Answers to Exercises and
Solutions to Problems
Answers to Exercises
Chapter 1
1.1. (a) 5, 1, 0, 7; (b) 5, 3, 2, 8; (c) 3, 0, 0, 4; (d) 2, -1, 1, 6.
1.2. (a) t7; (b) 2t3; (c) 0.
1.3. There are seven polynomials. (a) -00, 0, 0, 0, 0, 0; (b) 4, 0, 0, 3, 0,
3/16; (c) 2, 3, 0, 1, 3, 13/4; (d) 2, 0, -3, 8, 0, 7/2; (h) 3, 0, 3, -4, 0, -1;
(i) 4, 1, 0, 8, 1, -f; (k) 5, 0, 0, 6, 0, 9/16.
1.7. 650 078 260 327 823.
1.8. 731 765 148 134 177 451 740.
1.10. degpo q = degq op = (degp)(degq).
1.12. (a) Let p(t) = a,P +. .. + alt + as be such a polynomial. Then
a0 = p(0) = 0. For any real nonzero c, u,cn-’ +. .. + al = p(c)/c = 0.
(We do not know that the left side vanishes at c = 0 without further
development; in order to avoid this issue, we need to make a more elaborate
argument at this point.) Suppose, if possible, al # 0. Choose c such that
0 < c< 1 and
24a21+..*+ l%l> < bll.
Then
(^1) U73C n-1 + u,~1C”-2+...+ulI
(^2) Iall - [lu,p-l + lu,-llc”-2 + --a+ ~u2~c3
L la11 - c[bnl+ bn-1l+. ..+ b211
Iall - (1/2)l~ll > 0,
a contradiction. Hence al = 0, and, for all c # 0, u,c”-~ +. .+ + uz =
p(c)/c2 = 0. c on t inue on to show in turn that uz, us, a4,... a, all vanish.
Thus, p(t) must be the zero polynomial.
With more background, other proofs can be given. For example, the
conditions p(0) = p(1) =.. + = p(n) = 0 leads to a system of n + 1 linear
equations in the unknowns a~, al,... , a, for which the solution is unique.