276 Answers to Exercises and Solutions to Problems(h) t2(t + 1)2 + (t + 1) = (t3 + t2 + l)(t + 1).
(i) (3t + l)(t3 - t2 - 1).
(j) (t - 1)3(2t + 3)2.
(k) (t2 + l)(t - 1)2(t + 1)2.
(1) (t - 6)(t + 2)(t + 3).
(m) (t - 5)(t2 - 2t + 3).
(n) (t2 + t + l)(t3 - t2 -t - 1).
(0) (t2 - 2)(t3 - 3t2 + 2t - 7).
(p) (t4 + lots + l)(t4 - lot2 + 1).
2.8. Suppose that f = gh. Write g = ‘u + v where u is homogeneous and
deg v < deg u = deg g. Then gh = uh + vh and deg vh < deg uh = deg f.
Since f is homogeneous, we must have vh = 0, whence v = 0 and g is
homogeneous. Similarly, h is homogeneous.
2.9. No. A counterexample is z2y2+2ty-t2-y2 = (ty+z--y)(xy-z+y).
2.10. (a) (u - b)(a - 5b).
(b) (x - Y)(X + Y)(Y - z).
(cl (x - Y)(X - Z)(Y - z).
(d) (x + y + z)(z2 + y2 + z2 - zy - yz - IX).
(e) (a - b)(a - c)(b - c).
(0 (x - Y)(Y - z>(z - xl.
(g) Substituting z = ax + by yields a(1 + u)x3 + b(1 + b)y3 +
(a + 2ab + b2 + l)xg + (a2 + 2ab + b + 1)z2y. No choice of a, b will make
all coefficients vanish simultaneously. The polynomial is irreducible.2.11. (b) Prom (a), p,(x, y, z) is divisible by (z - y)(y - z)(z - z). Since
degp, < 3 when n = 0, 1, the only way p, can be divisible by a polynomial
of degree 3 is for it to vanish identically.
(d) qa(x, Y, z) = 1; 93(x, Y, z) = 2 + Y + z; a(x, Y, z) = x2 + y2 + z2 +
zy+yz+zz.
(e) pn(x, y, 0) = xy(y - x)(ynB2 + xfw3 +... + xnB2).
(f) 4(x, y, z) = z”-2 + (x + y)z”-3 + (x” + xy + y2)z”+ +
(x3 + x2y + xy2 + y7)z”-5 + + (x”-2 + x”--3y +. * f + y-2).
2.12.
(x + y + z)3 - x3 - y3 - z3 = 3(X + y)(y + Z)(Z + x)
(x+y+z)5-x5-y5- z5 = 5(t+y)(y+z)(z+x)(x2+y2+z2+xy+yz+zx)
(x + y+z)7-x7-y7-z7
= 7(t+y)(y+z)(z+x)(x4+...+2t3y+*..+3x2y2+***
+ 5t2yz + * * .)
= 7(x+y)(y+Z)(Z+x)[(x2+Y2+f2+xY+YZ+42
+ xyz(2 + y + z)].