January 27, 2005 11:58 l24-appb Sheet number 1 Page number 733 black
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APPENDIX B
Solving Polynomial Equations
EXERCISE SET B
1. (a) q(x)=x2+4x+2,r(x)=−11x+6
(b) q(x)=2x2+4,r(x)=9
(c) q(x)=x3−x2+2x−2,r(x)=2x+1
2. (a) q(x)=2x2−x+2,r(x)=5x+5
(b) q(x)=x3+3x2−x+2,r(x)=3x−1
(c) q(x)=5x3−5,r(x)=4x2+10
3. (a) q(x)=3x2+6x+8,r(x)=15
(b) q(x)=x3−5x2+20x−100,r(x) = 504
(c) q(x)=x4+x3+x2+x+1,r(x)=0
4. (a) q(x)=2x2+x−1,r(x)=0
(b) q(x)=2x3−5x2+3x−39,r(x) = 147
(c) q(x)=x6+x5+x4+x3+x2+x+1,r(x)=2
5. x0 1 −3 7
p(x)−4−3 101 5001
6. x1−1 3 −3 7 −7 21 −21
p(x)−24 −12 12 0 420 −168 10416 −7812
7. (a) q(x)=x2+6x+13,r = 20 (b) q(x)=x2+3x−2,r =−4
8. (a) q(x)=x4−x3+x2−x+1,r =−2(b) q(x)=x4+x3+x2+x+1,r =0
9. Assume r=a/b a and bintegers with a>0:
(a) bdivides 1, b=±1; adivides 24, a=1,2,3,4,6,8,12,24;
the possible candidates are {±1,±2,±3,±4,±6,±8,±12,±24}
(b) bdivides 3 so b=±1,±3; adivides −10 so a=1,2,5,10;
the possible candidates are {±1,±2,±5,±10,±1/3,±2/3,±5/3,±10/3}
(c) bdivides 1 so b=±1; adivides 17 so a=1,17;
the possible candidates are {±1,±17}
10. An integer zero cdivides −21, so c=±1,±3,±7,±21 are the only possibilities; substitution of
these candidates shows that the integer zeros are −7,−1,3
11. (x+ 1)(x−1)(x−2) 12. (x+ 2)(3x+ 1)(x−2)
13. (x+3)
3(x+1) 14. 2x4+x3−19x2+9
15. (x+ 3)(x+ 2)(x+1)
2(x−3) 17. −3 is the only real root.
18. x=−3/2,2±√3 are the real roots. 19. x=−2,−2/3,−1±√3 are the real roots.
