes). So x = -2, or x2 +1 = 0. Ifx2 + 1= 0, then x2 = -1
=*•
then x= 2±V4~*W =1± ^§^ =1±y/3i. Thus, the zeros are -2, ±i, 1±y/3i.
x = ±i. Ifx2 - 2x + 4 = 0,
SECTION 3.5 Complex Zeros and the Fundamental Theorem ofAlgebra
265
>3. P(x) = x4 - 6x3 +13x2 - 24x +36 has possible rational zeros ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18. P(x) has 4
variations in signand P (-x) has no variation in sign.
111 —6 13 -24 36
1 -5
T^5
211 —6 13 -24 36
8 -16
8 -16
2 -8
20~
1 -4
311 -6 13 -24 36
10 -28
5 -14
3 -9
8
1-3
12 -36
4 -12
0 => x = 3 is azero.
Continuing:
3|1 -3 4 -12
3
1
0
12
04
0=4>x = 3isa zero.
P (x) = (x - 3)2 (x2 + 4) = (x - 3)2 (x - 2i) (x + 2i). Therefore,the zeros are 3(multiplicity 2) and ±2i.
54. P (x) = x4 - x2 + 2x + 2 has possible rational zeros ±1, ±2.
111 0 -1 2 2
—111 0-12 2
110 2
11
0 2 4 1isan upper bound.
-111 -1 0 2
-110-2
-12-2
1-1020
1-2 20
P(x) =(x +l)2 (x2 - 2x +2). Using the quadratic formula on x2 - 2x +2, we have x= 2±^* =Z&i =1±i.
Thus, the zeros of P (x) are -1,1 ± i.
55. P(x) = 4x4 + 4x3 + 5x2 +4x + 1has possible rational zeros ±1, ±\, ±\. Since there is no variation in sign, all real
zeros (if there are any) are negative.
—1 I 4
45
-4
41
0-5
—^ I 4
1
405-12
4
5
4
1
-2
-1
-2
-1
42420
P(x) = (x+ \) (4x3 + 2x2 + 4x + 2). Continuing:
•i I 4
2 4
2
4
-2
0
-2
0
4
0
=*• x = - 3 is a zero again.
P (x) = (x + i)2 (4x2 + 4). Thus, the zeros of P (x) are -\ and ±i.
56. P (x) = 4x4 + 2x3 - 2x2 - 3x - 1has possible rational zeros ±1, ±5, ±\. P has one variation in sign, so P has one
positive real zero.
14
2
-2
-3
-1
1
0
6
46
4
=>• 1 is a zero.
P (x) = (x - 1)(4x3 + 6x2 + 4x+ 1). Continuing:
—1 I 4
6
-4
4 1
-2
—5 I 4
2
42_23
6
4
-2
-2
-1
4420
=> -\ isa zero.
p (x\ _ (x _ 1) (x _|_ i) (4X2 +4x + 2). Using the quadratic formula on 4x2 + 4x + 2, we find
x = -4±/i6=52- = -I ± \i. Thus, P has zeros 1, -\, -\ ± \i.
266
CHAPTER 3 Polynomial and Rational Functions
57. P(x) =x5 - 3x4 +12x3 - 28x2 +27x - 9has possible rational zeros ±1, ±3, ±9. P(x) has 4variations in sign and
P (—x)
sion
P
(—x) has
has 11 variation
variation in
insign.
1 | 1 -3 12 -28
1
1 | 1 -2 10 -18
27 -9
1
-2
10
-18
9
-2
10
-18
9
0
9
=>x = lisazero.
l I i _i 9 _9
1-19-9
1-1
9-9
io
0 =>x = lisazero.
10 9
0 =• x = 1 is a zero.
P(x) = (x - l)3 (x2 +9) = (x - l)3 (x - 3i) (x +3i). Therefore,the zeros are 1(multiplicity 3) and ±3i.
58. P (x) = x5 - 2x4 + 2x3 - 4x2 + x - 2has possible rational zeros ±1, ±2.
1| 1 -2
2-4
1-1
1-1
1-2
1-3
2| 1 -2 2 -4 1 -2
-2
1-3 -2 -4
2
~l
0
4
Q~2
0
0 i"
2
0~ =>x = 2isazero.
P(x) =(x - 2) (x4 +2x2 +1) =(x - 2) (x2 +l)2 =(x - 2) (x - i)2 (x +z)2. Thus, the zeros ofP(x) are 2, ±z.
59. (a) P(x) =x3 - 5x2 +4x - 20 =x2 (x - 5) +4(x - 5) = (x - 5) (x2 +4)
(b) P (x) = (x - 5)(x - 2i) (x + 2i)
60. (a) P(x) = x3-2x-4
1| 1 0 -2 -4
2| 1 0 -2 -4
11-1
2
4
11-1-5
12
2
P (x) = x3 - 2x - 4= (x - 2) (x2 + 2x + 2)
(b) P (x) = (x - 2) (x + 1 - i) (x + 1 + i)
61. (a)P(x) =x4+8x2-9=(x2-l)(x2 +9)=(x-l)(x +l)(x2 +9)
(b) P (x) = (x - 1)(x + 1)(x - 3i)(x + 3i)
62. (a) P (x) = x4 + 8x2 + 16 = (x2 + 4)2
(b) P(x) = (x-2i)2(x + 2i)2
63. (a)P(x)=x6-64=(x3-8)(x3+8) =(x-2)(x2+2x +4)(x +2)(x2-2x +4)
(b) P(x) = (x - 2) (x +2) (x +1- iy/3) (x +1+iy/3) (x - 1- iy/3) (x - 1+zV3)
64. (a)P(x) =x5-16x =x(x4-16)=x(x2-4)(x2+4)=x(x-2)(x +2)(x2 +4)
(b) P(x) = x(x-2)(x + 2)(x-2i)(x + 2i)
4
SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra
>5. (a) x4 - 2x3 - llx2 + 12x = x (x3 - 2x2 - llx + 12) = 0. We first find the
-> \
267
:
bounds for our viewing rectangle.
1
-2
-11
12
5
1
3
4
32
-4
1
—6
13
-50
\
/"M
=» x = 5 is an upperbound.
=» x = -4 is a lower bound.
We graph P (x) = x4 - 2x3 - llx2 + 12x in the viewing rectangle [-4,5] by
[-50,10] and see that it has 4real solutions. Since this matches the degree of
P (x), P (x) hasno imaginary solution.
(c)x4 - 2x3 - llx2 + 12x + 40 = 0. We graph
T (x) = x4 - 2x3 - llx2 + 12x + 40 in the viewing
rectangle [-4,5] by [-10,50], and see that T has no
(b) x4 - 2x3 - llx2 + 12x - 5 = 0. We use the same
bounds for ourviewing rectangle, [-4,5] by [-50,10],
and see that R(x) = x4 - 2x3 - llx2 + 12x - 5has
2 real solutions. Since the degree of R (x) is 4, R (x)
real solution. Since the degreeof T is 4, T must have
must have 2 imaginary solutions.
4 imaginary solutions.
i_i_j
1
^_^
£20
\
1_
I
.
66. (a) 2x + 4i = 1 «•
(b) x2 - ix = 0
«*•
3
r
:
;
2x = 1- 4i & x = \- 2*.
x (x - i) = 0 & x = 0, i.
(c) x2 + 2ix - 1= 0 «»
(x + i)2 = 0<^ x = -i.
(d) ix2 - 2x + i = 0. Using the quadratic formula, we get
„ _ 2±V(-2)a-4(«)(t) _ 2±s/8 _ 2*2^2 _ \±^2 - (l ± J$\ l-i) = (-1 ± \/2) i.
X—
2t
—
2t
2t
«
\
v
/ s
'
\
I
67. (a)P(x)=x2-(l +i)x+ (2 + 2i). SoP(2i) = (2z)2-(l+0(2i) + 2+ 2z = -4-2i + 2+ 2+ 2i = 0,
andP(l-i) = (l-i)2-(l+i)(l-0 + (2 + 2i) = 1- 2i - 1- 1 - 1+ 2+ 2i = 0.
Therefore, 2i and 1 - i are solutions ofthe equation x2 - (1 + i)x + (2 + 2i) = 0. However,
P(-2i) = (-2i)2 - (1 +z) (-20 + 2 + 2i = -4 + 2i - 2+ 2+ 2i = -4 + 4i, and
P(l + i) = (1 + if - (1 +0 (1 + i) +2+ 2i = 2+ 2i. Since, P(-2i) # 0and P(l + i) # 0, -2i and 1+i are
not solutions.
(b) This does not violate the Conjugate Roots Theorem because the coefficients ofthe polynomial P(x) are not all real.
68. (a) Since i and 1 + i are zeros, -i and 1 - i are also zeros. So
P(x) = C(x-i)(x + i)(a:-[l +i])(a:-[l-i]) = C,(x2 + l)(x2-2x + 2)
= C(x4-2x3 +2x2+x2-2x +2) = C(x4 - 2x3 +3x2 - 2x +2)
Since C = 1, the polynomial is P (x) = x4 - 2x3 + 3x2 - 2x + 2.
(b) Since i and 1 + i are zeros,
P(x) = C(x - i) (x - [i +1]) = C(x2 - xi - x- xi - 1+i) = C[x2 - (1 +2i)x - 1+i]
Since C = 1,the polynomial isP (x) = x2 - (1 + 2i) x - 1 + i.
268
CHAPTER 3 Polynomial and Rational Functions
69. Since Phas real coefficients, the imaginary zeros come in pairs: a±bi (by the Conjugate Roots Theorem), where b£ 0.
Thus there must be an even number of imaginary zeros. Since P is ofodd degree, it has an odd number ofzeros (counting
multiplicity). It follows that P has at least one real zero.
70. x4 -1 = 0 v-:(x2 -1) (x2 +1;
**• (x-l)(x
+ l)(.x- + 0(x-i) =0
<*
x = ±1, ±i. Sothere
are four fourth roots of1, two that are real and two that are complex. Consider P (x) = x" - 1, where n is even. P has one
change in sign so Phas exactly one real positive zero, namely x = 1. Since P (-x) = P (x), P also has exactly one real
negative zero, namely x = -1. Thus P must have n - 2 complex roots. As a result, xn = 1has two real n th zeros and
n —2 complex roots.
x -1 = 0 ^ (x - 1) (x2 +x+ 1) =0<& x= 1, =±&&. So there is one real cube zero of unity and two complex
roots. Now consider Q(x) = xk - 1, where A- is odd. Since Qhas one change in sign, Qhas exactly one real positive zero,
namely x = 1. But Q(-x) = -x - 1has no changes in sign, so there are no negative real zeros. As a result, xk = 1has
one real A: th zero and k - 1 complex roots.
3.6
1. r(x) =
Rational Functions
x-2
r(x)
X
X
X
r(x)
X
r(x)
1.5
-3
2.5
5
10
1.25
-10
0.833
1.9
-19
2.1
21
50
1.042
-50
0.962
100
1.020
-100
0.980
1000
1.002
-1000
0.998
1.99
-199
1.999
2.01
-1999
201
2.001
(b) 7- (x) —> -oo as x —> 2 and r(x)
(c) /• has horizontal asymptote y -=
2. r(x) =
r(x)
2001
oo as x
9+
1.
4x + 1
x-2
(a)
r(x)
(a)
X
r(x)
X
-14
2.5
22
10
5.125
-10
1.9
-86
2.1
94
50
4.188
-50
3.827
1.99
-896
2.01
100
4.092
-100
3.912
1.999
-8996
2.001
1000
4.009
-1000
3.991
X
(b) r (x) —> -oo as x —• 2 and '•(•'••)
(c) r has horizontal asymptote y •= 4.
3. r (x) =
r(x)
1.5
X
904
9004
r(x)
3.25
oo as x
3x- 10
(x - 2):
X
r(x)
X
r(x)
X
r(x)
X
r(x)
1.5
-22
2.5
-10
10
0.3125
-10
-0.2778
1.9
-430
2.1
-370
50
0.0608
-50
-0.0592
-39,700
100
0.0302
-100
-0.0298
-3,997,000
1000
0.0030
-1000
-0.0030
1.99
1.999
-40,300
2.01
-4,003,000
2.001
(b) 7- (x) -> -oo as x -^ 2.
(c) ;• has horizontal asymptote y
(I.
SECTION 3.6 Rational Functions
4. r (x) =
269
3x2 + l
(x-2)2
(a)
r(x)
X
r(x)
X
X
r (x)
X
r{x)
1.5
31
2.5
79
10
4.703
-10
2.09
1.9
1183
2.1
1423
50
3.256
-50
2.774
131,203
100
3.124
-100
2.884
13,012,003
1000
3.012
-1000
2.988
128,803
1.99
1.999
12,988,003
2.01
2.001
(b) r (x) -+ oo as x —• 2.
(c) r has horizontal asymptote y = 3.
5. r{x) =
X 1 When x = 0, we have r (0) = -\, so the y-intercept is -\. The numerator is 0when x = 1, so the
x + 4
x-intercept
is 1
x-interc
6. s (x) =
When x = 0, we have s(0) = 0, so the y-intercept is 0. The numerator is zero when 3x = 0or x = 0, so
3x
x - 5'
the x-intercept is 0.
7. t{x) = " x
When x=0, we have t(0) = ^ =|, so the y-intercept is ±. The numerator is 0when
—6
c2 _ x_ 2- (x _ 2) (x + 1) = 0or when x= 2or x= -1,so the x-intercepts are 2and -1
_JL
8. r (x) = x2 + 3x-4'
2
When x = 0, we have r (0) = —A
= -J, so the y-intercept is -\. The numerator is never zero, so
-4
there is no x-intercept.
9 r(x) = x "9. Since 0is not in the domain of r(x), there is no y-intercept. The numerator is 0when
.2
v
x2
x2 _ g_ (x _ 3) (x + 3) = oor when x = ±3, so the x-intercepts are ±3.
10 r(x) =*3 +8. when x=0, we have r(0) =f =2, so the y-intercept is 2. The x-intercept occurs when x3 +8=0
K'
x2 + 4
«• (x +2) (x2 - 2x +4) =0 <s> x=-2 or x=1±i\/3, which has only one real solution, so the x-intercept is
-2.
11. From the graph, the x-intercept is 3, the y-intercept is 3, the vertical asymptote is x= 2, and the horizontal asymptote ,s
y = 2.
12. From the graph, the x-intercept is 0, the y-intercept is 0, the horizontal asymptote is y=0, and the vertical asymptotes are
x = —1 andx = 2.
13. From the graph, the x-intercepts are -1 and 1, the y-intercept is about \, the vertical asymptotes are x= -2and x=2,
and the horizontal asymptote is y = 1.
14. From the graph, the x-intercepts are ±2, the y-intercept is -6, the horizontal asymptote is y=2, and there are no vertical
asymptotes
3
3/x
15. r(x) = ——. There is avertical asymptote where x+2 =0 <* x= -2. We have r(x) = ^-^ = Y+^fx) ~*
as x -* ±oo, so the horizontal asymptote is y = 0.
ie
c(„\ - 2x + 3 - 2+
(3/x) -» 2as x -> ±oo. The horizontal asymptote is y = 2. There is avertical asymptote when
lb. »lij- a,_1
i_(i/x)
x - 1 = 0 •&• x = 1, so the vertical asymptote is x = 1.
270
CHAPTER 3 Polynomial and Rational Functions
17. t (x) = —z
X
2
x* —x —6
X
= t
2
1
-r-
-7- =
(x —3J (x + 2)
1
=
~
i _ i.
_
x
x2
• 1 as x —• ±oo. Hence, the horizontal asymptote is y = 1.
The vertical asymptotes occur when (x - 3) (x + 2) = 0
&•
x = 3 or x = -2, and so the vertical asymptotesare
x = 3 and x = -2.
1 __L
18. r(x) = x +zx
* + l = l +x9± +g2J_ -> 0as x-• ±oo. Thus, the horizontal asymptote is y= 0. Also, y= 2^X
~2j
(x + 1)2
X
X2
so there is a vertical asymptote when x + 1= 0 <*
x = -1, so the vertical asymptote is x = -1.
6
JL
19' S^ ~ x^T2" There is no vertical ^ymP1016 sjnce x2 +2is never 0. Since s(x) = g6 = a:2 _, 0as
*£ "T* «
i
^
l + ~2
x —»• ±oo, thehorizontal asymptote is y = 0.
3
2
20. t(x) = fr-1)^-2) = *2-3x +2 _ 1~ x + ~2
(x - 3)(x - 4)
x2 - 7x + 12
7
12 ~> lasx -> ±oo, so the horizontal asymptote is y = 1.
x
x2
Also, vertical asymptotes occur when (x - 3) (x - 4) =0 => x=3,4, so the two vertical asymptotes are x=3and
x = 4.
6x — 2
21 r(X) = x2 +5x-6' AVertical "V™!*** occ^ when x2 +5x - 6= (x +6) (x - 1) =0&x=1or x= -6.
Because the degree ofthe denominator is greater than the degree ofthe numerator, the horizontal asymptote is y=0.
*S[X) ~ x2 +2x +5 =
2 5 ~* 3** x~* ±0°' so the horizontal asymptote is y= 3. Also, vertical asymptotes
x
occur when x2 + 2x + 5 = 0
^
u
x2
=>
=»
rr — -2±->/3=2o _ , , „. c. ^ .,
.
,
x
*
_ -l ± 22. Since there is no real zero, there are no vertical
asymptotes.
x2 + 2
23' V= ~x~^T- AVertical **™Ptote o^urs when x- 1=0&x= 1. There are no horizontal asymptotes because the
degree ofthe numerator is greater than the degree ofthe denominator.
24 r( ) = x3 + 3x2 -
a2 (x + 3)
[X) x2 - 4 ~ (x - 2) (x +2)" Because the degree ofthe numerator is greater than the degree ofthe denominator,
the function has no horizontal asymptotes. Two vertical asymptotes occur at x=2and x=-2. By using long division, we
4x + 12
see that r (x) =x+3+—=—J
so y=x+3is aslant asymptote.
SECTION3.6 Rational Functions
n Exercises 25-32, let / (x) = -.
5t r (j.) = _L_ =/ (x _ i). From this form we see that the graph ofr is obtained
from the graph of/ by shifting 1 unit to the right. Thus r has vertical asymptote
x = 1 and horizontal asymptote y = 0.
>6. r (x) =
= / (x+ 4). From this form we see that the graph ofr is obtained
x + 4
from the graph of/ by shifting 4 units to the left. Thus r has vertical asymptote
x = -4 and horizontal asymptote y = 0.
27 s(x) = 3 = 3f —^—| = 3/ (x + 1). From this form we see that the graph
v;
x+1
\x + lj
ofs is obtained from the graph of/ by shifting 1unit to the left and stretching
vertically by afactor of3. Thus s has vertical asymptote x = -1 and horizontal
asymptote y = 0.
28. s(x) = ~2 =-2 (—^— J=-2/(x - 2). From this form we see that the
graph of s is obtained from the graph of/ by shifting 2units to the right, stretching
vertically by afactor of2, and then reflecting about the x-axis. Thus r has vertical
asymptote x = 2 and horizontal asymptote y = 0.
29. t (x) =
v 7
x —2
= 2+ —
= / (x - 2) + 2 (see long
x
division below). From this form weseethat the graph oft is
obtained from thegraph of/ byshifting 2 units totheright
and2 units vertically. Thust hasvertical asymptote x = 2
and horizontal asymptote y = 2.
x-2
2x -
3
2x -
2
271
272
CHAPTER 3 Polynomial and Rational Functions
30. t(x)
= ^—3=2—
V'
x+2
x + 2 =3-9(-V)
\x + 2J
= -9/(x + 2) +3
X '
x+2
3x -
3
3x + 6
From thisform wesee thatthegraph of* is obtained from the
graph of/ by shifting 2 units tothe left, stretching vertically
by a factor of9,reflecting about the x-axis, and then shifting
3 units vertically. Thus t has vertical asymptote x = -2 and
-9
horizontal asymptote y = 3.
31" r{x) =f +3=*" x~+~3 =-/(a; +3) +1(see long
division below). From this form we see that the graph ofr is
x+ 3
x + 2
obtained from the graph of/ by shifting 3 units tothe left,
reflect about the x-axis, and then shifting vertically 1 unit.
x + 3
-1
Thusr has vertical asymptote x = -3 and horizontal
asymptote y = 1.
S2..(«)-=Zi.2—L.-2-P-}
x-4
x-4
\x-4J
= -/(x - 4) + 2
x-4
2x -
9
2x -
From this form weseethatthegraph oft is obtained from the
graph of/ by shifting 4 units tothe right, reflecting about the
8
-1
x-axis, and then shifting 2 units vertically. Thus t has vertical
asymptote x = 4 and horizontal asymptote y = 2.
4x
4
33' V= x+2' When X=°'y = ~2' so the ^-intercept is -2. When y=0,
4x- 4 = 0 «=> x = 1, so the x-intercept is 1. Since the degree ofthe
numerator and denominator are the same the horizontal asymptote is y = * = 4. A
vertical asymptote occurs when x= -2.As x-*• -2+, y= 4x ~4
x + 2
_
4x
-oo, and
4
asx->-2 ,y = —— -> oo.
x + 2
T4
t- \
2x + 6
2(x + 3)
J*, r{x) - _6x +3 = _3 f2x - 1)' When x=°»we have 2/ =2, so the
y-intercept is 2. When y = 0, we have x + 3 = 0 & x = -3, sothe
x-intercept is -3. Avertical asymptote occurs when 2x-l=0«-x=i.
Because the degree ofthe denominator and the numerator are the same, the
horizontal asymptote is y=^ = -|.
SECTION 3.6 Rational Functions
4-3x
35. s (x) = -——.
When x = 0, y = f, so the y-intercept is f. The x-intercepts
i
-"
x-f 7
occur when y = 0 & 4 - 3x= 0 *>
x = §. Avertical asymptote
occurs when x = -7. Since thedegree of thenumerator and denominator arethe
same the horizontal asymptote is y
36. s (x) =
l-2x
2x + 3'
-1
±-3.
When x = 0,y = k,so the y-intercept is |. When y = 0,we
have 1 - 2x = 0 <& x = ~, so the x-intercept is \. Avertical asymptote occurs
when 2x+ 3 = 0 <&
x=-|,and because the degree ofthe denominator
and the numerator are the same, the horizontal asymptote is y = -1.
37. r(x) = —^-2. When x=0, y= ^ =2, and so the y-intercept is 2. Since the
(x-3)
numerator can never be zero, there is no x-intercept. There is a vertical asymptote
whenx-3 = 0 <* x = 3, and because the degree of the asymptote is y = 0.
38. r (x) = X~ 2. When x = 0, we have y= -2, so the y-intercept is -2. When
(x + 1)
y = 0, we have x - 2 = 0 ^ x = 2, so the x-intercept is 2. Avertical
asymptote occurs when x = -1, and because the degree ofthe denominator is
greater than the degree ofthe numerator, the horizontal asymptote is y = 0.
39. s (x) =
4x-8
-. Whenx = 0, y
(x-4)(x + l)
~"~
"'"
(~4)(1)
= 2, so the y-intercept is 2.
When y = 0,4x - 8 = 0 4> x = 2,so the x-intercept is2. The vertical
asymptotes are x = -1 and x = 4, and because the degree ofthe numerator is less
than the degree ofthe denominator, thehorizontal asymptote is y = 0.
-2-'-'--*
273
274
CHAPTER 3 Polynomial and Rational Functions
X + 2
9
4°* S^ =(x +3)(x-l)' When X=°*y =Z3' so the intercept «s —| When
y = 0,we have x + 2 = 0
<*
x = -2, so the x-intercept is-2. Avertical
asymptote occurs when (x + 3) (x - 1) = 0 <*x = -3 andx = 1. Because the
degree ofthe denominator isgreater than the degree ofthe numerator, the
horizontal asymptote is y = 0.
41. s(x) =
6
x2 - 5x -6' When x= °»?/ = Zg = _1» so the y-intercept is -1.
Since the numerator is never zero, there is no x-intercept. The vertical asymptotes
occur when x2 - 5x - 6 = (x+ 1) (x- 6) & x = -1 and x = 6, and because
the degree ofthe numerator is less less than the degree ofthe denominator, the
horizontal asymptote is y = 0.
42. 5 (x) =
2x-4
2 (x - 2)
x2 -(. x_ 2 ~ (x - 1) (x +2)' When x=°'y = 2' so the intercept
is 2. When y = 0, we have 2x - 4 = 0 «» x = 2, so the x-intercept is 2. A
vertical asymptote occurs when (x - 1)(x + 2) = 0 <& x = 1and x = -2.
Because the degree ofthe denominator is greater than the degree ofthe numerator,
the horizontal asymptote is y = 0.
3x + 6
fi
1
43' *^ = x2 +2x-8' When x=°»2/ =Zg =~4'so^Hmwcept is -f.
When y = 0,3x+ 6 = 0 <s>
x = -2, so the x-intercept is -2. The vertical
asymptotes occur when x2 + 2x - 8 = (x- 2) (x + 4) = 0 «*•
x = 2and
x = -4- Since the degree ofthe numerator is less than the degree ofthe
denominator, the horizontal asymptote isy = 0.
x —2
x —2
44' l^ =x2-4x =x(x - 4) *Since x=°is not in the domain of*(x)» thereis
no y-intercept. When y = 0, we have x - 2= 0<* x = 2, so the x-intercept is 2.
Avertical asymptote occurs when x (x - 4) = 0
<&
x = 0 and x = 4.
Because the degree ofthe denominator is greater than the degree ofthe numerator,
the horizontal asymptote is y = 0.
L
SECTION 3.6 Rational Functions
I5# r(x) _ if.—11&L+|1
when x=0, y=§, so the y-intercept is §. When
(x + 1) (x —o)
y = 0, (x- l)(x + 2) = 0 =• x =-2,1, so, the x-intercepts are-2 and 1.
The vertical asymptotes are x = -1 and x = 3, and because the degree ofthe
numerator and denominator are the same the horizontal asymptote is y = \ =1.
46. r (x) = 2x(x +2)
v'
whgn x=Qwe haye ^= 0, so the graph passes
(x - 1) (x - 4)
through the origin. Also, when y = 0, we have 2x(x + 2) = 0 <* x = 0,-2, so
the x-intercepts are 0 and -2. There are two vertical asymptotes, x = 1and
x = 4. Because the degree ofthe denominator and numerator are the same, the
horizontal asymptote is y = f = 2.
47. r (x) =
x2 - 2x + 1
(x-1) =fx—l\ whenx =o,y = l,sothe
x2 + 2x+ 1 ~ (x+ l)2
y-intercept is 1. When y = 0, x = 1, so the x-intercept is 1. Avertical asymptote
occurs atx + l = 0^x = -l. Because the degree ofthe numerator and
denominator are the same the horizontal asymptote isy = } = 1.
4x2
4x2
48. r (x) = x2-2x-3 ~ (x-3)(x + l) When x = 0, we have y = 0, so the
graph passes through the origin. Vertical asymptotes occur at x = -1 and x = 3.
Because the degree ofthe denominator and numerator are the same, the horizontal
asymptote isy = \ —4.
49 r(x) =*«»+ 10*-12 =2(x-l)(x +6) whena; =
43-rW
X2+X-Q
(x-2)(x + 3)
y= 2(~1)(6) =2, so the y-intercept is 2. When y=0, 2(x - 1) (x +6) =0
(—2)(3)
=>
x = -6,1, sothe x-intercepts are -6 and 1. Vertical asymptotes occur when
(x- 2) (x+ 3) = 0«4> x = -3 or x = 2. Because the degree ofthe numerator
and denominator are the samethe horizontal asymptote is y = j = 2.
5
lt^
275
276
CHAPTER 3 Polynomial and Rational Functions
2x2 + 2x - 4 _ 2(x + 2) (x - 1)
50. r(x)
x2 + x
x (x + 1)
Vertical asymptotes occur at x = 0
\J\
and x = —1. Since x cannot equal zero, there is no y-intercept. When y = 0, we
have x = —2 or 1, so the x-intercepts are —2 and 1. Because the degree of the
denominator and numerator are the same, the horizontal asymptote isy = \ = 2.
r
r
r,1
51. y =
x2 - x - 6
(x - 3) (x + 2)
x2 + 3x
x (x 4- 3)
The x-intercept occurs when y = 0 «=>
(x - 3) (x + 2) = 0 => x = -2, 3, so the x-intercepts are -2 and3. There
is no y-intercept because y is undefined when x = 0. Thevertical asymptotes are
x = 0 and x = -3. Because the degree of the numerator and denominator are the
same, the horizontal asymptotes is y = j = 1.
52. r (x) =
x2+3x _
x (x + 3)
x2 - x - 6 ~ (x - 3) (x + 2) When x = 0, we have y = 0, so the
graph passesthroughthe origin. Wheny = 0, we havex = 0 or -3, so the
x-intercepts are 0 and -3. Vertical asymptotes occurat x = -2 and x = 3.
Becausethe degree of the denominatorand numeratorare the same, the horizontal
asymptote is y = y = 1.
53. r (x) =
3x2 + 6 _
3(x2 + 2)
When x = 0, y = —2, so the
x2 - 2x - 3 (x - 3) (x + 1)
y-intercept is -2. Since the numerator can never equal zero, there is no
x-intercept. Vertical asymptotes occur when x = -1,3. Because thedegree of the
numerator and denominator are the same, the horizontal asymptote is.y = f = 3.
54. r (x) =
5x2+5
x2 + 4x + 4
5 (x2 + 1)
(x + 2)2
Whenx
0, we have y = -, so the
y-intercept is f. Since x2 + 1 > 0 for all real x, y never equals zero, and there is
no x-intercept. The vertical asymptote is x = -2. Because thedegree of the
denominator and numerator arethe same, the horizontal asymptote occurs at
y = f = 5.
x
SECTION 3.6 Rational Functions
277
x2 - 2x + 1 _ (x - 1);
-. Since x = 0 is not in the domain of s (x),
x3 - 3x2 ~ x2 (x - 3)'
there is no y-intercept. Thex-intercept occurs when y = 0 •«>
15. S (x) =
x2 _ 2X + i = (x - l)2 = 0
=$•
x = 1,sothe x-intercept is 1. Vertical
asymptotes occur when x = 0, 3. Since the degree ofthe numerator is less than the
7i
degree ofthe denominator, thehorizontal asymptote isy = 0.
56. r (x) =
x
— x
_ _ _x_(x—1)_ When x_ Q haye y= o, so the
x3 - 3x - 2
x3 - 3x - 2
y-intercept is 0. When y = 0, we have x2 (x- 1) = 0, so the x-intercepts are 0
and 1. Vertical asymptotes occur when x3 - 3x- 2 = 0. Since x - 3x- 2 = 0
when x = 2, we can factor (x - 2) (x+ l)2 = 0, so the vertical asymptotes occur
1 V
atx = 2andx = -l. Because the degree ofthe denominator and numerator are
1.
the same, the horizontal asymptote is y = \ = 1.
57. r (x) =
x
When x = 0,y = 0,so the graph passes through the origin. There
x-2
is avertical asymptote when x- 2= 0&x = 2, with y—oo as x —2+, and
y _» _oo as x -• 2~. Because the degree ofthe numerator is greater than the
degree ofthe denominator, there is no horizontal asymptotes. By using long
4
•, so y =
division, we see that y = x -I- 2+ ^—2,
x + 2 is a slant asymptote.
58> rtx\ = e1±_2£ _ x(x +2). when x=0, we have y=0, so the graph passes
v
x—1
V-
x—1
i
ll
through the origin. Also, when y = 0, we have x = 0or -2, so the x-intercepts
^£
^^^
are -2 and 0. The vertical asymptote isx = 1.There isno horizontal asymptote,
•—•^"'
and the line y = x + 3 is aslant asymptote because by long division, we have
^
^X
'
1
|
' -i^*"""
^1
x — 1
ll
ll
ll
eg r(x\ = x* ~2x ~8 = (rE~4)(a: +2). The vertical asymptote is x=0, thus,
K'
X
x
there is no y-intercept. Ify = 0, then (x - 4) (x+ 2) = 0 => x = -2,4, so the
x-intercepts are -2 and 4. Because the degree ofthe numerator is greater than the
degree ofthe denominator, there are no horizontal asymptotes. By using long
g
division, we see that y = x - 2 - -, soy = x - 2 isa slant asymptote.
x
X
278
CHAPTER 3 Polynomial and Rational Functions
—x2 x (3 —x)
60. r (x) = 3x
2x_2
= 2) _ A. When x=0, we have y=0, so the graph passes
through the origin. Also, when y = 0, we have x = 0or x = 3, so the x-intercepts
are 0and 3. The vertical asymptote is x = 1. There is no horizontal asymptote, and
the line y = -\x + 1 is aslant asymptote because by long division we have
y= -Ix
+ l + ^—.
*
x - 1
at
f \ x2 + 5x + 4 (x + 4) (x + 1)
61. r(x)
= x_3 = v x'23 •Whenx =0,y =-j, so the
y-intercept is - §. When y = 0, (x+ 4) (x+ 1) = 0 <*•
x = -4, -1, so the
two x-intercepts are -4 and -1. Avertical asymptote occurs when x = 3,with
y -> oo as x -> 3+, and y -» -oo as x -• 3~. Using long division, we see that
y = x+ 8+
62. r (x) =
28
x3+4
x3+4
2x2 + x - 1
0 + 4
y =
0 + 0-1
- "S
q. soy = x + 8 isa slant asymptote.
(2x-l)(x + l) When x = 0, we have
-4, so the y-intercept is -4. Since x3 + 4 = 0
x= - ^4, the x-intercept is x = - &4. There are vertical asymptotes where
(2x - 1) (x + 1) = 0 =» x = i or x = -1. Since the degree ofthe numerator is
greater than the degree ofthe denominator, there is no horizontal asymptote. By
long division, we have y=|x- \ +2-±f±Z_ so the line y=\x - ±is a
slant asymptote.
fi4*
( \ —xZ + x2
x2 (x + 1)
V{X) ~ x2-4 = (x-2)(x +2)' When X=°'y =°'so the SraP11 P38868
through the origin. Moreover, when y = 0, we have x2 (x + 1) = 0 =>
x = 0,-1, so the x-intercepts are 0and -1. Vertical asymptotes occur when
x = ±2; as x -» ±2", y = -co and as x -+ ±2+, y-+ oo. Because the degree
ofthe numerator is greater than the degree ofthe denominator, there is no
iV
T
horizontal asymptote. Using long division, we see that y= x+ 1+ 4x + 4 so
x2 —4'
y = x + 1 is a slant asymptote.
RA
. .
2x3 + 2x =
64. r (x) = —
2x(x2
+! l)L
i
x2 - l~ ~ (x-l)(x+l)' Whe" X=°'We have y= °» so the ^P11
passes through the origin. Also, note that x2 + 1> 0, for all real x, so the only
x-intercept is0. There are two vertical asymptotes atx = -1 and x = 1.There is
no horizontal asymptote, and the line y= 2x is aslant asymptote because by long
division, we havey = 2x +
4x
x2-l
2>'
SECTION 3.6 Rational Functions
65. / (x) =
2x2 + 6x + 6
x + 3
, g (x) = 2x. / has vertical asymptotex = -3.
.
•10
-10
^*
10
»
. f(x)= X+6X -,9(x) =-x+4. / has vertical asymptotes x=0and x=2.
66.
^
20
^^V. 10
1 ^^J°
...
67 y^ = x -2x +16^ 9^ =I2j has verticai asymptote x=2.
x-2
68 y(x) _ -x +2x ^ 2x^ ^^ _ x_x2 j has vertical asymptote x=1.
(x - 1)
w
279
280
CHAPTER 3 Polynomial and Rational Functions
69. /(x) =
2x2 - 5x
has vertical asymptote x = -1.5, x-intercepts 0 and 2.5, y-intercept 0, local maximum (-3.9, -10.4),
2x + 3
and local minimum (0.9, -0.6). Using longdivision, we get / (x) = x - 4 +
12
From the graph, we see that the end
2x + 3
behavior off (x) is likethe end behavior of g (x) = x - 4.
x —
4
2x +3 | 2x2 - 5x
«
t
2x2 + 3x
•10
-
8x
-
8x -
^^
•20
10
-10 ^•»*
10
20
1-20
12
12
70. f(x) =
x4 - 3x3 + x2 - 3x + 3
x2-3x
has vertical asymptotes are x = 0, x = 3, x-intercept 0.82, andno y-intercept. The
local minima are (-0.80,2.63) and (3.38,14.76). The local maximum is (2.56,4.88). By using long division, we see that
/ (x) =x +1+ x2 _3x- From the second graph, we see that the end behavior off(x) is the same as the end behavior
ofg(x) = x2 + 1.
J2.
+
1
\
x4 - 3x3 + x2 - 3x + 3
xJ-3x
*°
x4 - 3x3
\
Ox3 + x2 - 3x
.1
•«
20
-4
-2
2
4
6
1
•20
x2 - 3x
-40
•60
71• ^W= x^T haS Vertical ^""P*046 x= 1> ^-intercept 0, y-intercept 0, and local minimum (1.4,3.1).
Thusy = x2 +
-$—-. From the graph we see that the end behavior off (x) is like the end behavior ofg(x) = x2.
x"5-!
•4
V
-2
1/
*
•4
•10 J
r
VV
*
-i
2
4
•10 '
Graph of /
Graph of / andg
x4
72. / (x) = -5—- has vertical asymptotes x = ±1.41, x-intercept 0, and y-intercept 0. The local maximum is (0,0). The
local minima are (-2,8) and (2,8). By using long division, we see that / (x) = x2 +2+ 24_—. From thesecond graph,
we see that the end behavior off (x) is the same as the end behavior ofg (x) = x2 + 2.
x2
+ 2
x2 -2 Ix4 + Ox3 + Ox2 + Ox + 0
x4
- 2x2
2x2
2x2
-
4
\J" 1/
•vj V
60
40
20
:
-. * -. l[°
| 2
4
6
I
SECTION 3.6 Rational Functions
73. / (x) =
x4 —3x3 + 6
—
x-3
281
has verticalasymptote x = 3, x-intercepts 1.6 and 2.7, y-intercept —2, local maxima (—0.4, —1.8)
and (2.4,3.8), and local minima (0.6,-2.3) and (3.4,54.3). Thus y = x3 +
x-3
. From the graphs, we see that the end
behavior of/ (x) is like the endbehavior ofg (x) = x3.
J*
x4 - 3x3 + 6
x-3
x4 - 3x3
74. r (x) =
4 + xl - x4
-x4 + x2 + 4
has vertical asymptotes x « ±1.41, x-intercepts ±1.6, and y-intercept —1. The
x2 - 1
(x - 1) (x + 1)
local maximum is (0, —0.4) and there is no local minimum
Thus y = —x +
x2 - 1
x2-l
From thegraphs, weseethattheendbehavior of / (x) is like theendbehavior of g (x) = —x2.
-x4 + Ox3 + x2 + Ox + 4
—x
+ x'
+ 4
75. (a)
(b) p (t) = -ttt = 3000 - TT~7- So as i -> oo, we
t+1
havep (t) -» 3000.
t + 1
(b) c (t) = -=—-. Sincethe degree of the denominator is
t2 + 2'
larger than the degree of the numerator, c(t) —• 0 as
t —• oo.
77. c(t) =
5t
t2 + l
(a) The highestconcentration of drug is 2.50 mg/L, and it is reached 1
hour after the drug is administered.
(b) The concentration of the drug in the bloodstreamgoes to 0.
(c) Fromthe first viewing rectangle, we see that an approximate solution
is near t = 15. Thus we graph y =
U
i2 + l
and y = 0.3 in the
viewing rectangle [14,18] by [0,0.5]. So it takes about 16.61 hours
for the concentration to drop below 0.3 mg/L.
282
CHAPTER 3 Polynomial and Rational Functions
78. Substituting for Rand g, we have h(v) =
(6-4 x10 )v
2 (9.8) (6.4 x 106) - v2
The vertical
asymptote isv « 11,000, and itrepresents the escape velocity from the earth's
gravitational pull: 11,000 m/s « 1900 mi/h.
<000
I000O 12003 I40OO
79. P(V) =P0 (-*>-) => p{v)=U0(JV_\
\so-vJ
v'
\332-vJ
Ifthe speed ofthe train approaches the speed ofsound, the pitch ofthe whistle
becomes very loud. This would beexperienced asa "sonic boom"— aneffect
seldom heard with trains.
80. (a) Ix +Iy =IF
y
F
x
xF
^
^ = -—BUsir|g F = 55»we gety =
x — r
:
x~F
y
xF
55x
—. Since y > 0, we
x — 55
use the viewing rectangle [0,1000] by[0,250].
(b) y approaches 55 millimeters.
(c) y approachesoo.
81. Vertical asymptote x=3: p(x) = ^—-. Vertical asymptote x=3and horizontal asymptote y= 2: r(x) = 2x
x
Vertical asymptotes x = 1and x = -1, horizontal asymptote 0,and x-intercept 4: a (x) =
-
-
(x-l)(x + l)'
— O
Of course,
other answers are possible.
x6 + 10
82. r(x) - ^ +8a,2 _|_ 15 has no x-intercept since the numerator has no real roots. Likewise, r(x) has no vertical
asymptotes, since the denominator has no real roots. Since the degree ofthe numerator is two greater than the degree ofthe
denominator, r (x) has nohorizontal orslant asymptotes.
tni»\~f~\
- 3x- 6 = -i
3(x-2)(x
n, + 1),for x ^ 2. Therefore,
83.
(a) r (x) = 3x2 ——
q. + l)L= 3(x
X
z
x
— Z
r (x) = 3x+ 3,x ^ 2. Since 3(2) + 3 = 9, the graph is the line y = 3x+ 3 with
the point (2,9) removed.
SECTION3.6 Rational Functions
283
lb) s(x) = x2+x~20 = (x-4)(x +5) =x_AJoxxlk _5. Therefore, s(x) =x- 4, x^ -5. Since
\ i
\ >
x+ 5
x+ 5
(_5) _ 4= -9, the graph is the line y= x- 4with the point (-5,-9) removed.
t(x) = 2*2 ~x- 1 = (2x +l)(x-l) = 2x +lforx ^ LTherefore, f(x) = 2x +1, x# 1. Since
^ '
x —1
x—1
2(1) + 1= 3, the graph is the line y = 2x + 1with the point (1,3) removed.
u(x) =
x-2
x2-2x
x~2 = I, forx ^ 2. Therefore, u(x) = -tx ± 2. Whenx = 2, - = -, so the graph is
x(x-2)
x
^
x
x
2
the curve y = - with the point (2, i) removed,
x
34. (a) Let / (x) = \.
Then r(x) = (x, —2)
* 2=/ (x - 2). From this form we see that
x*
the graph ofr is obtained from the graph off by shifting 2units to the right. Thus
r hasvertical asymptote x = 2 andhorizontal asymptote y = 0.
(b)s(x) =
2x2 + 4x + 5
x2 + 2x+ l
= 2 +
" ' (x + 1)2
= 3y(x + i) + 2. From this form we see that the graph of3 is obtained from
the graph off by shifting 1unit to the left, stretching vertically by afactor of 3, and shifting 2units vertically. Thus r
has vertical asymptote x = -1 and horizontal asymptote y = 2.
x2+2x + l I 2x2+4x+5
2x2+4x+2
I
1
i
_*^~r-=-=-_
'
i
I
1
—•—»•
•*
284
CHAPTER 3 Polynomial and Rational Functions
(c) Using long division, we see that p (x) =
2-3x^
x2 — 4x + 4
= -3 +
-12s+ 14
x2 - 4x + 4
which cannot be graphed by transforming
/ (x) = —r-. Using long division on q we have:
-3
-3
x2 - 4x +4 |-3x2+ 0x+ 2
.
. .
-3x2+12x-12
-3x2+12x-12
-12x+14
12
12
12x - 3x"
Sog(x) = —
:—— =
x2-4x + 4
x2 - 4x + 4 | -3x2+12x+ 0
' (x-2)2
= 12/ (x —2) —3. From this Ibrm we see that the graph ofg is obtained
from the graph of/ by shifting 2 units to the right, stretching vertically by a factor of 12, and then shifting 3 units
vertically down. Thus the vertical asymptote is x = 2 and the horizontal asymptote is y = —3. We showy = p(x) just
to verify that we cannot obtain p (x) from y — —.
y
i
^
\f
y = q(x)
y = V (x)
i
Chapter 3 Review
2. P (x) = 2x:i - 16 = 2(x - 2) (x2 + 2x+ 4)
1. P(x) = -x3 + 64
v
(0.641
\(4,0)
10
1
3. P(x) = 2(x-1)4 -32
\
'
4. P (x) = 64 - (x - 3)4
(3-2V2.0)