SECTION 3.3 Real Zeros of Polynomials 227...P\x)=C(x +1) (x -2)2 =C(x3 -3x2 +4). Since P(0) =4we...

$: SECTION 3.3 Real Zeros of Polynomials 227...P\x)=C(x +1) (x -2)2 =C(x3 -3x2 +4). Since P(0) =4we have 4=C[(0)3 - 3(0)2 +4] 4* 4=AC 4* C= 1and P(x) = x3 - 3x2 + 4. 65. 1he y-intercept$
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 arex + 2, x + 1, x, x - 1, andx - 2. ThusP(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 thatP (x) = C(x - 1) (x + 2) (x - 3) = C(x3 - 2x2 - 5x + 6) = Cx3 - 2Cx2 - 5Cx + 6C . Since the coefficient ofx2 is to be 3, -2C= 3so C= -§. Therefore, P (x) = -f (x3 - 2x2 - 5x + 6) = -f x3 + 3x2 + f x - 9is thepolynomial.

62. Since the zeros of the polynomial are 1, -1, 2, and |, it follows thatP (x) = C(x - 1) (x + 1) (x - 2) (x - §) = C (x2 - 1) (x2 - \x + 1). To ensure integercoefficients, 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 ofmanypolynomials with the desired zeros and integer coefficients. We can choose C tobeany even integer.)

63. The y-intercept is 2 and the zeros of the polynomial are -1, 1, and 2. It follows thatP(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 thatP\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= AC4* 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. Itfollows that P (x) = C(x + 2)2(x-l)2 = C (x4 + 2x3 - 3x2 - Ax + 4). Since P(0) = 4we have 4 = C[(0)4 + 2(0)3 - 3(0)2 - 4(0) + 4] «• 4 = AC 4» C = 1. ThusP (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 followsthat P (x) = C(x+ 2) (x+ 1)(x - l)2 = C (x4 + x3 - 3x2 - x + 2). Since P (0) = 2 we have4= 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 isP(_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 + 4Soto calculate R (3), westartwith 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 Polynomials1. 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- 8has 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 - 3has 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 hence1 are ±1, ±2, ±4, ±8. P (x) has 3 variations in sign and

positive real zero.P(-x) = -x3 + 3x2 - 4 has 2 hence 1or3 positive real zeros.

variations insign and hence 0 or 2 negative real zeros. P (-x) = —x3 —7x2 - 14x - 8 has 0 variations insign

i I i o q a and hence no negative real zeros.

14 4 1 I 1 -7 14 -80 =>• x = 1 is a zero. 1—6 8

„. . , , , ,., . 1—6 8 0^-x = lisa zero.P (x) = x3 + 3x2 - 4 = (x - 1)(x2 + 4x+ 4)

= (x-l)(x +2)2 S°_. e tU 0 . P (x) = x3 - 7x2 + 14x - 8 = (x - 1) (x2 - 6x + 8)Therefore, the zeros are x = -2,1. v ' \ > \ i

= (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 insign and hence 1 positive realzero.P(-x) = -x3 + 3x - 2 has 2 variations insign and hence 0 or2 negative real zeros.

1 | 1 0 -3 -2 2 | 1 0 -3 -211-2 2 4 2

1 1 -2 -4 =» x = 1 is not a zero. 1 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 or2 negative real zeros.

1 | 1 4 -3 -18 2 | 1 4 -3 -1815 2 2 12 18

15 2 -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 2 | 1 -6 12 -81-5 7 2-8 8

1-5 7-1 => x = 1 is not a zero. 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.


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 12 2 I 1 -1 -8 120 -8 -12

: 0 -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 zerosare ±1, ±2, ±5, ±10.P (x) has 2 variations insign andhence0 or 2 positive real zeros.

p (_£) = -x3 - 4x2 + 7x + 10 has 1variation in signand hence 1 negative real zero.

1 I 1 -4 -7 10

±1, ±2,±3, ±6. P (x) has 2 variations in sign and hence

0or 2positive real zeros.P (-x) = -x3- 4x2 - x+6has 1 variation insign and hence 1 negative real zeros.

-4 16-1

-1 -6

-5 6 x + 1 is a factor.

So

P(x) = x3 - Ax2 +x+6= (x + 1) (x2 - 5x +6)= (x + l)(x-3)(x-2)

Therefore, the zerosare x = -1, 2,3.

-3 -10

•10 0 => x = 1 is a zero.

So

P (x) = x3 - 4x2 - 7x + 10 = (x - 1) (x2 - 3x - 10)= (x - 1)(x - 5) (x + 2)

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 nopositive real zeros.P (-x) = -x3 +3x2 - 6x +4has 3variations in sign and hence 1or 3negative real zeros.

-1 | 1 3 6 4-1 -2 -4

1 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 thediscriminant of this 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 1positivereal 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 -31-1-3 3 3 3

1 _i _3 _6 1 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 thediscriminant is b2 - Aac = (l)2 - 4(1) (1) = -3 < 0. Thus, the only real zero is x = 3.


21. Method 1: P(x) =x4 - 5x2 +4.The possible rational zeros are ±1, ±2, ±4. P(x) has 1variation in sign and hence 1positive real zero. P (-x) = x4 - 5x2 +4has 2variations in sign and hence 0or 2negative real zeros.

1 | 1 0 -5 0 41 1-4-4

1 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

±2

-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,

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 4

1-1-4 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 -4 410-4

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 andhence 1positive real zero. P(-x) =x4 - 6x3 +7x2 +6x - 8has 3variations in sign and hence 1or 3negative real zeros.

1 | 1 6 7-6-81 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 byfactoring thequotient, wehave:

-11 7 14

-1 -6 -8

1 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 +90has2 variations insign, P has0 or 2 negative real zeros.

1 | 1 -1 -23 -3 90 2| 1 -1 -23 -3 901 Q -23 -26 2 2 -42 -90

1 0 -23 -26 64 1 1 -21 -45 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 2variations insign, there are 0 or2 negative real zeros.

1 I 4 0 -25 0 36 2 I 4 0 -25 0 364 4 -21 -21 8 16 -18 -36

1 4 I2I ^21 15~ 4 8-9 -18 0 => x = 2is azero.

P (x) = (x - 2) (4x3 + 8x2 - 9x - 18)

2 I 4 8 -9 -18 2 8 -9 -18

2 5-2

10 -4 -20all positive, x = 2 isanupper bound.8 32 46

4 16 23 28 =>allF

il4 8 -9 -18

19

4

27

16

4 927

4

315

16

§ 4 8 -9 -186 21 18

4 14 12 0 =» 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; thenP(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 3 6 2 I 1 -1 -5 3 61 Q-5-2 2 2-6-6

10-5-24 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 -1 I l 1-3-3

12 27 0

1 4 9 24 =>x = 3isan upper bound. 10-30 =» x = -1 is azero.

P (x) = (x - 2) (x + 1) (x2 - 3). Therefore, the rational zeros are x = -1, 2.


27. P(x) - x4 +8x3 +24x2 +32x +16. The possible rational zeros are ±1, ±2, ±4, ±8, ±16. P(x) has no variations iisign and hence no positive real zero. P(-x) = x4 - 8x3 +24x2 - 32x +16 has 4variations in sign and hence 0or 2ornegative 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 6 12 8

-2 -8 -8

1 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, Pha1positive 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 -42

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 areX = —- ~ 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 1variation in sign, P has 1positive real zero. Since P(-x) = -8x3 +10x2 +x- 3has 2variations in sign, Phas 0or 2negative real zeros.

1 8 10 -1 -3 I1 2 8 10 -1 -3

8 18 17 4 7 3

8 18 17 14 =>x =lisan upper bound. 8 14 6 0 =• x= \ is azero.P(x) = 8x3 + 10x2 -x-3=(x-i) (8x2 +14x +6)

= 2(x-|)(4x2 + 7x + 3) = (2x-l)(x + l)(4x + 3) = 0Therefore, 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 0or2 negative real zeros.

-11 -15

x = 1 isnota zero. 4 20 49 132 =• x = 3 isnota zero.

x = 5 is nota zero. 4 20 49 132 =• x = 3 is nota zero

4 12 1

4

14

12

8

1

-11

-14

-15

20 140 645

4

14

28

8

129

-11

630

-15

2 5 -3

4 10 -6 -18 =*x=±isnotazero. 4 9 -f -^ =• x = J is not azero.

§ I 4 8 -11 -15

3 4 8 -11 -15

12 60 147

3

4

14

20

8

49

-11

132

-15

12 60 147

l

4

4

14

20

8

49

-11

132

-15

19

4

35

16

6 21 15

4 14 10 0 =>• x = § isazero.

Thus P(x) =4x3 +8x2 - llx- 15 = (x - |) (4x2 +14x +10). Continuing by factoring the quotient, whose possiblerational zeros are -1, -5, -|, -J, -§, and -f, we have

-1 I 4 14 10-4 -10

4 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 signand hence 1positive real zero. P(-x) = -6x3 + llx2 +3x - 2has 2variations in sign and hence 0or 2negative realzeros.

11-3-2 2 6 11 -3

17 14 12 46 86

17 14 12 =>x = lisnotazero. 6 23 43 84 =» x = 2 isnot a zero.

\ 6 11 -3 -2

6 14 4 0 =^x=§isa zero.

Thus, P (x) = 6x3 + llx2 - 3x - 2= (x - \) (6x2 + 14x + 4). Continuing:-1 I 6 14 4 _2 I 6 14

-6 -8 -12 -4

6 8-4 =>x = -lisnotazero. 6 2 0 =• x = -2 isa zero.

Thus, P (x) = (x - §) (x + 2) (6x + 2) has zeros J, -2, and -§.


35. P(x) =2x4 - 7x3 +3x2 +8x - 4. The possible rational zeros are ±1, ±2, ±4, ±J. P(x) has 3variations in sign ahence 1or 3positive real zeros. P(-x) = 2x4 +7x3 +3x2 - 8x - 4has 1variation in sign and hence 1negative real ze

1I2 -7 3 8-4 I | 2 -7 3 8 -42-5-2 6 1-3 0 4

2-5-2 6 2 =»x = lisnotazero. ~2 ^6 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 84 -4 -8

2 —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,thzeros 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 2variations in sign, Phas 0or 2positive real zeros. Since P(-x) =6x4 +7x3 - 12x2 - 3x +2has 2variations in sign,has 0 or 2 negative realzeros.

1 I6 -7 -12 3 2 2 I 6 -7 -12 3 2-1 -13 -10 12 10 -4 -2

6 -1 -13 -10 -8 6 5-2-10 => x = 2isazero.

P(x) = 6x4 - 7x3 - 12x2 +3x +2= (x - 2) (6x3 +5x2 - 2x - l). Continuing by factoring the quotient, we firstnote 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 iisignand hence 1 or 3 negative real zeros.

1 | 1 3 -9 -31 0 3614-5 -36 -36

14-5 -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 quotiemwe have:

1 |j_ 4 -5 -36 -36 2 I1 4 -5 -36 -360 -36 2 12 14 -44

0 -36 -72 16 7 -22 -80

3 | 1 4 -5 -36 -363 21 48 36

1 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, sowecontinue byfactoring 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 2variations insign, there are0 or 2 negative real zeros.

ill -4 -3 22 -4 -24 2 | 1 -4 -3 22 -4 -241 -3 -6 16 12 2 -4 -14 16 24

1 -3 -6 16 12 -12 1-2-7 8 12 0 =» x = 2 isa zero.

P(x) = (x - 2) (x4 - 2x3 - 7x2 + 8x + 12)2 I 1 -2 -7 8 12

0 -14 -12

1 0-7 -6 0 => x = 2 is a zero again.

P(x) = (x-2)2(x3-7x-6)

2 | 1 0 -7 -6 311 0 -7 -62 4-6 3 9 6

12-3 -12 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, ±\,±|, ±|, ±\. SinceP(x) has 2variations in sign, there are 0or 2positive real zeros. Since P(-x) =-3x5 - 14x4 +14x3 +36x2 - 43x +10has3 variations insign, there are 1 or 3 negative real zeros.

1 | 3 -14 -14 36 43 10 2 | 3 -14 -14 36 43 103 -11 -25 11 54 6 -16 -60 -48 -10

3 -11 -25 11 54 64 3 -8 -30 -24 -5 0 =* x = 2 isazero.

P(x) = (x- 2) (3x4 - 8x3 - 30x2 - 24x - 5)2 I 3 -8 -30 -24 -5 5 I 3 -8 -30 -24 -5

6 _4 _68 -184 15 35 25

3 -2 -34 -92 -189 3 7 5 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 morepositive zeros.

-1 | 3 7 5 1-3 -4 -1

3 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 arex = -1,-J,2,5.


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 3or 5 positive real zeros. SinceP(-x) =2x6 +3x5 - 13x4 - 29x3 +27x2 - 32x - 12 has 3variations in sign, there are 1or 3negative real zeros.

1 I 2 -3 -13 29 -27 32 -122 -1 -14 15 -12 20

2 -1 -14 15 -12 20

2 | 2 -3 -13 29 -27 32 -124 2 -22 14 -26 12

2 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 7 -13 6

4 10 -2 10 -6

2 5-1 5-3 o =>• x = 2 isazero again.

P(x) = (x - 2) (2x4 +5x3 - x2 +5x - 3). We continue with the quotient, first noting 2is no longer apossiblerational solution:

3 | 2 5-1 5-36 22 42 94

2 11 21 47 91 => x = 3 isanupper bound.We know thatthere isat least 1 more positive zero.

\ | 2 5 -1 5 -312 13

2 6 2 6 0 =>• x = ^ isazero.

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 1positivereal zero. P (-x) = -x3 +4x2 - 3x - 2has 2variations in sign and hence 0or 2negative real zeros.

1 | 1 4 3 -2 —1 I 1 4 3-28 -1 -3 0

5 8 6 =£• x = 1 isanupper bound. 13 0—2

-2 | 1 4 3-2-2 -4 2

1 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 ,-

x - 2(7) - —2 = —2 = -1 ± V2.Therefore, thezeros arex = -2, -1 + y/2, -1 - x/2.


2. P(x) =x3 - 5x2 +2x +12. The possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±12. P(x) has 2variations in signand 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 2 121-4-2 2 -6 -8

-4 -2 10 1 "3 -4

3 | 1 -5 2 123 -6 -12

1 _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) 2±>/2o = 2±2^5 = x± /q Therefore, the zeros are x = 3,1 ± \/5.X — OM\ 2 2 v2(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 0or 2positive real zeros. P (-x) = x4 +6x3 +4x2 - 15x +4has 2variations in sign and hence 0or 2negative real zeros.

1 | 1 -6 4 15 4 2 | 1 -6 4 15 42 -8 -8 14

1 ~A ~A 7 18~~1 -5 -1 14

15

1 -5 -1

4 | 114 18

-6 4 4

4 -8 -16 -4

1 -2 -4 -1 0 => 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 -2 -4 -14 8 16 -13 1

1 2 4 15 => x = 4 isanupper bound. 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 2positive real zeros. P (-x) = x4 - 2x3 - 2x2 + 3x + 2has 2variations in sign and hence 0or 2negative real zeros.

1 | 1 2 -2 -3 213 1-2

1 3 1 -2 0 => x = 1 is a zero.

P (x) = (x- 1) (x3 + 3x2 + x - 2). Continuing with the quotient:1I1 3 1 -2 -1 I1 3 1 -2 -2 I1 3 1 -2

14 5 -1 -2 1 -2-2

14 5 3 =*>x = lisan 12-1-1 11-10 =• 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 henc1or 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 14 -3 -91-685 3 -12 6 9

1-6854 1-4230 =»x = 3isazero.

So P(x) = (x - 3) (x3 - 4x2 +2x +3). Since the constant term of the second term is 3, ±9 are no longer possiblezeros. Continuing by factoring the quotient, we have: 3 I 1 -4 2 3

3 -3 -3

1-1—1 0 ^ x = 3 isa 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 andhence 1or 3positive real zeros. P (-x) = -x5 - 4x4 + x3 + 10x2 - 2x - 4has 2variations in sign and hence 0or 2negative real zeros.

1 ll -4 -1 10 2 -4 2 I 1 -4 -1 10 2 -41-3-468 2-4 -10 0

1-3-4684 1 -2 -5 020 => x = 2 isa zero.

So P (x) = (x- 2) (x4 - 2x3 - 5x2 + 2) . Since the constant term of the second factoris 2, ±4 are no longer possible zeros. Continuing by factoring the quotient, we have:

2| 1 -2 -5 0 2 -1 | 1 -2 -5 0 22 0 -10 -20 -l 3 2-2

1 0 -5 -10 18 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 isa 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 2positive real zeros. P (-x) = -4x3 - 6x2 + 1has 1variation in sign and hence 1negative real zero.

1 [4-6 0 1 I I 4 -6 0 14-2-2 2 -2 -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, **&.


J. P(x) = 3x3 - 5x2 - 8x - 2. The possible rational zeros are ±1, ±2, ±|, ±|. P(x) has 1variation in sign and hence 1positive real zero. P(-x) = -3x3 - 5x2 +8x - 2has 2variations in sign and hence 0or 2negative real zeros.

13 -5 -8 -2

6 2 -12

3

13

1

-5

-6

-8

-14

-2

2 -220

3

13 -5 -8 -2

3 -2 -10

3

13

-2

-5

-10

-8

-12

-2

14

3

28

9

3 _4 _M _M 3 _3 _io -f

Thus wehave tried allthepositive rational zeros, sowetrythenegative zeros.

—1 I 3 -5 -8 -2 -2 I 3 -5 -8 -2-3 8 0

-8

3 -8 0

-i 1 33 1 d

-2

-5 -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, wehave. x= -(-2)±V(-2)2-4(i)(-2) = 2±y/i2 = 2±2^1 = i ±^ Therefore, the zeros are x= -§, 1± V3.

-6 22 -28

3 -11 14 -30

19. P(x) = 2x4 + 15x3 + 17x2 + 3x - 1. The possible rational zeros are ±1, ±|. P (x) has 1variation in sign and hence 1positive real zero. P (-x) = 2x4 - 15x3 + 17x2 - 3x - 1has 3variations in sign and hence 1or 3negative real zeros.

1

? 12 15 17 3 -1

1 825

2

31

4

2 16 2531

2

27

4=^- x = \ isan upper bound.

1 12 [ 2 15 17 3 -1

-1 -7 -5 1

2 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 -6 1

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.


50. P (x) = 4x5 - 18x4 - 6x3 + 91x2 - 60x + 9. The possible rational zeros are ±1, ±3, ±9, ±± ±# ±2 ±i ±2 ±9P (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 91 -60 9 3| 4 -18 -6 91 -60 94 -14 -20 71 1 12 -18 -72 57 -9

x = 3 is a zero.4 -14 -20 71 11 10 4 -6 -24 19-3 0 =

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

4 6 -6 1 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 1 h I4 6 -6 1

12 54 144 4 -1

4 18 48 1445 x = 3 is an upper bound. 8 -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 onthe 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,±3, ±4, ±6, ±12.

ill -3 -4 121 -2 -6

1

11-2

-3

-6

-4

6

12

2 -2 -12

1 -1 -6 0 => x = 2 is a zero

(b)

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 + 6has possible rational zeros ±1, ±2,±3, ±6.

1 I -1 -2 5 6-1 -3 2

2

-1

-1

-3

-2

2

5

8

6

-2 -8 -6

-1 -4 -3 0 ^ x = 2 is a

(b)

SoP(x) = (x-2)(-x2-4x-3) = -(x-2)(x2+4x + 3) = -(x-2)(x + l)(x + 3).TherealzerosofPare 2, -1, -3.

t. (a) P (x) = 2x3 - 7x2 + 4x + 4has possible rational zeros ±1, ±2,

±4,±i.

112 —7 4 4 212 —7 4 42 -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:

2 I 2 -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 -\.


(b)

4. (a) P(x) = 3x3 + 17x2 + 21x - 9has possible rational zeros ±1, ±3, (b)±9,±i,±|.

1 | 3 17 21 -93 20 41

3 20 41 32 =» x = 1 is an upper bound.

i 3 17 21 -9

3 18 27 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 Aa, 3.

55. (a) P (x) = x4 - 5x3 + 6x2 + 4x- 8 has possible rational zeros ±1,±2, ±4, ±8.

1 I 1 -5 6 4-81 -4 2 6

2

1

1

-4

-5

2

6

6

4

-2

-8

2 -6 0 8

1-3 0 4 x = 2 is a zero.

(b)

So P (x) = (x - 2) (x3 - 3x2 + 4) and the possible rational zeros are restricted to -1, ±2, ±4.2 | 1 -3 0 4

2 -2 -4

1 -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


56. (a) P (x) = -x4 + 10x2 + 8x - 8has possible rational zeros ±1, ±2, (b)±4, ±8.

11 —1 0 10 8 -8-1 -1 9 17

-1 -1

41-1 0

9 17 9

10 8 -8

-4 -16 -24 -64

21 —1 0 10 8 -8-2 -4 12 40

-1 -2 6 20 32

—1 -4 —6 -16 -72 =*• x = 4 isan upper bound.

-1 I -1 0 10 8 -8 -2 I -1 0 10 8 -8

1 -1 -9 2 -4 -12 8

-119-1-7 -12 6 -4

So P (x) = (x + 2) (-x3 + 2x2 + 6x - 4). Continuing, we have:

-2 | -1 2 6-42-8 4

0 =^ x = —2 is a zero.

—1 4 —2 0 =£• 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) =^ =^£=2±V2 so the real zeros ofP are-2,

57. (a) P (x) = x5 - x4 - 5x3 + x2 + 8x + 4has possible rational zeros (b)±1, ±2, ±4.

11 -1 -5 8 4

1 0-5-4 4

1 -1 -5

0 -5 -4

1 8 4

2 -6 -10 -4

4 8

1 1 -3 -5 -2 0 => x = 2 is a zero.

So P (x) = (x - 2) (x4 + x3 - 3x2 - 5x - 2), and the possiblerational zeros are restricted to -1, ±2.

2 I 1 1 -3 -5 -26

we have

2±\/2.

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 3 3 1•1 -2 •1

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.

8. (a) P (x) = x5 - x4 - 6x3 + 14x2 - llx + 3has possible rationalzeros ±1, ±3.

ill -1 -6 14 -11 31 0 -6 8 -3

1 0-6 8-3 0

So P(x) = (x- 1) (x4 - 6x2 + 8x- 3):

1 I 1 0 -6 8-31 -5

x = 1 is a zero.

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 31 2 -3


(b)

1 2 -3 0 =*• 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- 3has 2variations in sign, P has 2or0 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 3or 1 positive real zeros. SinceP(_£) = _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. SinceP(-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. SinceP(-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. SinceP(_x) _ _x5 _ 4x3 - x2 - 6x has no variation in sign, P has no negative real zero. Therefore, P has atotal of1or 3real zeros (since x = 0 isa zero, but isneither 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 0positivereal zeros. Since P (-x) has no variation in sign, the polynomial has no negative real zeros. Therefore, P has 6, 4, 2, or 0real zeros.

65. P (x) = 2x3 + 5x2 + x - 2;a = -3, b= 1

-3 -2

-6 -12

2 -1 4 -14 alternating signs

1 I 2 5 1 -22 7 8

2 7 8 6 all nonnegative

Therefore a = -3 and b = 1 are lower and upper bounds.

lower bound.

upper bound.


66. P(x) = x4 - 2x3 - 9x2 + 2x + 8;a = -3, b= 5

3 1 -2 -9 2 8

-3 15 -18 48

5 1

1

1

-5

-2

6

-9

-16

2

56

8

56 Alternating signs

5 15 30 160

1 3 6 32 168 All nonnegative

Therefore a = -3 and b= 5 arelower and upper bounds.

67. P(x) = 8x3 + 10x2 - 39x+ 9;a = -3, 6= 2

-3 I 8 10 -39-24 42 -9

8

2 I 8

-14 3

10 -39

0 alternating signs

lower bound.

upper bound.

lower bound.

16 52 26

8 26 13 35 allnonnegative =>• upper bound.

Therefore a = -3 andb= 2 are lower and upper bounds. Note thatx = -3 isalsoa zero.

68. P(x) = 3x4 - 17x3 + 24x2 - 9x+ 1;a = 0,b= 6

0 I 3 -17 24 -9 10 0 0 0

3 -17 24 -9

3 -17 24 -9

1 Alternating signs

1

lower bound.

18 6 180 1026

3 1 30 171 1027 All nonnegative =>• upper bound.

Therefore a = 0, b= 6are lower and upper bounds, respectively. Note, since P (x) alternates in sign, by Descartes' Ruleof Signs,0 is automatically a lowerbound.

69. P (x) = x3 - 3x2 + 4 and use the Upper and Lower Bounds Theorem:

-11 -3 0

0 alternating signs

-1 4 -4

1 -4 4 0

11 -3 0 4

lower bound.

3 0 0

1 0 0 4 all nonnegative =>• upperbound.

Therefore -1 isa lower bound (and azero) and 3 is an upper bound. (There are many possible solutions.)


70. P (x) = 2x3 - 3x2 - 8x+ 12 and using the Upper and Lower Bounds Theorem:-2 I 2 -3 -8 12

-4 14 -12

2-7 6 0 Alternating signs => x = -2 is a lower bound (and azero).

3 | 2 -3 -8 126 9 3

2 3 1 15 All nonnegative => x = 3 isanupper bound.(Thereare manypossiblesolutions.)

71. P(x) = x4 - 2x3+ x2 - 9x + 2

1 -2 1 -9 2

1 -1 0 -9

1 -1 0 -9 -7

3 | 1 -2 1 -9 2

2 1 -2 1 -9

2 0 2-14

0 1-7 -12

3 3 12 9

1 14 3 11 all positive =$• upper bound.

-1 | 1 -2 1 -9 2-1 3 -4 13

1 —3 4 -13 15 alternating signs =$>• lower bound.

Therefore -1 isa lower bound and3 isan upper bound. (There aremany possible solutions.)

72. SetP (x) = x5 - x4+ 1.

1 I 1 -1 0 0 0 110 0 0 0

1 0 0 0 0 1 All nonnegative =£• x = 1 is an upperbound.

-1 | 1 -1 0 0 0 1-12-2 2 -2

1-22-22-1 Alternating signs =*• x =-1 is a lowerbound.

(There are many possible solutions.)

73. P (x) = 2x4 + 3x3 - 4x2 - 3x + 2.

1 I 2 3 -4 -3 21 -2

2 5 1-2 0 ^x = lisazero.

P(x) = (x- 1) (2x3 + 5x2 + x - 2)

-1 | 2 5 1-2-2 -3 2

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.


74. P (x) = 2x4 + 15x3 + 31x2 + 20x + 4. The possible rational zeros are ±1,±2,±4, ±±. Since all ofthe coefficients anpositive, there are no positive zeros. Since P (-x) = 2x4 - 15x3 + 31x2 - 20x + 4 has 4variations in sign, there are 0,2or 4 negative real zeros.

—112 15 31 20 4 -212 15 31 20 4-2 -13 -18 -2 -4 -22 -18 -4

2 13 18 2 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 9 2 -J 12 11 9 2-4 -14 10 -8 -12 12 -1 -5 -2

2 7 -5 12 2 3 -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, thezeros are -2, -±, and -5*vT7;

75. Method 1: P(x) = 4x4 - 21x2 +5has 2variations in sign, so by Descartes' rule of signs there are either 2or 0positivezeros. Ifwe replace x with (-x), the function does not change, so there are either 2or 0negative zeros. Possible rationalzeros are ±1, ±\, ±\, ±5, ±|, ±|.By inspection, ±1 and ±5 are not zeros, so we must look for non-integer solutions:

i | 4 0 -21 0 52 1 -10 -5

4 2 -20 -10 0 =>x = iisazero

P (x) = (x- §) (4x3 + 2x2 - 20x - 10), continuing with the quotient, we have:

-3 | 4 2-20 -10-2 0 10

4 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 havex2 =|, 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 2variations in sign, there are 0or 2positive real zeros. Since Q(-x) = 6x4 + 7x3 - 8x2 - 5x has 1variation in sign, thereis 1 negative real zero.

1 | 6 -7 -8 5 5| 6 -7 -8 56 -1 -9 30 115 535

6 -1 -9 -4 6 23 107 540 All positive => upper bound.

h I 6 -7 -8 53 -2 -5

6 -4 -10 0 => x = | isazero.

P(x) =x(2x - 1) (3x2 - 2x - 5) =x(2x - 1) (3x - 5) (x +1). Therefore, the zeros are 0, -1, \ and f.


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 1variation in sign, and hence 1 negative real zero.

1 | 1 -7 9 23 -50 241-6 3 26 -24

1-6 3 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 3 26 -241 -5 -2 24

1 -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 24 3 | 1 -5 -2 241-4-6 2 -6 -16 3 -6 -24

1 -4 -6 18 1-3-8 8 1 -2 -8 0 => x = 3 isa 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 4positive real zeros. Since2 * 4* 4' Q 8 *

p (_x) = _8x5 - I4x4 + 22x3 + 57x2 + 35x + 6has 1variation in sign, there is 1negative real zero._1 | 8 -14 -22 57 -35 6 -2 | 8 -14 -22 57 -35 6

-8 22 0 -57 92 -16 60 -76 38 -6

~^ Z22 0 57 ^92 98~ 8 -30 38 -19 3 0 =* x = -2 is azero.

P (x) = (x + 2) (8x4 - 30x3 + 38x2 - 19x + 3). All the other real zeros are positive.

1 I 8 -30 38 -19 38 -22 16 -3

8 -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 J I 8 -22 16

8 -14 2 4 -9

8 -14 2-1 8 -18 7

Since / (5) > 0 > / (1), there must be azero between \ and 1. We tryI 8 -22 16 -34

6 -12 3

8 -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.


79. P (x) = x3 - x - 2. The only possible rational zeros ofP (x) are ±1 and ±2.

1 | 1 0 -1 -21 1 0

0 -1 -2 -1 0 -1 -2

110-2 1234 1

Since therow that contains -1 alternates between nonnegative and nonpositive, -1 isa lower bound and there isnoneed totry —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

-1

1 0

0 -2

1

2 0

2 -1

All nonnegative x = - is an upper bound.

-1 1

-2

2 -3

-3 2

-2 4

_i2

Alternating signs

2-10 1

x = —1 is a lower bound.

-1

2 -2

1 -I _I1 2 4

1

81

Therefore, there is no rational zero.

P(x) =3x3 - x2 - 6x +12 has possible rational zeros ±1, ±2, ±3, ±4, ±6, ±12, ±|, ±|, ±4.

1

2

-1

-2

3 -1 -6 12

3 2 -4 8

3 5 4 20

3 -4 -2 14

3 -7 8 -4

all positive => x = 2 isan upper bound

alternatingsigns =$> x = -2 is a lower bound - ±

Therefore, there is no rational zero.

82. P(x) = x50 - 5x25 + x2 - 1. The only possible rational zeros ofP(x) are ±1. SinceP(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 arational 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 The possible rational solutions are ±1, ±2, ±4.observing the graph ofP, the rational zeros are x = -2, By observing the graph ofthe equation, the solutions of2' 3- the given equation are x = ±1, ±2.

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

3 -5 2

3

100

9

13

/ 10

/ J

•* -i h -I °

/ *5/ -10

' -t)

i r*~^j 4 M\J•20

.*-? » ;

SECTION 3.3 RealZeros of Polynomials 249

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

graph ofP, the zeros are x = -§, —1,1,4.

86. P (x) = 3x3 + 8x2 + 5x+ 2,[-3,3] by [-10,10]

The possible rational solutions are ±1, ±2,±|, ±|. Byobserving thegraph of theequation, theonlyrealsolution

of the given equation is x = —2.

to

X

K 1•2 •! '• 10

•20hi

10

' -10

1 2 J

37. x4 - x - 4 = 0. Possible rational solutions are ±1, ±2, ±4.

-1

0 0-1-4 10 0-1-4

1 1 1 0 2 4 8 14

1 1 1 0 -4 1 2 4 7 10

11 0 0 -1 -4 -2 1 0 0 -1 -4

-1 1 -1 2 -2 4 -8 18

1 -1 1 -2 -2 1 -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 twosolutions. 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.

20

'2 >S**S^ -2 1 / 2

01

/IS2

I"15* US 1*

-01/

88. 2x3 - 8x2 + 9x- 9 = 0. Possible rational solutions are ±1, ±3,±9, ±£, ±f, ±f.

1 I 2 -8 9-9 3 I 2 -8 9-92 -6

2-6 3-6

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 aquadraticexpression, we can use the quadratic formula to locate the other possible solutions:

6

2 -2

_ 2dV2^24(2)(32, whicn ^6 not reai. So the only solution is x=3.

-6

3 x = 3 is a zero.

»

20

-1 -2 °

/-JO

' -*0

,, J S 4 6


89. 4.00x4 + 4.00x3 - 10.96x2 - 5.88x+ 9.09= 0.

lU 4 -10.96 -5.88 9.09

-2

8 -2.96 -8.84

4 8 -2.96 -8.84 0.25

4 4 -10.96 -5.88 9.09

2[4 4 -10.96 -5.88 9.098 24 26.08 40.40

4 12 13.04 20.2 49.49 =• x = 2 is an upper bound.

-314 4 -10.96 -5.88 9.09-8 8 5.92 -0.08 -12 24 -39.12 135

4 -4 -2.96 0.04 9.01 4 -8 13.04 -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 solutionx « -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 thex-axis. Thus there isnosolution here. Therefore, theonly solution isx « -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= 0is an upper bound.

—211 2 0.96 5 10 4.8 —311 2 0.96 5 10 4.8-2 0 -1.92 -6.16 -7.68

1 0 0.96 3.08 3.84 -2.88

Therefore, we graph P(x) = x5 + 2x4 +0.96x3 +5x2 + lOx +4.8 in the viewing rectangle [-3,0] by [-10,5] andsee that there are three possible solutions. In the viewing rectangle [-1.75, -1.7] by [-0.1,0.1], we find the solutionx « -1.71. In the viewing rectangle [-1.25, -1.15] by [-0.1,0.1], we find the solution x « -1.20. In the viewingrectangle [-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.

-3 3 -11.88 20.64 -91.92

1 -1 3.96 -6.88 30.64 -87.12 =» x = -3 is a lower bound.

Ol

0• 174 •172\^r •17

-01

Ol

0• 124 •1 22 -11-Vjl .116

-01

01

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 haveQ(z) > 0(being asum ofpositive terms). Thus, P (z) = (z - b) •Q(z) + r > 0, since the sum ofapositive numberand a nonnegative number.

(b) In part (a), we showed that ifbsatisfies the conditions ofthe first part ofthe Upper and Lower Bounds Theorem andz > 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 6 is an upper bound forPi (x) = P (-x). Thus, as in Part (a), we can write Pi (x) = (x- b) •Q(x) + r, where r > 0and the coefficientsofQare all nonnegative, and P (x) = Pi (-x) = (-x-b)-Q (-x) +r = (x+ b)-[-Q (-x)) + r. Since thecoefficients 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.


)2. P (x) = x5 - x4 - x3 - 5x2 - 12x - 6 has possible rational zeros ±1, ±2, ±3, ±6. Since P (x)has 1variation in sign,there is1positive real zero. Since P (-x) = -x5 - x4 + x3 - 5x2 + 12x - 6 has 4 variations insign, there are 0,2,or4negative real zeros.

3 1 -1

-1 -5 -12 -6

0 -1 -6 -18

1 0-1-6 -18 -24

-1 -5 -12 -6

3 6 15 30 54

1 2 5 10 18 48 =$> 3 is an upper bound.

P (x) = (x+ 1) (x4 - 2x3 + x2 - 6x - 6), continuing with the quotient we have

-1 I 1 -2 1 -6 -6

-1

-1 -4 10

-1 -5 -12 -6

2 2 2 -6 -36

1 1 1 -3 -18 -42

ll -1 -1 -5 -12 -6

-1 2 -16 6

1 -2 1 -6 -6 0 =>x = —1 is

1 -3 4 -10 4 =>• -1 is a lower bound.

Therefore, there is1rational zero, namely -1. Since there are 1, 3or5real zeros, and we found 1rational zero, there mustbe 0, 2or4 irrational zeros. However, since 1zero must be positive, there cannot be 0 irrational zeros. Therefore, there isexactly 1 rational zeroand2 or 4 irrational zeros.

93. Let r be the radius ofthe silo. The volume ofthe hemispherical roof is \ (f 7rr3) = §7rr3. The volume ofthe cylindricalsection 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 inthe 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 of the 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 isx = 106.08. Therefore, the dimensions of the rectangle are 47 ft by 106 ft.

500

400

200

too

O 20 40 M SO 100 120

ot

(0602 10*04 10*0* KM 01 10*1

•Ol


95. h (t) = 11.60* - 12.41*2 + 6.20*3

- 1.58*4 + 0.20i5 - O.Olt6

is shown in the viewingrectangle

[0,10] by [0,6].

(a) It startedto snowagain.

(b)No,h(t)<A.

(c) The function h(t) is shown in the viewing rectangle [6,6.5] by[0,0.5]. The x-intercept ofthe function isa little less than 6.5,whichmeans that the snow melted justbefore midnight onSaturday 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 so0 < x < 10.

-30 200 -3751

5 -125 375

1 ~ 25 75 0 =>• x = 5 is a zero.

x3 - 30x2 + 200x - 375 = (x - 5) (x2 - 25x +75) = 0. Using the quadratic formula, we find theother 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

cylinder is Vx = irr2 • (cylinder height) = 207rr2, and the volume ofthe cone is

V2 = ±7rr2/i = |7rr2 (2r) = §7rr3. Since the total volume is ^E, it follows

thatf7rr3 +207rr2 = ^2ZE3 3

ofSigns, there is 1positive zero. Since r is between 2.76 and 2.765 (see the table),the radius should be2.76 m(correct totwo decimals).

«*• r3 + 30r2 - 250 = 0. By Descartes' Rule

r r3 + 30r2 - 250

1 -219

2 -122

3 47

2.7 -11.62

2.76 -2.33

2.77 1.44

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 of the 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 of the diagonalbetween 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


(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, thenheight = 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 thevolume isV = lb2 = 2200. Solving the first equation for Iand substituting this value into the second equation yieldsI= 108 - Ab => V= (108 - 46) 62 = 2200 & 463 - 10862 + 2200 = 0 <& A(63 - 2762 + 550) = 0. NowP (6) = 63 - 2762 + 550 has two variations in sign, so there are 0or 2positive real zeros. We also observe that sincel> 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 0 550 2 I 1 -27 0 550-26 -26

-26 -26 524

-27

5

-22

0

-110

-110

550

-550

0

2

-25

-50

-50

6 = 5 is a zero.

-100

450

„/,x ,. rs /t2 nm ,i/VV -m. *u „^ I. 22±-x/484-4(l)(-U0) _ 22±v^924 _ 22±30.397 ji-pp (b) = (6-5) (62 - 226 - 110). The other zeros are 6 = Y ^ —2— — 2 • inepositive 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 ofthepolynomial 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-real zeroes, then the polynomial must have even degree.

5x2 + 6. If a polynomialwith integercoefficientshas no


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 nv2 _i_ a v _l fl3 i ^2 2a2 a3 , , __ a6- X - aX + yX + —+aX - —X + —+bX- —+c

- 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. Settinga= 6,6 = 9, and c= 4, we have: X3+(9 - 62) X+(32 - 18 +4) = X3-27X+18.

fom,ula,,= ^ +̂ +(^)! +̂ -^+(^! =̂ l+^T=_1_1 =_2.102. (a) Usingthe cubic

-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 have1 I 1 0 -3 2

1 1 -2

11-2 0

\2So x3 - 3x + 2= (x - 1) (x2 + x- 2) = (x - l)2 (x + 2) = 0 & x = -2,1.Sincethis factors easily, the factoring method waseasier.

(b) Using the cubic formula,

-(-54) /(-54)2 (-27);2 V 4 "*" 27

- ?~("54) Il/("54)2 I("27)3 I?~(~54) J\| 2 +V 4 + 27 +\~ V

=^ +̂ 2^^+^-72^^7^=^7+^27=3+3=66 1 0 -27 -54

36

9

54

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 0-27 -54-1 1 26

1 -1 -26 -28

-2 1 0 -27 -54 -3 1 0 -27 -54

-2 4

1 -2 -23

46

-8

-3

1 -3

Sox3-27x-54 = (x+ 3)(x2-3x-18) = (x-6)(x + 3)2=0 ^ x = -3,6.Sincethis factors easily, the factoring method waseasier.

9

-18

54

0

11.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

SECTION 3.4 Complex Numbers

(c) Using the cubic formula,

"T +12

£ 2!•1 + 27

./-4

4 + 27

-2 + V4TT+ y-2-VT+T= y_2+V5+ \f-2-V5From 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.

1 I 1 0 3 4 -ill 0 3 4114 -11-4

114 4 => 1 is an upper bound. 1-14 0 =>

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

x = —1 is a zero.

- 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 Numbers5 - li: real part 5, imaginary part -7.

-2 - 5/ 2 5-, 2 • -o = ~3 - T\r- real Par' -f. imaginary part -^.

2. -6 + Ai: real pan —6, imaginary part 4.

4 + li4. —-—= 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. 10. 2- /=5=2- i-JZ: real part 2, imaginary part -y/§.(2-5i) + (3 +4i) = (2 +3) + (-5+4)i =5-i 12. (2 + M) + (4-6i)= (2 +4) + (5 - 6)i =6-t(-6 + 60 + (9 - i) = (-6 + 9)+ (6- 1)i = 3+ 5i(3 - 2/) + (-5 - }<) = (3 - 5) + (-2 - J) i = -2 - |<3i + (6 - 40 = 6 + (3i - Ai) = 6 - 7/

(J-W +(i +M =(i +i) +H +i)< =i(7 - }i) - (5 + |<) = (7 - 5) + (-1 - |) i = 2- »(-4 + 0 - (2 - 50 = -4 + i - 2+ ft = (-4 - 2) + (1 + 5) i = -6 + 6/(-12 +8i) - (7 + 40 = -12 + 8i- 7- 4i = (-12- 7) + (8 - A)i = -19 + 4i6i - (4 - i) = 6i - 4 + i = (-4) + (6 + 1)i = -4 + li

ii-(i-¥) = -i + ll-H)]i = -\ + li(0.1 - 1.10 - (1.2 - 3.60 = (0.1 - 1.2) + [-1.1 - (-3.6)] i = -1.1 + 2.5/4 (-1 + 20 = -4 + 8z

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/(3 - Ai) (5 - 12Z) = 15 - 36/ - 20/ + 48/2 = (15 - 48) + (-36- 20) i = -33 - 56/(I+120 (I+24/) = | +16/ +2/ +288/2 =(| - 288) +(16 +2) i =-^ +18/

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SECTION 3.3 Real Zeros of Polynomials 227...P\x)=C(x +1) (x -2)2 =C(x3 -3x2 +4). Since P(0) =4we...

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Transcript of SECTION 3.3 Real Zeros of Polynomials 227...P\x)=C(x +1) (x -2)2 =C(x3 -3x2 +4). Since P(0) =4we...