SECTION 3.3 Real Zeros of Polynomials
227
60. Since thezeros are x = -2, x = -1, x = 0, x = 1, and x = 2, the factors are x + 2, x + 1, x, x - 1, andx - 2. Thus
P(x) = c(x + 2) (x + 1) x(x - 1) (x - 2). If we let c = 1, then P (x) = a:5 - 5x3 + 4x.
61. Since the zeros of the polynomial are 1, -2, and 3, it follows that
P (x) = C(x - 1) (x + 2) (x - 3) = C(x3 - 2x2 - 5x + 6) = Cx3 - 2Cx2 - 5Cx + 6C . Since the coefficient of
x2 is to be 3, -2C = 3so C= -§. Therefore, P(x) = -f (x3 - 2x2 - 5x + 6) = -f x3 + 3x2 + f x - 9is the
polynomial.
62. Since the zeros of the polynomial are 1, -1, 2, and |, it follows that
P (x) = C(x - 1) (x + 1) (x - 2) (x - §) = C (x2 - 1) (x2 - \x + 1). To ensure integer
coefficients, we choose C to be any nonzero multiple of 2.
When C =
2, we have
P (x) = 2(x2 - 1) (x2 - §x+ l) = (x2 - 1) (2x2 - 5x + 2) = 2x4 - 5x3 + 5x - 2. (Note: This is just one ofmany
polynomials with the desired zeros and integer coefficients. We can choose C to beany even integer.)
63. The y-intercept is 2 and the zeros of the polynomial are -1, 1, and 2. It follows that
P(x) = C(x + 1) (x - 1) (x - 2) = C(x3 - 2x2 - x+ 2). Since P(0) = 2we have 2= C[(0)3 - 2(0)2 - (0) + 2]
4*
2 = 2C & C = 1and P (x) = (x + 1) (x - 1) (x - 2) = x3 - 2x2 - x + 2.
64. The y-intercept is 4 and the zeros of the polynomial are -1 and 2 with 2 being degree two. It follows that
P\x)=C(x + 1) (x - 2)2 = C(x3 - 3x2 +4). Since P(0) = 4we have 4= C[(0)3 - 3(0)2 +4]
4* 4= AC
4* C = 1 and P(x) = x3 - 3x2 + 4.
65. 1he y-intercept is 4 and the zeros of the polynomial are -2 and 1 both being degree two. It
follows that P (x) = C(x + 2)2(x-l)2 = C(x4 + 2x3 - 3x2 - Ax + 4). Since P(0) = 4
we have 4 = C[(0)4 + 2(0)3 - 3(0)2 - 4(0) + 4]
«• 4 = AC 4» C = 1. Thus
P (x) = (x + 2)2 (x - l)2 = x4 + 2x3 - 3x2 - 4x+ 4.
661. The -(/-intercept is 2 and the zeros of the polynomial are -2, -1, and 1 with 1 being degree two. It follows
that P (x) = C(x + 2) (x + 1)(x - l)2 = C (x4 + x3 - 3x2 - x + 2). Since P (0) = 2 we have
4= C[(0)4 + (0)3 - 3(0)2 - (0) + 2]
«* 2= 2C so C- 1and P(x) = x4 + x3 - 3x2 - x+ 2.
57. A. By the Remainder Theorem, the remainder when P (x) = 6x10()0 - 17x562 + 12x + 2G is divided by x + 1 is
P(_l) = 6(-l)1000 -17(-l)562 + 12(-l)+26 = 6-17- 12 + 26 = 3.
B. If x - 1 is a factor of Q(x) = x567 - 3x400 + x9 + 2, then Q(l) must equal 0.
Q(l) = (l)567 - 3(l)400 + (l)9 +2 = 1-3 + 1+ 2 = 1^0, so rc-1 is not a factor.
68. R(x) = x5 - 2x4 + 3x3 - 2x2 + 3x + 4 = (x4 - 2x3 + 3x2 - 2x + 3)x + 4 = ((x3 - 2x2 + 3x - 2) x + 3) x +
4 = (((x2 - 2x +3) x- 2) x + 3)x + 4 = ((((x - 2) x + 3) x - 2)x + 3) x + 4
So to calculate R (3), we startwith 3, then subtract 2, multiply by3, add 3, multiply by 3, subtract 2, multiply by 3, add3,
multiply by 3, and add 4, to get 157.
3.3
Real Zeros of Polynomials
1. P (x) = x3 - 4x2 + 3 has possible rational zeros ±1 and ±3.
2. Q (x) = x4 - 3x3 - 6x + 8 has possible rational zeros ±1, ±2, ±4, ±8.
3. P(x) = 2x5 + 3x3 + 4x2 - 8 has possible rational zeros ±1, ±2, ±4, ±8, ±\.
4. S (x) = 6x4 - x2 + 2x+ 12 has possible rational zeros ±1, ±2, ±3, ±4, ±6, ±12,±\, ±|, ±^, ±§, ±|, ±\.
5. T(x) = 4x4 - 2x2 - 7 has possible rational zeros ±1, ±7, ±\, ±|, ±\, ±\.
6. U(x) = 12x5 + 6x3 - 2x- 8 has possible rational zeros ±1, ±2,±4,±8, ±\, ±\, ±|, ±f, ±f, ±\, ±\, ±^.
228
CHAPTER 3 Polynomial and Rational Functions
7. (a) P (x) = 5x3 —x2 —5x+ 1has possible rational zeros ±1,±jr.
(b) From thegraph, theactual zeroes are —1, ^, and 1.
8. (a) P (x) = 3x3 + 4x2 - x - 2has possible rational zeros ±1,±2,±|, ±|.
(b) From the graph, the actual zeroes are —1 and |.
9. (a) P (x) = 2x4 - 9x3 + 9x2 + x - 3 has possible rational zeros ±1, ±3, ±±, ±|.
(b) From the graph, the actual zeroes are —|, 1, and 3.
10. (a) P (x) = 4x4 —x3 —4x+ 1has possible rational zeros ±1,±^,±\.
(b) From the graph, the actual zeroes are \ and 1.
11. P (x) = x3 + 3x2 - 4. The possible rational zeros are
12. P (x) = x3 —7x2 + 14x—8. The possible rational zeros
±1, ±2, ±4. P (x) has 1 variation in sign and hence 1
are ±1, ±2, ±4, ±8. P (x) has 3 variations in sign and
positive real zero.P(-x) = -x3 + 3x2 - 4 has 2
variations insign and hence 0 or 2 negative real zeros.
i Ii o q
a
hence 1 or3 positive real zeros.
P (-x) = —x3 —7x2 - 14x - 8 has 0 variations insign
14
4
1 I 1 -7
0
„. .
,
and hence no negative real zeros.
,
=>• x = 1 is a zero.
,
,.,
= (x-l)(x +2)2
zeros are x =
-8
1—6
.
1—6
P (x) = x3 + 3x2 - 4 = (x - 1)(x2 + 4x+ 4)
_. e tU
Therefore, the
14
8
8
0 ^ - x = lisa zero.
S°
0.
-2,1.
P (x)
= x3 - 7x2 + 14x - 8 = (x
- 1)> (x2
- 6x + 8)i
v '
\
\
= (x - 1) (x - 2) (x - 4)
Therefore, the zeros are x = 1, 2,4.
\
i
13. P (x) = x3 - 3x - 2. The possible rational zeros are ±1, ±2. P (x) has 1 variation in sign and hence 1 positive real
zero.P(-x) = -x3 + 3x - 2 has 2 variations insign and hence 0 or 2 negative real zeros.
1 | 1 0 -3 -2
2 | 1 0 -3 -2
11-2
1
1
-2
-4
=» x = 1 is not a zero.
1
2
4
2
2
1
0
=> x = 2 is a zero.
P (x) = x3 - 3x - 2 = (x- 2) (x2 + 2x + 1) = (x - 2) (x + l)2. Therefore, the zeros are x = 2, -1.
14. P (x) = x3 + 4x2 - 3x - 18. The possible rational zeros are ±1, ±2, ±3, ±6, ±9, ±18. P (x) has 1 variation insign and
hence 1 positive real zero. P (-x) = -x3 + 4x2 + 3x - 18has 2 variations insign and hence 0 or 2 negative real zeros.
1 | 1 4 -3 -18
15
15
2
2 | 1 4 -3 -18
2
2
12
18
-16
16
9
0
=• x = 2 is a zero.
P (x) = x3 + 4x2 - 3x - 18 = (x- 2) (x2 + 6x + 9) = (x- 2) (x + 3)2. Therefore, the zeros are x = -3, 2.
15. P (x) = x3 - 6x2 + 12x- 8. The possible rational zeros are ±1, ±2, ±4, ±8. P (x) has 3 variations insign and hence 1
or3 positive real zeros.P (-x) = —x3 - 6x2 - 12x- 8 has novariations insign and hence 0 negative real zeros.
1 | 1 -6 12 -8
1-5
1-5
2 | 1 -6 12 -8
7
7-1
2-8
=> x = 1 is not a zero.
8
1-440
=> x = 2 is a zero.
P (x) = x3 - 6x2 + 12x - 8 = (x- 2) (x2 - 4x + 4) = (x- 2)3. Therefore, the zero is x = 2.
SECTION 3.3 Real Zeros of Polynomials
229
16. P(x) =x3 - x2 - 8x +12. The possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±12. P(x) has 2variations in sign and
' hence 0or 2positive real zeros. P(-x) =-x3 - x2 +8x +12 has 1variation in sign and hence 1negative real zero.
1 I 1 -1 -8
:
0
12
2 I 1 -1 -8
12
0
-8
-12
-8
4
11-60
=^ x = 2 isa zero.
P{x) =x3 _x2 _8x +12 =(x - 2) (x2 +x- 6) =(x - 2) (x +3) (x - 2). Therefore, the zeros are x=-3,2.
17. P(x) =x3 - 4x2 +x+6. The possible rational zeros are 18. P(x) =x3 - 4x2 - 7x +10. The possible rational zeros
±1, ±2, ±3, ±6. P (x) has 2 variations in sign and hence
are ±1, ±2, ±5, ±10. P (x) has 2 variations insign and
0or 2positive real zeros.P (-x) = -x3 - 4x2 - x+ 6
hence0 or 2 positive real zeros.
has 1 variation insign and hence 1 negative real zeros.
p (_£) = -x3 - 4x2 + 7x + 10 has 1variation in sign
and hence 1 negative real zero.
-1
-4
16
-1
-6
-5
1I1
-4
x + 1 is a factor.
6
-7
10
-3
-10
0
•10
=> x = 1 is a zero.
So
So
P(x) = x3 - Ax2 + x+ 6= (x + 1) (x2 - 5x +6)
P (x) = x3 - 4x2 - 7x + 10 = (x - 1) (x2 - 3x - 10)
= (x + l)(x-3)(x-2)
= (x - 1) (x - 5) (x + 2)
Therefore, the zeros are x = -1, 2,3.
Therefore, the zeros are x = -2,1,5.
19. P(x) =x3 +3x2 +6x +4. The possible rational zeros are ±1, ±2, ±4. P(x) has no variation in sign and hence no
positive real zeros.P (-x) = -x3 +3x2 - 6x +4has 3variations in sign and hence 1or 3negative real zeros.
-1 | 1 3 6 4
1
-1
-2
-4
2
4
0
=>x + 1 is a factor.
So P(x) = x3 +3x2 +6x +4= (x +1) (x2 +2x +4). Now, Q(x) =x2 +2x +4has no real zeros, since the
discriminant ofthis quadratic is b2 - Aac = (2)2 - 4(1) (4) = -12 <0. Thus, the only real zero is x= -1.
20. P(x) =x3 - 2x2 - 2x - 3. The possible rational zeros are ±1, ±3. P(x) has 1variation in sign and hence 1positive
real zero. P(-x) = -x3 - 2x2 + 2x - 3has 2variations in sign and hence 0or 2negative real zeros.
1| 1 -2 -2 -3
3| 1 -2 -2 -3
1-1-3
1
_i
_3
_6
1
3
3
3
1
1
0
=> x = 3 isa zero.
So P(x) = x3 - 2x2 - 2x - 3= (x - 3) (x2 +x+1). Now, Q(x) = x2 +x+1has no real zeros, since the
discriminant is b2 - Aac = (l)2 - 4(1) (1) = -3 < 0. Thus, the only real zero is x = 3.
230
CHAPTER 3 Polynomial and Rational Functions
21. Method 1: P(x) =x4 - 5x2 +4.The possible rational zeros are ±1, ±2, ±4. P(x) has 1variation in sign and hence 1
positive real zero. P (-x) = x4 - 5x2 + 4has 2variations in sign and hence 0or 2negative real zeros.
1 | 1 0 -5
1
1
0
4
1-4-4
1
-4
-4
0
=*• x = 1 is a zero.
Thus P (x) = x4 - 5x2 +4= (x - 1) (x3 +x2 - 4x - 4). Continuing with the quotient we have:
-ill
1-4-4
-1
0
10-40
=> x = —1 is a zero.
P(x) =x4-5x2 +4=(x-l)(x +l)(x2 -4) =(x-l)(x +l)(x-2)(x +2). Therefore, the zeros arex =±l,
±2
Method 2: Substituting u = x2, the polynomial becomes P(u) = u2 - 5u + A, which factors:
u2-5tx +4=(w-l)(w-4) =(x2-l)(x2-4),soeitherx2 =lorx2=4.Ifx2 =l,thenx =±l;ifx2=4,
then x = ±2. Therefore, the zeros are x = ±1, ±2.
22. P(x) =x4 - 2x3 - 3x2 +8x - 4. Using synthetic division, we see that (x - 1) is afactor ofP(x):
1 | 1 -2 -3 8 -4
1-1-4
1-1-4
4
4
0
=*• x = 1 is a zero.
We continue by factoring the quotient, and we see that (x - 1) is again afactor:
1 | 1 -1 -4
4
10-4
1
0
—4
0
=$• x = 1 is a zero.
P(x)=x4-2x3-3x2 +8x-4= (x-l)(x-l)(x2-4) = (x-l)2(x-2)(x +2)
Therefore, the zerosare x = 1, ±2.
23. P(x) =x4 +6x3 +7x2 - 6x - 8. The possible rational zeros are ±1, ±2, ±4, ±8. P(x) has 1variation in sign and
hence 1positive real zero. P(-x) =x4 - 6x3 +7x2 +6x - 8has 3variations in sign and hence 1or 3negative real zeros.
1 | 1 6 7-6-8
1
7
14
8
14
8
0
^ x = 1 is a zero
and there are no other positive zeros. Thus P(x) = x4 +6x3 +7x2 - 6x - 8= (x - 1) (x3 +7x2 +14x +8).
Continuing by factoring thequotient, wehave:
-11
1
7
14
-1
-6
-8
6
8
0
=>• x = —1 is a zero.
SoP(x)=x4 +6x3+7x2-6x-8 =(x-l)(x +l)(x2 +6x +8)=(x-l)(x+l)(x +2)(x +4). Therefore,
the zeros are x = -4, -2, ±1.
SECTION 3.3 Real ZerosofPolynomials
231
4. P(x) =x4 - x3 - 23x2 - 3x +90. The possible rational zeros are ±1, ±2, ±3, ±5, ±6, ±9, ±10, ±15, ±18, ±30, ±45,
±90. Since P(x) has 2variations in sign, Phas 0or 2positive real zeros. Since P(-x) =x4 +x3 - 23x2 +3x +90
has2 variations insign, P has0 or 2 negative real zeros.
1| 1 -1 -23
1
1
-3
90
Q -23
-26
0 -23
-26
2| 1 -1 -23
-3
90
2
-42
-90
1 -21
-45
2
64
1
0 => x = 2is azero.
P (x) = (x - 2) (x3 + x2 - 21x - 45). Continuing with the quotient we have:
3 | 1 1 -21 -45
~
5| 1 1 -21 -45
3
12
-27
5
30
45
4
Ig
_72
1 6
9
0
=> x = 5 isa zero.
P(x) = (x - 2) (x - 5) (x2 +6x +9) = (x - 2) (x - 5) (x +3)2. Therefore, the zeros are x= -3,2, 5.
>5. P(x) =4x4 - 25x2 +36 has possible rational zeros ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36, ±5, ±J, ±5, ±4' ±5>
±f. Since P(x) has 2variations in sign, there are 0or 2positive real zeros. Since P(-x) =4x4 - 25x2 +36 has 2
variations insign, there are 0 or 2 negative real zeros.
1 I 4 0 -25
0
36
4
-21
-21
4 I2I
^21
4
1
15~
2 I 4 0 -25
8
16
0
36
-18
-36
4 8-9 -18
0 => x = 2is azero.
P (x) = (x - 2) (4x3 + 8x2 - 9x - 18)
2 I4
4
8
-9
-18
8
32
46
16
23
28
il4
-9
8
4
all positive, x
=>allF
10
= 2 isanupper bound.
§
-18
4
8
-9
-18
6
21
18
4
14
12
0
27
-18
5-2
-4
-20
16
4
315
16
27
4
9
-9
2
9
1
8
2
=» x = § is a zero.
P(x) =(x - 2) (2x - 3) (2x2 +7x +6) =(x - 2) (2x - 3) (2x +3) (x +2). Therefore, the zeros are x=±2, ±§.
Note: Since P(x) has only even terms, factoring by substitution also works. Let x2 = u; then
P(u) =Au2 - 25u +36 = (u - A) (Au - 9) = (x2 - 4) (4x2 - 9), which gives the same results.
26. P(x) =x4 - x3 - 5x2 +3x +6. The possible rational zeros are ±1, ±2, ±3, ±6. Since P(x) has 2variations in sign, P
' has 0or 2positive real zeros. Since P(-x) =x4 +x3 - 5x2 - 3x +6has 2variations in sign, Phas 0or 2negative real
zeros.
1 I 1 -1 -5
1
3
6
Q-5-2
10-5-24
2 I 1 -1 -5
2
3
6
2-6-6
11-3-30
=> x = 2 is azero.
P (x) = (x - 2) (x3 + x2 - 3x - 3). Continuing with the quotient we have:
3 I 1 1 -3 -3
12
27
-1 I l
1-3-3
0
1 4
9 24 =>x = 3isan upper bound.
1 0 - 3 0 =» x = -1 is azero.
P (x) = (x - 2) (x + 1) (x2 - 3). Therefore, the rational zeros are x = -1, 2.
232
CHAPTER 3 Polynomial and Rational Functions
27. P(x) - x4 +8x3 +24x2 +32x +16. The possible rational zeros are ±1, ±2, ±4, ±8, ±16. P(x) has no variations ii
sign and hence no positive real zero. P(-x) = x4 - 8x3 +24x2 - 32x +16 has 4variations in sign and hence 0or 2or
negative real zeros.
-1 [l 8 24 32 16
_2 |1 8 24 32 16
-1 -7 -17 -15
-2 -12 -24 -16
1 7 17
15
1 =>x = -lisnotazero.
1
6 12
8
0~ => x = -2 is azero.
Thus P(x) =x4 +8x3 +24x2 +32x +16 =(x +2) (x3 +6x2 +12x +8). Continuing by factoring the quotient, w<
have
-2 J 1
1
6
12
8
-2
-8
-8
4
4
0
=> x = -2 is a zero.
Thus P (x) = (x + 2)2 (x2 + 4x + 4) = (x + 2)4. Therefore, the zero is x = -2
28. P(x) =2x3 +7x2 +4x - 4. The possible rational zeros are ±1, ±2, ±4, ±±. Since P(x) has 1variation in sign, Pha
1positive real zero. Since P(-x) = -2x3 +7x2 - 4x - 4has 2variations in sign, Phas 0or 2negative real zeros.
1 | 2 7 4 -4
i I 2 7 4 -4
2
13
2 9 13
9 =*x =lis an upper bound.
2 8 8
0 => x= \ is azero.
P(*) = (x - i) (2x2 +8x +8) =2(x - 1) (x2 +4x +4) =2(x - J) (x +2)2. Therefore, the zeros are x=-2,
2*
29. Factoring by grouping can be applied to this exercise. 4x3 + 4x2 - x - 1 = 4x2 (x +1) (x +1) = (x +1) (4x2 - 1) = (x +1) (2x +1) (2x - 1). Therefore, the zeros are x= -1,±1.
30. We use factoring by grouping: P(x) = 2x3 - 3x2 - 2x + 3 = 2x (x2 - l) -
3(x2 - 1) =(x2 - 1) (2x - 3) =(x - 1) (x +1) (2x - 3). Therefore, the zeros are x=§, ±1.
31. P(x) =4x3 - 7x +3. The possible rational zeros are ±1, ±3, ±1, ±|,±1, ±|. Since P(x) has 2variations in sign,
there are 0or 2positive zeros. Since P(-x) = -4x3 +7x +3has 1variation in sign, there is 1negative zero.
5 I 4 0 -7
3
-3
4 2 -6
0 =>x=|isa zero.
2
P(^) = (^-i)(4x2 +2x-6)=(2x-l)(2x2+x-3)=(2x-l)(x-l)(2x +3)=0. Thus, the zeros are
X =
—-
~
1
2' 2> i-
32. P(x) =8x3 +10x2 - x- 3. The possible rational zeros are ±1, ±3, ±\, ±|,±i,±f, ±1, ±|. since P(x) has 1
variation in sign, P has 1positive real zero. Since P(-x) = -8x3 + 10x2 +x- 3has 2variations in sign, Phas 0or 2
negative real zeros.
1
1
8
10
-1
-3
8
18
17
8 18 17 14 =>x = lisan upper bound.
P(x)
I
2
8
10
-1
-3
4
7
3
8 14
6
0 =• x= \ is azero.
= 8x3 + 10x2 -x-3=(x-i) (8x2 + 14x +6)
= 2(x-|)(4x2 + 7x + 3) = (2x-l)(x + l)(4x + 3) = 0
Therefore, the zeros are x = -1, -| ±
SECTION 3.3 RealZerosof Polynomials
233
p/xj _ 43.3 +ga,2 _ llx _15 Tne p0SSible rational zeros are ±1, ±3, ±5, ±|, ±\, ±f, ±4, ±5» ±4- ^ (x) nas *
variation in sign and hence 1positive real zero. P(-x) = -4x3 +8x2 + llx - 15 has 2variations in sign, so P has 0or
2 negative real zeros.
-11
4
12
1
12
1
-14
8
-11
-15
20
140
645
4
28
129
630
14
8
-11
-15
4
14
5
4 10
-6
-11
-15
12
60
147
4
4
20
20
49
49
132
132
14
8
-11
-15
12
60
147
20
20
49
49
132
132
x = 1 isnota zero.
3
x = 5 is nota zero.
4
4
l
2
8
4
3
-15
4
14
4
35
16
4 9 -f
-^
9
1
-3
-18 =*x=±isnotazero.
=• x = 3 is nota zero
-15
-11
8
=• x = 3 is nota zero.
=• x = J is not azero.
§ I 4 8 -11 -15
6
21
4 14
10
15
0 =>• x = § isa zero.
Thus P(x) = 4x3 +8x2 - llx- 15 = (x - |) (4x2 +14x +10). Continuing by factoring the quotient, whose possible
rational zeros are -1, -5, -|, -J, -§, and -f, we have
-1 I 4
4
14
10
-4
-10
10
0
=»x= -lis a zero.
Thus P (x) = (x - |) (x + 1) (4x + 10) has zeros §, -1,and -§.
4. P(x) =6x3 +llx2 - 3x - 2. The possible rational zeros are ±1, ±2, ±J, ±±, ±J, ±|. P(x) has 1variation in sign
and hence 1positive real zero. P(-x) = -6x3 + llx2 + 3x - 2has 2variations in sign and hence 0or 2negative real
zeros.
11-3-2
17
14
17
14
12
2
6
=>x = lisnotazero.
\
6
6
11
-3
-2
6
14
4
0
11
12
-3
46
86
23
43
84 =» x = 2 isnot a zero.
=^x=§isa zero.
Thus, P (x) = 6x3 + llx2 - 3x - 2= (x - \) (6x2 + 14x + 4). Continuing:
-1 I 6 14
-6
6
4
_2 I 6
-8
8-4
=>x = -lisnotazero.
Thus, P(x) = (x - §) (x + 2) (6x + 2) has zeros J, -2, and -§.
6
14
-12
-4
2
0
=• x = -2 isa zero.
234
CHAPTER 3 Polynomial and Rational Functions
35. P(x) =2x4 - 7x3 +3x2 +8x - 4. The possible rational zeros are ±1, ±2, ±4, ±J. P(x) has 3variations in sign a
hence 1or 3positive real zeros. P(-x) = 2x4 +7x3 +3x2 - 8x - 4has 1variation in sign and hence 1negative real ze
1I2 -7
3 8-4
2-5-2
2-5-2
I | 2 -7
6
6
3 8 -4
1-3
2 =»x = lisnotazero.
~2 ^6
0
4
0~~8^
0~ =* x = ±is azero.
Thus P(x) =2x4 - 7x3 +3x2 +8x - 4= (x - |) (2x3 - 6x2 +8). Continuing by factoring the quotient, we have:
2 | 2 -6 0 8
2
4
-4
-8
—2
—4
0
=> x = 2 is a zero.
^(^) = (^-^)(x-2)(2x2-2x-4)=2(x-i)(x-2)(x2-x-2)=2(x-I)(x-2)2(x +l).Thus,th
zeros are x = |, 2, -1.
36. P(x) =6x4 - 7x3 - 12x2 +3x +2. The possible rational zeros are ±1, ±2, ±i,±|, ±§, ±J. Since P(x) has 2
variations in sign, Phas 0or 2positive real zeros. Since P(-x) =6x4 +7x3 - 12x2 - 3x +2has 2variations in sign,
has 0 or 2 negative real zeros.
1 I 6 -7 -12
6
-1
3
2
-1
-13
-10
-13
-10
-8
2 I 6 -7 -12
12
6
3
10
2
-4
-2
5-2-10
=> x = 2 isa zero.
P(x) = 6x4 - 7x3 - 12x2 +3x +2= (x - 2) (6x3 +5x2 - 2x - l). Continuing by factoring the quotient, we first
note that the possible rational zeros are-l,±i,±±,±±.We have:
5 I 6 5 -2 -1
6
8
2
0
=>x = ~isa zero.
P(x) = (x-2)(x-i)(6x2 +8x +2)=2(x-2)(x-i)(3x2 +4x +l)=(x-2)(x-I)(x+l)(3x +l).
Therefore, the zeros are x = -1, -|, ±, 2.
37. P(x) =x5 +3x4 - 9x3 - 31x2 +36. The possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±8, ±9, ±12, ±18. P(x)
has 2variations in sign and hence 0or 2positive real zeros. P (-x) = -x5 + 3x4 + 9x3 - 31x2 + 36 has 3variations ii
sign and hence 1 or 3 negative real zeros.
1 | 1 3 -9 -31
0
14-5
14-5
-36
36
-36
-36
-36
0
=^ x = 1 is a zero.
So P(x) = x5 +3x4 - 9x3 - 31x2 +36 = (x - 1) (x4 +4x3 - 5x2 - 36x - 36). Continuing by factoring the quotiem
we have:
1 |j_ 4 -5 -36 -36
0
2 I 1 4 -5 -36 -36
0
-36
2
12
14
-44
-36
-72
16
7
-22
-80
3 | 1 4 -5 -36 -36
1
3
21
48
36
7
16
12
0
=» x = 3 is a zero.
SECTION3.3 Real Zeros of Polynomials
235
So P (x) = (x - 1) (x - 3) (x3 + 7x2 + 16x + 12). Since we have 2positive zeros, there are no more positive zeros, so
wecontinue by factoring thequotient with possible negative zeros.
-1 | 1
7 16
12
-2 11
7
16
12
-1
-6
-10
-2
-10
-12
16
10
2
15
6
0
=*• x = -2 is a zero.
ThenP(x) = (x-l)(x-3)(x +2)(x2+5x +6)=(x-l)(x-3)(x +2)2(x +3).Thus,thezerosarex = l,3,
-2, -3.
38. P(x) = x5 - 4x4 - 3x3 +22x2 - 4x - 24 has possible rational zeros ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24. Since P(x)
has 3variations in sign, there are 1or 3positive real zeros. Since P(-x) = -x5 - 4x4 + 3x3 + 22x2 + 4x - 24 has 2
variations in sign, there are0 or 2 negative real zeros.
ill -4 -3
1
22 -4 -24
1
-3
-6
16
12
-3
-6
16
12
-12
2 | 1 -4 -3
2
22 -4 -24
-4
-14
16
24
1-2-7
8
12
0
=» x = 2 isa zero.
P(x) = (x - 2) (x4 - 2x3 - 7x2 + 8x + 12)
2 I1
-2
1
-7
8
12
0
-14
-12
0-7
-6
0
=> x = 2 is a zero again.
P(x) = (x-2)2(x3-7x-6)
2 | 1 0 -7
2
-6
3 11 0 -7 -6
4-6
12-3
-12
3
9
6
13
2
0
=4> x = 3 is a zero.
P(x) = (x - 2)2 (x - 3) (x2 + 3x + 2) = (x - 2)2 (x - 3) (x + 1) (x + 2) = 0. Therefore, the zeros are x = -1, ±2,
3.
39. P(x) =3x5 - 14x4 - 14x3 +36x2 +43x +10 has possible rational zeros ±1, ±2, ±5, ±10, ±\,±|, ±|, ±\. Since
P(x) has 2variations in sign, there are 0or 2positive real zeros. Since P(-x) =-3x5 - 14x4 +14x3 +36x2 - 43x+10
has3 variations in sign, there are 1 or 3 negative real zeros.
1 | 3 -14 -14
3
36 43 10
3
-11
-25
-11
-25
11
11
2| 3 -14 -14
36
43
10
6
-16
-60
-48
-10
-8
-30
-24
-5
0
54
54 64
3
=* x = 2 isa zero.
P (x) = (x- 2) (3x4 - 8x3 - 30x2 - 24x - 5)
2 I 3 -8 -30 -24
3
-5
6
_4
_68
-184
-2
-34
-92
-189
5 I 3 -8 -30 -24 -5
3
15
35
7
5
25
10
=* x = 5 isa zero.
p (x) = (x - 2) (x - 5) (3x3 + 7x2 + 5x + 1). Since 3x3 + 7x2 + 5x + 1has no variation in sign, there are no more
positive zeros.
-1 | 3
3
7
5
1
-3
-4
-1
4
1
0
=$> x = -1 is a zero.
P (x) = (x - 2) (x - 5) (x + 1) (3x2 + 4x + 1) = (x - 2) (x - 5) (x + 1) (x + 1) (3x + 1). Therefore, the zeros are
x = -1,-J,2,5.
236
CHAPTER 3 Polynomial and Rational Functions
40. P(x) = 2x6 - 3x5 - 13x4 + 29x3 - 27x2 +32x - 12 has possible rational zeros ±1, ±2, ±3, ±4,
±6, ±12, ±|, ±|. Since P(x) has 5variations in sign, there are 1or 3 or 5 positive real zeros. Since
P(-x) =2x6 +3x5 - 13x4 - 29x3 +27x2 - 32x - 12 has 3variations in sign, there are 1or 3negative real zeros.
1 I 2 -3 -13
2
32 -12
2
-1
-14
15
-12
-1
-14
15
-12
20
2 | 2 -3 -13
2
29 -27
29 -27
20
32 -12
4
2
-22
14
-26
12
1
-11
7
-13
6
0
=>x = 2isazero.
P (x) = (x - 2) (2x5 + x4 - llx3 + 7x2 - 13x + 6). We continue with the quotient:
2 | 2 1 -11
4
7 -13
10
-2
2 5-1
6
10
-6
5-3
o =>• x = 2 isa zero again.
P(x) = (x - 2) (2x4 +5x3 - x2 +5x - 3). We continue with the quotient, first noting 2is no longer apossible
rational solution:
3 | 2 5-1
2
5-3
6
22
42
94
11
21
47
91
=> x = 3 isanupper bound.
We know thatthere is at least 1 more positive zero.
\ | 2 5 -1 5 -3
12
2 6
13
2 6
0
=>• x = ^ isa zero.
P(x) = (x - 2)2 (x - i) (2x3 +6x2 +2x +6). We can factor 2x3 + 6x2 + 2x + 6by grouping;
2x3 +6x2+2x+6=(2x3 +6x2)+(2x +6) =(2x +6)(x2 +l).SoP(x) =2(x-2)2(x-l)(x +3)(x2 +l).
Since x2 + 1has no real zeros, the zeros ofP are x = -3, 2, -.
41. P(x) =x3 +4x2 +3x - 2. The possible rational zeros are ±1, ±2. P(x) has 1variation in sign and hence 1positive
real zero. P (-x) = -x3 +4x2 - 3x - 2has 2variations in sign and hence 0or 2negative real zeros.
1 | 1 4 3 -2
—1 I 1
8
5
8
6
4
-1
=£• x = 1 is an upper bound.
-2 | 1
1
4
13
3-2
-3
0
0—2
3-2
-2
-4
2
2
-1
0
=*• x = -2 is a zero.
So P(x) = (x + 2) (x2 + 2x - 1). Using the quadratic formula on the second factor, we have:
_ -2±V22-4(1)(-1) _ -2±>/8 _ -2±2v/2
, . IR x. f tU
n ,
n
,2(7)
- —2 = —2
= -1 ± V2. Therefore, thezeros arex = -2, -1 + y/2, -1 - x/2.
x-
SECTION 3.3 RealZerosof Polynomials
237
2. P(x) =x3 - 5x2 +2x +12. The possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±12. P(x) has 2variations in sign
and hence 0or 2positive real zeros. P (-x) = -x3 - 5x2 - 2x +12 has 1variation in sign and hence 1negative real zero.
1 | 1 -5
2 12
2 | 1 -5
1-4-2
2
-6
"3
-4
-4
-2
10
1
3 | 1 -5
1
2
2 12
-8
12
3
-6
-12
_2
-4
0
=> x = 3 is a zero.
So P(x) = (x - 3) (x2 - 2x - 4). Using the quadratic formula on the second factor, we have:
„. —_ -(-2)±V(-2)2-4(i)(-4)
OM\
X
2±>/2o
= 2±2^5
= x± v /q Therefore, the zeros are x = 3,1 ± \/5.
2
2
2(1)
13. P(x) = x4 - 6x3 +4x2 +15x +4. The possible rational zeros are ±1, ±2, ±4. P(x) has 2variations in sign and hence 0
or 2positive real zeros. P (-x) = x4 +6x3 + 4x2 - 15x + 4has 2variations in sign and hence 0or 2negative real zeros.
1 | 1 -6
1
4 15
4
1
-5
-1
14
-5
-1
14
18
4| 1
1
4
15
4
4
-8
-16
-4
-2
-4
-1
0
-6
1
2 | 1 -6
4 15
2
-8
~A
~A
-8
4
14
7 18~~
=> x = 4 is a zero.
So P (x) = (x - 4) (x3 - 2x2 - 4x - 1). Continuing by factoring the quotient, we have:
4| i -2 -4 -1
1
-1 | 1 -2 -4 -1
4
8
16
2
4
15 => x = 4 isanupper bound.
-13
1
1-3-1
0
=• x = -1 isa zero.
So P (x) = (x - 4) (x + 1) (x2 - 3x - 1). Using the quadratic formula on the third factor, we have:
x= -(-3)±V(-3)2-4(D(-i) = 3±^TJ Therefore, the zeros are x=4, -1, &^.
44 p(x) = x4 +2x3 - 2x2 - 3x +2. The possible rational zeros are ±1, ±2. P(x) has 2variations in sign and hence 0or 2
positive real zeros. P (-x) = x4 - 2x3 - 2x2 + 3x + 2has 2variations in sign and hence 0or 2negative real zeros.
1 | 1 2 -2 -3 2
13
1
3
1
1-2
-2
0
=> x = 1 is a zero.
P (x) = (x - 1) (x3 + 3x2 + x - 2). Continuing with the quotient:
1I1 3 1 -2
14
14 5
-1 I1 3 1 -2
5
-1 -2
3 =*>x = lisan
1
12-1-1
-2 I1 3 1 -2
-2-2
1 1 - 1 0 =• x = -2 isa zero,
upper bound.
So P (x) = (x- 1) (x+ 2) (x2 + x - 1). Using the quadratic formula on the third factor, we have:
-l±y/l'-4 (!)(-!) i±V5
2(1)
2
^
2
238
CHAPTER 3 Polynomial andRational Functions
45. P(x) =x4 - 7x3 +14x2 - 3x - 9. The possible rational zeros are ±1, ±3, ±9. P(x) has 3variations in sign and henc
1or 3positive real zeros. P(-x) = x4 + 7x3 + 14x2 + 3x - 4has 1variation in sign and hence 1negative real zero.
1| 1 -7 14 -3 -9
3| 1 -7
1-685
3
1-6854
14 -3 -9
-12
6
9
1-4230
=»x = 3isazero.
So P(x) = (x - 3) (x3 - 4x2 +2x +3). Since the constant term ofthe second term is 3, ±9 are no longer possible
zeros. Continuing by factoring the quotient, we have:
3 I 1 -4
2
3
3
-3
-3
1-1—1
0
^ x = 3 is a zero again.
So P(x) = (x - 3)2 (x2 - x - 1). Using the quadratic formula on the second factor, we have:
x= -(-i)±V(-y-4(D(-i) =i^ ^ ^ the m arex =3> ^
46. P(x) =x5 - 4x4 - x3 +10x2 +2x - 4. The possible rational zeros are ±1, ±2, ±4. P(x) has 3variations in sign and
hence 1or 3positive real zeros. P (-x) = -x5 - 4x4 + x3 + 10x2 - 2x - 4has 2variations in sign and hence 0or 2
negative real zeros.
1 ll -4 -1 10 2 -4
2 I 1 -4 -1
1-3-468
2-4
1-3-4684
1
-2
10 2 -4
-10
-5
0
020
=> x = 2 isa zero.
So P (x) = (x- 2) (x4 - 2x3 - 5x2 + 2) . Since the constant term of the second factor
is 2, ±4 are no longer possible zeros. Continuing by factoring the quotient, we have:
2| 1 -2 -5
1
0
2
2
0
-10
-20
0
-5
-10
18
-1 | 1 -2 -5 0 2
-l
3
2-2
1-3-220
=*x = -lisazero.
So P (x) = (x- 2) (x+ 1) (x3 - 3x2 - 2x - 2). Continuing by factoring the quotient, we have:
-1 I 1 -3 -2
2
-1
4
-2
1-420
=>-x = -1 is a zeroagain.
So P(x) = (x - 2) (x + l)2 (x2 - 4x + 2). Using the quadratic formula on the second factor, we have:
x=-<-4>W(-<>2-4(2M =i^ = 4^2 =2± ^.Therefore, the zeros are x=-1,2,2± V2.
47. P (x) = 4x3 - 6x2 + 1. The possible rational zeros are ±1, ±±, ±±. P (x) has 2variations in sign and hence 0or 2
positive real zeros. P (-x) = -4x3 - 6x2 + 1has 1variation in sign and hence 1negative real zero.
1 [4-6
0
1
4-2-2
I I 4 -6
2
0
-2
1
-1
4 -2 -2 -1
4-4-2
0 =* x = § is azero.
So P(x) = (x- i) (4x2 - 4x - 2) . Using the quadratic formula on the second factor, we have:
x= -(-4)±V(-ff-4(4)(-a =4_^M = 4±4^ = i^ Therefore, the zeros are x=4, **&.
SECTION 3.3 RealZerosof Polynomials
239
J. P(x) = 3x3 - 5x2 - 8x - 2. The possible rational zeros are ±1, ±2, ±|, ±|. P(x) has 1variation in sign and hence 1
positive real zero. P (-x) = -3x3 - 5x2 +8x - 2has 2variations in sign and hence 0or 2negative real zeros.
13
3
13
-5
-8
-2
3
-2
-10
-2
-10
-12
-5
-8
-2
3
3
_4
3
13
28
9
4
1
13
_M
_M
3
-5
-8
-2
6
2
-12
1
-6
-14
-5
-8
-2
2
-2
20
3
_3
_io
-f
-5
-8
-2
-6
22
-28
-11
14
-30
Thus wehave tried allthepositive rational zeros, sowetrythenegative zeros.
—1 I 3 -5 -8
-3
-2 I 3
0
8
-8
3
-2
3
-2
0
-i3 11 3d
-5
-8
-2
-1
2
2
3-6-6
0
So P(x) = (x + |) (3x2 - 6x - 6) = 3(x + |) (x2 - 2x - 2). Using the quadratic formula on the second factor, we
have. x= -(-2)±V(-2)2-4(i)(-2) = 2±y/i2 = 2±2^1 = i ± ^ Therefore, the zeros are x= -§, 1± V3.
19. P(x) = 2x4 + 15x3 + 17x2 + 3x - 1. The possible rational zeros are ±1, ±|. P (x) has 1variation in sign and hence 1
positive real zero. P (-x) = 2x4 - 15x3 + 17x2 - 3x - 1has 3variations in sign and hence 1or 3negative real zeros.
1
?
12
15
17
3
-1
8
25
2
31
25
31
2
27
4
1
2
16
1 1 2
2 [
2
4
=^- x = \ isan upper bound.
15
17
3
-1
-1
-7
-5
1
14
10
-2
0
So P(x) = (x + i) (2x3 + 14x2 + lOx - 2) = 2(x + ±) (x3 +7x2 +5x - l).
-1 | 1 7 5-1
1
-1
-6
1
6
-1
0
=» x = -1 is a zero.
So P (x) = (x + i) (2x3 + 14x2 + lOx - 2) = 2(x + \) (x + 1) (x2 + 6x - l) Using the quadratic formula on the
third factor, we have x= -W^ff-'W-l) = =^M = =^m = _3 ±,/To. Therefore, the zeros are x= -1,
-i,-3±V/10.
240
CHAPTER 3 Polynomial andRational Functions
50. P (x) = 4x5 - 18x4 - 6x3 + 91x2 - 60x + 9. The possible rational zeros are ±1, ±3, ±9, ±± ±# ±2 ±i ±2 ±9
P (x) has 4variations in sign and hence 0or 2or 4positive real zeros. P(-x) = -4x5 - 18x4 + 6x3 + 91x2 + 60x + <
has1 variation insign and hence 1 negative real zero.
1| 4 -18 -6
4
4
-14
91 -60 9 3| 4 -18 -6
-14
-20
-20
71
71
11
1
10
12
-6
4
-18
-24
91 -60
9
-72
57
19-3
-9
0
=
x = 3 is a zero.
So P(x) = (x - 3) (4x4 - 6x3 - 24x2 + 19x - 3). Continuing by factoring the quotient, we have:
3 I 4 -6 -24
19 -3
12
18
-18
6
-6
1
4
0
=$» x = 3 isa zero again.
So P (x) = (x - 3)2 (4x3 + 6x2 - 6x + 1). Continuing by factoring the quotient, we have:
3 |_4
6 -6
4
1
12
54
144
18
48
1445
h I 4 6 -6
x = 3 is an upper bound.
8
1
4
-1
-2
0
x = £ is a zero.
So P(x) = (x - 3)2 (x - I) (4x2 +8x-2)= 2(x - 3)2 (x - \) (2x2 +4x - l). Using the quadratic formula on
the second factor, we have x=^ y ^ ==^^ =-1 ±^. Therefore, the zeros are x=±, 3, -1 ±^.
51. (a) P (x) = x3 - 3x2 - 4x + 12 has possible rational zeros ±1, ±2,
(b)
±3, ±4, ±6, ±12.
ill
-3
-4
12
1
-2
-6
-2
-6
6
11
-3
-4
12
2
-2
-12
1
-1
-6
0
1
=> x = 2 is a zero
So P (x) = (x - 2) (x2 - x - 6) = (x - 2) (x + 2) (x - 3). The real zeros ofP are -2,2,3.
52. (a) P (x) = -x3 - 2x2 + 5x + 6 has possible rational zeros ±1, ±2,
(b)
±3, ±6.
1 I -1
-2
-1
-3
2
-1
-3
2
8
2
-1
-1
5 6
-2
5
6
-2
-8
-6
-4
-3
0
^
x = 2 is a
SoP(x) = (x-2)(-x2-4x-3) = -(x-2)(x2+4x + 3) = -(x-2)(x + l)(x + 3).TherealzerosofP
are 2, -1, -3.
SECTION 3.3 RealZerosof Polynomials
t. (a) P (x) = 2x3 - 7x2 + 4x + 4 has possible rational zeros ±1, ±2,
241
(b)
±4,±i.
112 —7
4 4
212 —7
4 4
2
-5 -1
4
-6 -4
2 -5 -1 -3
2-3-2
0 =» x = 2 is a zero.
So P (x) = (x- 2) (2x2 - 3x- 2).Continuing:
2I2
-3
-2
2
1
0
=*> x = 2 is a zero again.
Thus P(x) = (x - 2)2 (2x + 1). The real zeros of P are 2and -\.
4. (a) P(x) = 3x3 + 17x2 + 21x - 9has possible rational zeros ±1, ±3,
(b)
±9,±i,±|.
1 | 3 17 21 -9
3
20
41
3
20
41
32
i
3
17
21
3
18
27
=» x = 1 is an upper bound.
-9
0
=$• x = 5 is a zero.
SoP(x) = (x-i)(3x2 +18x +27)=3(x-i)(x2+6x +9)=3(x-|)(x +3)2. The real zeros of Pare
-3
a, A
3.
55. (a) P (x) = x4 - 5x3 + 6x2 + 4x- 8 has possible rational zeros ±1,
(b)
±2, ±4, ±8.
1I1
2
-5
6
4-8
1
-4
2
6
1
-4
2
6
-2
1
-5
6
4
-8
2
-6
0
8
1-3
0
4
x = 2 is a zero.
So P (x) = (x - 2) (x3 - 3x2 + 4) and the possible rational zeros are restricted to -1, ±2, ±4.
2 | 1 -3
1
0
4
2
-2
-4
-1
-2
0
=^x = 2isa zero again.
P(x) = (x-2)2(x2-x-2) = (x - 2)2 (x - 2) (x +1) = (x - 2)3 (x +1). So the real zeros of Pare -land
242
CHAPTER 3 Polynomial and Rational Functions
56. (a) P (x) = -x4 + 10x2 + 8x - 8has possible rational zeros ±1, ±2,
(b)
±4, ±8.
11 —1
-1
41-1
0 10 8 -8
21 —1
-1
-1
9
17
-1
9
17
9
0
10
8
0 10 8 -8
-2 -4 12
-1 -2
6 20
40
32
-8
-4 -16 -24 -64
—1 -4
—6 -16 -72 =*• x = 4 is an upper bound.
-1 I -1 0 10
1
8 -8
-1
-2 I -1 0 10
-9
-119-1-7
8
-8
2
-4
-12
8
-12
6
-4
0
=^ x = —2 is a zero.
So P (x) = (x + 2) (-x3 + 2x2 + 6x - 4). Continuing, we have:
-2 | -1 2
—1
6-4
2-8
4
4
0
—2
=£• x = —2 is a zero again.
P (x) = (x + 2)2 (-x2 + 4x - 2). Using the quadratic formula on the second factor,
a=-4±^-4(-i)(-g) =^
we have
=^£=2±V2 so the real zeros ofP are-2, 2±\/2.
57. (a) P (x) = x5 - x4 - 5x3 + x2 + 8x + 4has possible rational zeros
(b)
±1, ±2, ±4.
11
-1
1
0
1
1
-1
1
-5
8
4
0-5-4
4
-5
-4
4
-5
1
8
4
2
-6
-10
-4
-3
-5
-2
0
8
=> x = 2 is a zero.
So P (x) = (x - 2) (x4 + x3 - 3x2 - 5x - 2), and the possible
rational zeros are restricted to -1, ±2.
2 I 1 1 -3
-5
-2
6
1
3
3
1
0
=^> x = 2 is a zero again.
So P (x) = (x - 2)2 (x3 + 3x2 + 3x + l), and the possible rational zeros are restricted to -1.
-ill
1
3
3
1
•1
-2
•1
2
1
0
=>• x = —1 is a zero.
So P (x) = (x- 2)2 (x + 1) (x2 + 2x + 1) = (x - 2)2 (x + l)3., and the real zeros ofP are -1 and 2.
SECTION 3.3 RealZerosof Polynomials
8. (a) P (x) = x5 - x4 - 6x3 + 14x2 - llx + 3has possible rational
243
(b)
zeros ±1, ±3.
ill
1
-1
-6
1
0
14 -11
3
8
-3
8-3
0
-6
0-6
x = 1 is a zero.
So P(x) = (x- 1) (x4 - 6x2 + 8x- 3):
1 I1 0
-6
1
8-3
-5
11-5
3
0
=^ x = 1 is a zero again.
So P(x) = (x - l)2 (x3 + x2 - 5x+ 3):
1 I 1 1 -5
3
1
2
-3
2
-3
0
1
=*• x = 1 is a zero again.
So P (x) = (x - l)3 (x2 + 2x - 3) = (x - l)4 (x + 3), and the real zeros of P are 1and -3.
59. P(x) =x3 - x2 - x- 3. Since P(x) has 1variation in sign, Phas 1positive real zero. Since P(-x) =-x3 - x2 +x- 3
has 2 variations in sign, P has 2 or0 negative real zeros. Thus, P has 1or3 real zeros.
50. P (x) = 2x3 - x2 + 4x - 7. Since P (x) has 3variations in signs, P has 3 or 1 positive real zeros. Since
P(_£) = _2x3 - x2 - 4x - 7has no variation in sign, there is no negative real zero. Thus, P has 1or 3real zeros.
61. P(x) = 2x6 + 5x4 - x3 - 5x - 1. Since P(x) has 1variation in sign, P has 1positive real zero. Since
P (-x) = 2x6 + 5x4 + x3 + 5x - 1has 1variation in sign, P has 1negative real zero. Therefore, P has 2real zeros.
62. P (x) = x4 + x3 + x2 + x + 12. Since P (x) has no variations in sign, P has no positive real zeros. Since
P(-x) = x4 - x3 +x2 - x+ 12 has 4variations in sign, P has 4, 2, or 0negative real zeros. Therefore, P (x) has 0, 2,
or 4 real zeros.
63. P (x) = x5 + 4x3 - x2 + 6x. Since P (x) has 2variations in sign, P has 2or 0 positive real zeros. Since
P(_x) _ _x5 _ 4x3 - x2 - 6x has no variation in sign, P has no negative real zero. Therefore, P has atotal of 1or 3
real zeros (since x = 0 isa zero, but is neither positive nor negative).
64. P (x) = x8 - x5 + x4 - x3 + x2 - x+ 1. Since P (x) has 6variations in sign, the polynomial has 6, 4, 2, or 0positive
real zeros. Since P (-x) has no variation in sign, the polynomial has no negative real zeros. Therefore, P has 6, 4, 2, or 0
real zeros.
65. P (x) = 2x3 + 5x2 + x - 2;a = -3, b= 1
-2
-3
-12
-6
2
1I2
2
-1
4
-14
5
1
-2
2
7
8
7
8
6
alternating signs
all nonnegative
Therefore a = -3 and b = 1 are lower and upper bounds.
lower bound.
upper bound.
244
CHAPTER 3 Polynomial andRational Functions
66. P(x) = x4 - 2x3 - 9x2 + 2x + 8;a = -3, b= 5
3
1
-2
-9
2
8
-3
15
-18
48
1
-5
6
-16
56
5 11
-2
-9
2
8
5
15
30
160
3
6
32
168
1
Alternating signs
lower bound.
All nonnegative
upper bound.
Therefore a = -3 and b= 5 arelower and upper bounds.
67. P(x) = 8x3 + 10x2 - 39x+ 9;a = -3, 6 = 2
-3 I 8
8
2 I8
10
-39
-24
42
-9
-14
3
0
10
-39
16
52
alternating signs
lower bound.
26
8 26
13 35 all nonnegative =>• upper bound.
Therefore a = -3 andb = 2 are lower and upper bounds. Note thatx = -3 is alsoa zero.
68. P(x) = 3x4 - 17x3 + 24x2 - 9x + 1;a = 0,b= 6
0 I 3 -17 24
3
3
-9
1
0
0
0
0
-17
24
-9
1
Alternating signs
-17
24
-9
1
18
6
180
1026
1
30
171
1027
3
All nonnegative
lower bound.
=>•
upper bound.
Therefore a = 0, b= 6are lower and upper bounds, respectively. Note, since P (x) alternates in sign, by Descartes' Rule
of Signs,0 is automatically a lowerbound.
69. P (x) = x3 - 3x2 + 4 and use the Upper and Lower Bounds Theorem:
-11
1
11
1
-3
0
-1
4
-4
-4
4
00
-3
0
alternating signs
lower bound.
4
3
0
0
0
0
4
all nonnegative
=>•
upper bound.
Therefore -1 isa lower bound (and a zero) and 3 is an upper bound. (There are many possible solutions.)
SECTION3.3 Real Zeros of Polynomials
245
70. P (x) = 2x3 - 3x2 - 8x+ 12 and using the Upper and Lower Bounds Theorem:
-2 I 2 -3 -8
12
-4
14
2-7
6
-12
0 Alternating signs
=>
x = -2 is a lower bound (and azero).
3 | 2 -3 -8 12
2
6
9
3
3
1
15 All nonnegative
=>
x = 3 isanupper bound.
(Thereare manypossiblesolutions.)
71. P(x) = x4 - 2x3 + x2 - 9x + 2
1
1
-2
1
1
-1
0
-1
0
-9
3| 1
1
-9
2
-2
1
-9
2
0
-7
0
1-7
-2
1
-9
2
3
3
12
9
14
3
11
-1 | 1 -2 1 -9
1
2
all positive
1
=$•
-9
2-14
-12
upper bound.
2
-1
3
-4
13
—3
4
-13
15
alternating signs
=$>•
lower bound.
Therefore -1 is a lower bound and3 is an upper bound. (There aremany possible solutions.)
72. SetP (x) = x5 - x4 + 1.
1 I 1 -1
1
0 0 0 1
10
0
0
0
0
0
0
1
0
-1 | 1 -1 0
0 0
-12-2
2
All nonnegative
=£•
x = 1 is an upperbound.
1
-2
1-22-22-1
(There are many possible solutions.)
Alternating signs
=*•
x =-1 is a lowerbound.
73. P (x) = 2x4 + 3x3 - 4x2 - 3x + 2.
1I2 3
2
-4
5
-3
2
1
-2
1-2
0
^ x = lisazero.
P(x) = (x- 1) (2x3 + 5x2 + x - 2)
-1 | 2
5
-2
2
1-2
-3
2
3-20
^-x = —1 is a zero.
P (x) = (x- 1) (x+ 1) (2x2 + 3x- 2) = (x- 1) (x+ 1) (2x - 1) (x+ 2). Therefore, the zeros are x = -2, A ±1.
246
CHAPTER 3 Polynomial and Rational Functions
74. P (x) = 2x4 + 15x3 + 31x2 + 20x + 4. The possible rational zeros are ±1, ±2, ±4, ±±. Since all ofthe coefficients an
positive, there are no positive zeros. Since P (-x) = 2x4 - 15x3 + 31x2 - 20x + 4 has 4 variations in sign, there are 0,2
or 4 negative real zeros.
—112 15 31 20 4
-212 15 31 20 4
-2 -13 -18 -2
2 13
18
2
-4 -22 -18 -4
2
2 11
9
2
0 => x =-2 is a zero.
P(x) = (x+ 2) (2x3 + llx2 + 9x + 2):
-212 11
9 2
-412 11
-4 -14 10
2
7
9 2
-J 12 11 9 2
-8 -12 12
-5 12
2
3
-1 -5 -2
-3 14
2 10
4
0 x = -Aisazero.
P(x) = (x +2) (2x +1) (x2 +5x +2). Now if x2 +5x +2= 0, then x= ~5±^252"4(1)(2) = -5*v^, Thus, the
zeros are -2, -±, and -5*vT7;
75. Method 1: P(x) = 4x4 - 21x2 + 5has 2variations in sign, so by Descartes' rule ofsigns there are either 2or 0positive
zeros. Ifwe replace x with (-x), the function does not change, so there are either 2or 0 negative zeros. Possible rational
zeros are ±1, ±\, ±\, ±5, ±|, ±|.By inspection, ±1 and ±5 are not zeros, so we must look for non-integer solutions:
i | 4 0 -21
4
0
5
2
1
-10
-5
2
-20
-10
0
=>x = iisazero
P (x) = (x- §) (4x3 + 2x2 - 20x - 10), continuing with the quotient, we have:
-3 | 4
4
2-20 -10
-2
0
10
0
-20
0
=>x = -±isazero.
P(x) =(x - ±) \x +ij (4x2 -20) =0. If4x2 -20 =0, then x=±y/E. Thus the zeros are x=±±,±y/E.
Method 2: Substituting u = x2, the equation becomes 4w2 - 21u + 5 = 0, which factors:
Au2 - 21u +5= (Au - 1) (u - 5) = (4x2 - l) (x2 - 5). Then either we have x2 = 5, so that x = ±^5, or we have
x2 =|, so that x= ±yj\ =±1. Thus the zeros are x=±±, ±VE.
76. P(x) = 6x4 - 7x3 - 8x2 +5x = x(6x3 - 7x2 - 8x +5). So x = 0is azero. Continuing with the quotient,
Q(x) =6x3 - 7x2 - 8x +5. The possible rational zeros are ±1, ±5, ±±, ±§, ±1 ±|, ±1, ±|. Since Q(x) has 2
variations in sign, there are 0or 2positive real zeros. Since Q(-x) = 6x4 + 7x3 - 8x2 - 5x has 1variation in sign, there
is 1 negative real zero.
1 | 6 -7 -8
5
6
-1
-9
6 -1
-9
-4
5| 6 -7 -8
6
30
115
23
107 540
h I 6 -7
-8
5
3
-2
-5
-4
-10
0
6
5
535
All positive
=>
upper bound.
=> x = | isa zero.
P(x) =x(2x - 1) (3x2 - 2x - 5) =x(2x - 1) (3x - 5) (x +1). Therefore, the zeros are 0, -1, \ and f.
SECTION3.3 Real Zeros of Polynomials
247
77. P (x) = x5 - 7x4 + 9x3 + 23x2 - 50x + 24. The possible rational zeros are ±1,±2,±3, ±4,±6,±8,±12, ±24. P (x)
has 4variations in sign and hence 0, 2, or 4positive real zeros. P (-x) = -x5 - 7x4 - 9x3 + 23x2 + 50x + 24 has 1
variation in sign, and hence 1 negative real zero.
1 | 1 -7
9 23 -50
1-6
1-6
3
24
3
26
-24
26
-24
0
=4» x = 1 is a zero.
P (x) = (x- 1) (x4 - 6x3 + 3x2 + 26x - 24); continuing with the quotient, we try 1again.
1 | 1 -6
1
3 26 -24
1
-5
-2
24
-5
-2
24
0
=> x = 1 is a zero again.
P (x) = (x - l)2 (x3 - 5x2 - 2x + 24); continuing with the quotient, we start by trying 1again.
1 | 1 -5 -2 24
2 | 1 -5 -2
1-4-6
1
-4
-6
2
18
24
-6
-16
1-3-8
8
3 | 1 -5 -2
1
24
3
-6
-24
-2
-8
0
=> x = 3 is a zero.
P (x) = (x - l)2 (x - 3) (x2 - 2x - 8) = (x - l)2 (x - 3) (x - 4) (x + 2). Therefore, the zeros are x = -2,1, 3, 4.
78. P(x) = 8x5 - 14x4 - 22x3 + 57x2 - 35x + 6. The possible rational zeros are ±1, ±2, ±3, ±6, ±-,
±3 ±i ±1 ±i 4.2 since P (x) has 4 variations in sign, there are 0, 2, or 4 positive real zeros. Since
p (_x) = _8x5 - I4x4 + 22x3 + 57x2 + 35x + 6has 1variation in sign, there is 1negative real zero.
2 *
4*
4'
Q
8
*
_1 | 8 -14 -22 57 -35
-8
22
0
-57
6
-2 | 8 -14 -22
92
-16
60
57 -35
-76
38
6
-6
~^ Z22
0 57 ^92 98~
8 -30
38 -19
3
P (x) = (x + 2) (8x4 - 30x3 + 38x2 - 19x + 3). All the other real zeros are positive.
1 I 8 -30
8
38 -19
0 =* x = -2 is azero.
3
8
-22
16
-3
-22
16
-3
0
=*• x = 1 is a zero.
P(x) = (x+ 2) (x- 1) (8x3 - 22x2 + 16x - 3).
1 I 8 -22
16 -3
8
8
-14
-14
J I 8 -22
16
4
-9
-18
7
2
2-1
8
Since / (5) > 0 > / (1), there must be azero between \ and 1. We try
I4
8
8
-22
16
-3
6
-12
3
-16
4
0
=> x = § isa zero.
P(x) =(x +2) (x - 1) (4x - 3) (2x2 - 4x +1). Now, 2x2 - 4x +1=0when x= 4±^%*i2)W =2±fi- Thus,
the zeros are 1, §, -2, and 2±fi.
248
CHAPTER 3 Polynomial and Rational Functions
79. P (x) = x3 - x - 2. The only possible rational zeros ofP (x) are ±1 and ±2.
1 | 1 0 -1
1
-2
1
0
-1
-2
-1
0
110-2
1234
1
0
-1
-2
-1
1
0
-1
0
-2
Since therow that contains -1 alternates between nonnegative and nonpositive, -1 isa lower bound and there is noneed to
try —2. Therefore, P (x) does not haveany rational zeros.
80. P (x) = 2x4 - x3 + x+ 2. The only possible rational zeros ofP (x) are ±1, ±2, ±|.
k I 2 -1
0 1 2
1
2
2
-1
All nonnegative
0
2
-1
1
-2
-3
2
-3
-2
4
_i
2
x = - is an upper bound.
Alternating signs
x = —1 is a lower bound.
2-10
1
-1
1
-I
-2
1
2
1
2
_I
4
Therefore, there is no rational zero.
81
P(x) =3x3 - x2 - 6x +12 has possible rational zeros ±1, ±2, ±3, ±4, ±6, ±12, ±|, ±|, ±4.
3
-1
-6
12
1
3
2
-4
8
2
3
5
4
20
-1
3
-4
-2
14
-2
3
-7
8
-4
all positive
=>
alternating signs
x = 2 isan upper bound
=$>
x = -2 is a lower bound
- ±
3
-1
-6
12
3
0
-6
10
3
1
16
3
76
9
3
3
-2
28
3
3
-2
16
3
124
9
3
-3
-4
44
3
-5
2
3
3
100
9
Therefore, there is no rational zero.
82. P(x) = x50 - 5x25 + x2 - 1. The only possible rational zeros of P(x) are ±1. Since
P(1) = (I)50 - 5(I)25 +(I)2 - 1= -4 and P(-1) = (-1)50 - 5(-1)25 + (-1)2 - 1= 6, P(x) does not have a
rational zero.
83. P(x) =x3 - 3x2 - 4x +12, [-4,4] by [-15,15]. The 84. P(x) =x4 - 5x2 +4, [-4,4] by [-30,30].
possible rational zeros are ±1, ±2,±3,±4,±6,±12. By
observing the graph ofP, the rational zeros are x = -2,
2' 3-
The possible rational solutions are ±1, ±2, ±4.
By observing the graph ofthe equation, the solutions of
the given equation are x = ±1, ±2.
13
/
10
/
•*
-i
h
/
/
J
-I
°
*5
-10
i
r*~^j
4
M\J
.*-?
•20
'
-t)
»
;
SECTION 3.3 Real Zeros of Polynomials
249
86. P (x) = 3x3 + 8x2 + 5x + 2, [-3,3] by [-10,10]
The possible rational solutions are ±1, ±2, ±|, ±|. By
15. P (x) = 2x4 - 5x3 - 14x2 + 5x+ 12, [-2,5] by
[-40,40]. The possible rational zerosare
±1,±2,±3,±4,±6,±12, ±|, ±|. By observing the
observing the graph of the equation, the only realsolution
graph ofP, the zeros are x = -§, —1,1,4.
of the given equation is x = —2.
10
to
X
•2
•!
'
• 10
•20
K 1
1
hi
'
2
J
-10
37. x4 - x - 4 = 0. Possible rational solutions are ±1, ±2, ±4.
1
-1
11
1
0
0-1-4
10
1
1
1
0
2
4
8
14
1
1
0
-4
2
4
7
10
1
0
0
-1
-4
-1
1
-1
2
-1
1
-2
-2
-2
1
1
0-1-4
0
0
-1
-4
-2
4
-8
18
-2
4
-9
14
x = 2 is an upper bound.
x = —2 is a lower bound.
Therefore, we graph the function P (x) = x4 - x - 4 in the viewing rectangle [-2,2] by [-5,20] and see there are two
solutions. In the viewing rectangle [-1.3, -1.25] by [-0.1,0.1], we find the solution x w -1.28. In the viewing rectangle
[1.5.1.6] by [-0.1,0.1], we find the solution x « 1.53. Thus the solutions are x « -1.28,1.53.
01
20
/
IS2
'2
>S**S^
-2
1
/
15*
I"
US
1*
/
2
-01
88. 2x3 - 8x2 + 9x- 9 = 0. Possible rational solutions are ±1, ±3, ±9, ±£, ±f, ±f.
1 I 2 -8
2
2-6
9-9
3 I 2 -8
-6
3-6
2
9-9
6
-6
-2
3
We graph P (x) = 2x3 - 8x2 + 9x - 9 in the viewing rectangle [-4,6] by
[-40,40]. Itappears that the equation has no other real solution. We can factor
2x3 - 8x2 + 9x- 9 = (x - 3) (2x2 - 2x+ 3). Since the quotient is a quadratic
expression, we can use the quadratic formula to locate the other possible solutions:
_ 2dV2^24(2)(32, whicn ^6 not reai. So the only solution is x=3.
x = 3 is a zero.
»
20
-1
-2
° ,,
/-JO
' -*0
JS
4
6
250
CHAPTER 3 Polynomial andRational Functions
89. 4.00x4 + 4.00x3 - 10.96x2 - 5.88x+ 9.09 = 0.
lU 4 -10.96 -5.88
8
4 8
-2
4
9.09
-2.96 -8.84
0.25
4 -10.96 -5.88
9.09
-8
8
4 -4
-2.96
2[4 4 -10.96 -5.88 9.09
-2.96 -8.84
24
4 12
13.04
-314
5.92 -0.08
0.04
8
20.2 49.49 =• x = 2 is an upper bound.
4 -10.96 -5.88
-12
9.01
26.08 40.40
24 -39.12
4 -8
13.04
9.09
135
-45 144.09 => x = -3 isa lower bound.
Therefore, we graph the function P (x) = 4.00x4 + 4.00x3 - 10.96x2 - 5.88x + 9.09 in the viewing rectangle [-3,2]
by [-10,40]. There appear to be two solutions. In the viewing rectangle [-1.6, -1.4] by [-0.1,0.1], we find the solution
x « -1.50. In the viewing rectangle [0.8,1.2] by [0,1], we see that the graph comes close but does not go through the
x-axis. Thus there is nosolution here. Therefore, theonly solution is x « -1.50.
20
•J
-2
1
0
/
1
2
•10
90. x + 2x + 0.96x3 + 5x2 + lOx + 4.8 = 0. Since all the coefficients are positive, there is no positive solution. So x = 0
is an upper bound.
—211 2 0.96
-2
1
5
10
4.8 —311 2 0.96
0 -1.92 -6.16 -7.68
0 0.96
3.08
-3
3.84 -2.88
5
10
4.8
3 -11.88 20.64 -91.92
1 -1 3.96
-6.88 30.64 -87.12 =» x = -3 is a lower bound.
Therefore, we graph P(x) = x5 + 2x4 + 0.96x3 +5x2 + lOx +4.8 in the viewing rectangle [-3,0] by [-10,5] and
see that there are three possible solutions. In the viewing rectangle [-1.75, -1.7] by [-0.1,0.1], we find the solution
x « -1.71. In the viewing rectangle [-1.25, -1.15] by [-0.1,0.1], we find the solution x « -1.20. In the viewing
rectangle [-0.85, -0.75] by [-0.1,0.1], we find the solution x « -0.80. So the solutions are x « -1.71, -1.20, -0.80.
Ol
0
-01
Ol
• 174
•172\^r
•17
0
01
• 124
•1 22
-11-Vjl
-01
.116
0
•014
•012/
Sj>t
-071
•076
•01 '
91. (a) Since z > b, we have z - b> 0. Since all the coefficients ofQ(x) are nonnegative, and since z > 0, we have
Q(z) > 0(being asum ofpositive terms). Thus, P (z) = (z - b) •Q(z) + r > 0, since the sum ofapositive number
and a nonnegative number.
(b) In part (a), we showed that ifbsatisfies the conditions ofthe first part ofthe Upper and Lower Bounds Theorem and
z > b, then P (z) > 0. This means that no real zero ofP can be larger than 6, so bis an upper bound for the real zeros.
(c) Suppose -6 is a negative lower bound for the real zeros ofP (x). Then clearly 6is an upper bound for
Pi (x) = P (-x). Thus, as in Part (a), we can write Pi (x) = (x- b) •Q(x) + r, where r > 0 and the coefficients
ofQare all nonnegative, and P (x) = Pi (-x) = (-x-b)-Q (-x) +r = (x+ b)-[-Q (-x)) + r. Since the
coefficients ofQ(x) are all nonnegative, the coefficients of-Q (-x) will be alternately nonpositive and nonnegative,
which proves thesecond partof the Upper and Lower Bounds Theorem.
SECTION 3.3 Real Zeros of Polynomials
251
)2. P (x) = x5 - x4 - x3 - 5x2 - 12x - 6 has possible rational zeros ±1, ±2, ±3, ±6. Since P (x) has 1 variation in sign,
there is1 positive real zero. Since P (-x) = -x5 - x4 + x3 - 5x2 + 12x - 6 has 4 variations insign, there are 0,2, or4
negative real zeros.
-1
0
1
3
1
1
-12
-5
-1
-5
-12
-6
2
2
2
-6
-36
1
1
-3
-18
-42
-6
-1
-6
-18
0-1-6
-18
-24
1
-1
-1
-1
-5 -12 -6
3
6
15
30
2
5
10
18 48 =$> 3 is an upper bound.
ll -1
-1
-1
2
-16
6
-2
1
-6
0
54
1
-5 -12 -6
-6
=>x
= —1 is
P (x) = (x+ 1) (x4 - 2x3 + x2 - 6x - 6), continuing with the quotient we have
-1 I 1
-2
1
-1
1
-3
4
-6
-6
-4
10
-10
4
=>• -1 is a lower bound.
Therefore, there is1rational zero, namely -1. Since there are 1, 3or5 real zeros, and we found 1rational zero, there must
be 0, 2or4 irrational zeros. However, since 1zero must be positive, there cannot be 0 irrational zeros. Therefore, there is
exactly 1 rational zeroand 2 or 4 irrational zeros.
93. Let r be the radius ofthe silo. The volume ofthe hemispherical roof is \ (f 7rr3) = §7rr3. The volume ofthe cylindrical
section is tt (r2) (30) = 307rr2. Because the total volume of the silo is 15,000 ft3, we get the following equation:
§7rr3 +307rr2 = 15000 «• §7rr3 +307rr2 - 15000 =0<* Trr3 +457rr2 - 22500 = 0. Using agraphing device,
we first graph the polynomial in the viewing rectangle [0,15] by [-10000,10000]. The solution, r « 11.28 ft., is shown in
the viewing rectangle [11.2,11.4] by [-1,1].
94. Given that x is the length ofaside ofthe rectangle, we have that the length ofthe diagonal is x+10, and the length ofthe other
side ofthe rectangle is yj(x +10)2 - x*. Hence xyf(x +10)2-x* =5000 =* x2 (20x +100) =25,000,000
& 2x3 + 10x2 - 2,500,000 = 0 & x3 + 5x2 - 1,250,000 = 0. The first viewing rectangle, [0,120] by
[-100,500], shows there is one solution. The second viewing rectangle, [106,106.1] by [-0.1,0.1], shows the solution is
x = 106.08. Therefore, the dimensions of the rectangle are 47 ft by 106 ft.
ot
500
400
200
(0602
too
O
20
40
M
SO
100
120
•Ol
10*04
10*0*
KM 01
10*1
252
CHAPTER 3 Polynomial and Rational Functions
95. h (t) = 11.60* - 12.41*2 + 6.20*3
(a) It startedto snowagain.
(b)No,h(t)<A.
- 1.58*4 + 0.20i5 - O.Olt6
is shown in the viewingrectangle
(c) The function h(t) is shown in the viewing rectangle [6,6.5] by
[0,10] by [0,6].
[0,0.5]. The x-intercept ofthe function isa little less than 6.5,which
means that the snow melted just before midnight on Saturday night.
96. The volume of the box is V = 1500 = x(20 - 2x) (40 - 2x) = 4x3 - 120x2 + 800x <s>
4x3 - 120x2 +800x - 1500 = 4(x3 - 30x2 +200x - 375) = 0. Clearly, we must have 20 - 2x > 0 and so
0 < x < 10.
1
1
-30
200
-375
5
-125
375
~ 25
75
0
=>• x = 5 is a zero.
x3 - 30x2 + 200x - 375 = (x - 5) (x2 - 25x +75) = 0. Using the quadratic formula, we find the
other zeros: x = 25±V625"4(1)(7il
= 25^325
_ 25*5v/i3
oin„ 25+5^ . ,n t. .
2
2
~
2
• i,nce — 2*— > 10, the two
answers are:
x = height = 5cm, width = 20 - 2(5) = 10 cm, and length = 40 - 2(5) = 30 cm; and x = m-b^
width = 20 - (25 - 5^ =5n/T3 - 5cm, and length =40 - (25 - 5>/l3) = 15 +5x/l3 cm.
97. Let r be the radius ofthe cone and cylinder and let hbe the height ofthe cone.
Since the height and diameter are equal, we get h = 2r. So the volume ofthe
r
r3 + 30r2 - 250
cylinder is Vx = irr2 •(cylinder height) = 207rr2, and the volume ofthe cone is
1
-219
V2 = ±7rr2/i = |7rr2 (2r) = §7rr3. Since the total volume is ^E, it follows
thatf7rr3 +207rr2 = ^2ZE «*• r3 + 30r2 - 250 = 0. By Descartes' Rule
2
-122
3
ofSigns, there is 1positive zero. Since r is between 2.76 and 2.765 (see the table),
2.76
-2.33
the radius should be2.76 m(correct totwo decimals).
2.77
1.44
3
3
2.7
47
-11.62
2.765
1.44
2.8
7.15
98. (a) Let xbe the length, in ft, ofeach side ofthe base and let hbe the height. The volume ofthe box is V= 2v^ = hx2,
and so hx2 =2x/2. The length ofthe diagonal on the base is Vx2 +x2 = VW, and hence the length ofthe diagonal
between opposite corners is \/2x2 + h2 = x+1. Squaring both sides ofthe equation, we have 2x2 +h? = x2 +2x +1
h2 = -x2 + 2x + 1 & h= V-x2 + 2x + 1. Therefore, 2^2 = hx2 = (yj-x2 + 2x + l) x5
«*>
(-x2 + 2x + 1) x4 = 8 & x6 - 2x5 - x4 + 8 = 0
SECTION 3.3 RealZeros of Polynomials
253
(b) We graph y = x6 - 2x5 - x4 + 8in the viewing rectangle [0,5] by [-10,10], and we see that there are two solutions.
In the second viewing rectangle, [1.4,1.5] by [-1,1], we see the solution x « 1.45. The third viewing rectangle,
[2.25,2.35] by [-1,1], shows thesolution x « 2.31.
Ifx = width = length = 1.45 ft, then height = \/-x2 + 2x+ 1 = 1.34 ft, and ifx = width = length = 2.31 ft, then
height = y/-x2 + 2x + 1 = 0.53 ft.
99. Let 6 bethe width of the base, and letI bethe length of the box. Then the length plus girth is I + 46 = 108, and the
volume is V = lb2 = 2200. Solving the first equation for Iand substituting this value into the second equation yields
I = 108 - Ab => V = (108 - 46) 62 = 2200 & 463 - 10862 + 2200 = 0<& A(63 - 2762 + 550) = 0. Now
P (6) = 63 - 2762 + 550 has two variations in sign, so there are 0or 2positive real zeros. We also observe that since
l> 0,6 < 27, so 6= 27 is an upper bound. Thus the possible positive rational real zeros are 1,2,3,10,11,22,25.
1 I 1 -27
-26
0
550
2 I 1 -27
0
550
-26
-26
2
-50
-100
-25
-50
450
-26
524
-27
0
550
5
-110
-550
-22
-110
0
,.
rs /t2
nm ,i/VV -m. *u
„^ I.
p„/,x
(b) = (6-5)
(62 - 226
- 110). The other zeros are
6=
6 = 5 is a zero.
22±-x/484-4(l)(-U0) _ 22±v^924 _ 22±30.397 ji-p
Y
^
—2— —
2
• ine
positive answer from this factor is 6« 26.20.Thus we have two possible solutions, 6 = 5or 6« 26.20. If6 = 5, then
/ = 108 - 4(5) = 88; if6« 26.20, then / = 108 - 4(26.20) = 3.20. Thus the length ofthe box is either 88 in. or 3.20 in.
100. (a) An odd-degree polynomial must have areal zero. The end behavior ofsuch apolynomial requires that the graph ofthe
polynomial heads offin opposite directions as x -♦ 00 and x -» -00. Thus the graph must cross the x-axis.
(b) There are many possibilities one ofwhich is P (x) = x4 + 1.
(C) P(x) = x (x - y/2) (x+ n/2) = x3 - 2x.
(d) P(x) = (x - V2) (x + V2) (x -y/S)(x + VS)=x4- 5x2 + 6. If a polynomial with integer coefficientshas no
real zeroes, then the polynomial must have even degree.
254
CHAPTER 3 Polynomial and Rational Functions
101. (a) Substituting X - - for x we have
x3+<zx2 +6x +c= (x-|)3+a(x-|)2 +&(x-^)+c
— y3
„ v2 _l a v . °3 .
/v2
2a __ , a2\ , , _,
ab
- x ~aX +yx+27+a(x -Yx+-9)+bx-Y+c
y3 - aX
nv2 +
_i_ yX
a v+
_l —+
fl3 i aX
^2 - —X
2a2 + —+bXa3 , , __ —+c
a6
-— X
- v3±/ n±„\v2, / "2
2a2
\
/a3 , a3
a6
\
- * +(-a+a>* +(k-T-x+i,j;f+(27 +-9--T+cj
=X° +(t-a*)X+(^-f+c)
(b) x3+6x2+9x+4 = 0. Setting a = 6,6 = 9, and c= 4, we have: X3+(9 - 62) X+(32 - 18 +4) = X3-27X+18.
102. (a) Usingthe cubic
fom,ula,,= ^ +^ +(^)!+^-^ +(^! =^l+^T=_1_1 =_2.
-2
1
0-3
-2
4
-2
1-210
So (x + 2) (x2 - 2x + 1) = (x+ 2) (x - l)2 = 0 =•
x = -2,1. Using the methods from this section, we have
1 I 1 0 -3
1
2
1
-2
11-2
0
So x3 - 3x + 2= (x - 1) (x2 + x - 2) = (x - l)2 (x + 2) = 0 & x = -2,1.
\2
Sincethis factors easily, the factoring method waseasier.
(b) Using the cubic formula,
-(-54) J/(-54)2 (-27);
- \|?~("54)
Il/("54)2
I?~(~54)
2 +V
4 +I("27)3
27 +\~
V
2
V 4 "*" 27
=^ +^2^^+^-72^^7^=^7+^27 =3+3=6
6
1
0
-27
-54
36
54
9
0
x3 - 27x - 54 = (x - 6) (x2 + 6x + 9) = (x - 6) (x + 3)2 = 0 =J>
x = -3, 6.
Using methods from this section,
-1 | 1
1
0-27 -54
-1
1
26
-1
-26
-28
-2
1
1
0
-27
-2
4
-54
46
-2
-23
-8
Sox3-27x-54 = (x+ 3)(x2-3x-18) = (x-6)(x + 3)2=0 ^
Sincethis factors easily, the factoring method was easier.
-3
1
1
x = -3,6.
0
-27
-54
-3
9
54
-3
-18
0
SECTION 3.4 Complex Numbers
(c) Using the cubic formula,
12
"T +
£ 2!
•1 + 27
./-4
4 + 27
-2 + V4TT+ y-2-VT+T= y_2+V5+ \f-2-V5
From the graphing calculator, we seethat P (x) = x:i + 3x + 4 has one zero.
Using methods from this section, P (x) has possible rational zeros ±1, ±2, ±4.
1I1 0 3 4
-ill
114
114
4
0 3
4
-11-4
=> 1 is an upper bound.
1-14
0
=>
x = —1 is a zero.
P(x) = x3 + 3x + 4 = (x + 1) (x2 - x + 4). Using the quadratic formula we have:
T _ -(-1)±n/(-1)2-4(1)(.|) _ 1±V/3I
- which is not a real number. Since it is not easy to see that
V-2 + y/E + \f -2 —y/E = —1, we see that the factoring method was much easier.
.4
Complex Numbers
5 - li: real part 5, imaginary part -7.
-2 - 5/
o
11.
2
5-,
2 •
2. -6 + Ai: real pan —6, imaginary part 4.
-
4 + li
= ~3 - T\r- real Par' -f. imaginary part -^.
4. —-—= 2+ p: real part 2, imaginary part |.
3: real part 3, imaginary part 0.
6. -J: real part -§, imaginary part 0.
-1i: real part 0, imaginary part - f.
8. i^3: real part 0, imaginary part y/5.
v/3 +>/=4 = \/3 +2/: real part \/3, imaginary part 2.
(2-5i) + (3 +4i) = (2 +3) + (-5+4)i =5-i
10. 2- /=5 =2- i-JZ: real part 2, imaginary part -y/§.
12. (2 + M) + (4-6i) = (2 + 4) + (5 - 6)i = 6-t
13.
(-6 + 60 + (9 - i) = (-6 + 9) + (6- 1)i = 3 + 5i
14.
(3 - 2/) + (-5 - }<) = (3 - 5) + (-2 - J) i = -2 - |<
15.
3i + (6 - 40 = 6 + (3i - Ai) = 6 - 7/
16.
(J-W +(i +M =(i +i) +H +i)< =i
(7 - }i) - (5 + |<) = (7 - 5) + (-1 - |) i = 2- »
17.
18.
(-4 + 0 - (2 - 50 = -4 + i - 2+ ft = (-4 - 2) + (1 + 5) i = -6 + 6/
19.
(-12 + 8i) - (7 + 40 = -12 + 8i- 7- 4i = (-12- 7) + (8 - A)i = -19 + 4i
20.
6i - (4 - i) = 6i - 4 + i = (-4) + (6 + 1)i = -4 + li
21.
ii-(i-¥) = -i + ll-H)]i = -\ + li
22.
(0.1 - 1.10 - (1.2 - 3.60 = (0.1 - 1.2) + [-1.1 - (-3.6)] i = -1.1 + 2.5/
23.
4 (-1 + 20 = -4 + 8z
24.
2i(\-i) =i-2f =2 + i
(1 - 0 (4 + 20 = 28 + Ui - Ai - 2r = (28 + 2) + (14 - 4) i = 30 + 10/
(5 - 3i) (1 + 0 = 5+ 5/ - 3/ - 3/2 = (5 + 3) + (5 - 3) i = 8 + 2/
25.
26.
27.
(3 - Ai) (5 - 12Z) = 15 - 36/ - 20/ + 48/2 = (15 - 48) + (-36 - 20) i = -33 - 56/
28.
(I +120 (I +24/) = | + 16/ +2/ +288/2 = (| - 288) +(16 +2) i = -^ +18/
255