135 MATH Assignment #10
Hand-In Problems None Recommended Problems 1. Text, page 262, #5 2. Text, page 262, #8 3. Text, page 262, #22 4. Text, page 262, #24 5. Text, page 263, #30 6. Text, page 263, #31 7. Text, page 263, #32 8. Text, page 263, #37 9. Text, page 263, #41 10. Text, page 263, #48 11. Text, page 263, #56 12. Text, page 264, #58 13. Text, page 264, #62 14. Text, page 265, #99 15. Text, page 267, #119 16. Text, page 269, #158
Fall 2007
MATH 135 Assignment #10 Solutions
Recommended Problems 1. Let f (x) = x3 + 2x + 2 and g(x) = x4 + x2 + x + 1. Then f (x) + g(x) = (x3 + 2x + 2) + (x4 + x2 + x + 1) = x4 + x3 + x2 + 3x + 3 = x4 + x3 + x2 f (x) g(x) = (x3 + 2x + 2) (x4 + x2 + x + 1) = x4 + x3 x2 + x + 1 = 2x4 + x3 + 2x2 + x + 1 f (x)g(x) = = = =
Fall 2007
(x3 + 2x + 2)(x4 + x2 + x + 1) x7 + 2x5 + 2x4 + x5 + 2x3 + 2x2 + x4 + 2x2 + 2x + x3 + 2x + 2 x7 + 3x5 + 3x4 + 3x3 + 4x2 + 4x + 2 x7 + x2 + x + 2
2. Let f (x) = x5 + 2x + 4 and g(x) = x7 + 5x3 + 4. Then f (x) + g(x) = (x5 + 2x + 4) + (x7 + 5x3 + 4) = x7 + x5 + 5x3 + 2x + 8 = x7 + x5 + 5x3 + 2x + 1 f (x) g(x) = (x5 + 2x + 4) (x7 + 5x3 + 4) = x7 + x5 5x3 + 2x + 0 = 6x7 + x5 + 2x3 + 2x f (x)g(x) = = = = 3. Using long division, ix
2
(x5 + 2x + 4)(x7 + 5x3 + 4) x12 + 2x8 + 4x7 + 5x8 + 10x4 + 20x3 + 4x5 + 8x + 16 x12 + 7x8 + 4x7 + 4x5 + 10x4 + 20x3 + 8x + 16 x12 + 4x7 + 4x5 + 3x4 + 6x3 + x + 2
+x2
ix2 + x 4 | x + 0x3 x4 ix3 ix3 ix3
+ + + +
(i 2) 0x2 2ix2 2ix2 x2 (1 2i)x2 (1 2i)x2
+
x+
1
+ x 2x + 3x + 1 + (i 2)x + (4 2i) (5 i)x + (2i 3)
Therefore, the quotient is ix2 + x + (i 2) and the remainder is (5 i)x + (2i 3). 4. 5x
3
+ 2x
2
+ 6x + 3
2x2 + 4x | 3x5 + 3x4 3x5 + 4x4 6x4 6x4
+ 3 + 0x3 + 5x3 + 2x3 + x3 x3 x3
+ + + + + +
4x2 6x2 5x2 3x2 2x2 6x2 3x2
+ 4x + 3 + + + + + 4x 5x 6x + 3 4x + 2 2x + 1
Therefore, the quotient is 2x2 + 4x + 3 and the remainder is 3x2 + 2x + 1. 5. Since g(x) = x (i), then by the Remainder Theorem, the remainder is f (i) = = = = i(i)6 + (1 2i)(i)5 + 5i(i)4 (i) + 4 i(1) + (1 2i)(i) + 5i(1) + i + 4 i i 2 + 5i + i + 4 2 + 4i
6. Since g(x) = x (1 4i), then by the Remainder Theorem, the remainder is f (1 4i) = = = = = (1 4i)3 + 3(1 4i) + 1 (1 4i)(1 4i)(1 4i) + 3 12i + 1 (15 8i)(1 4i) + 4 12i (47 + 52i) + 4 12i 43 + 40i
3i 2 4 3 1 . Now 2 3i 2
1 7. We rewrite g(x) as g(x) = x 2
= cis
4 3
so the remainder when
dividing f (x) by g(x) is equal to f (cis
) by the Remainder Theorem.
Using De Moivres Theorem, we calculate f cis
4 3
=
cis
4 3 72(4) 3
72
+ 2 cis + 2 cis
4 3 31(4) 3
31
1 1
= cis
= cis (96) + 2 cis 124 1 3 124 = 1 + 2 cis 3 40 1 = 2 cis 4 3
1 = 2 2 23i = 1 3i
Therefore, the remainder is 1
3i.
8. x4 + px + q is divisible by x2 + mx + 1 if the remainder after division is 0. Using long division, x2 + mx + 1 x2 | x4 + x4 + mx 0x3 mx3 (m)x3 (m)x3 + + + + (m2 1) 0x2 x2 (1)x2 m2 x2 2 (m 1)x2 (m2 1)x2 + + + + px + q
px mx (p + m)x + q 3 2 (m m)x + (m 1) 3 (p + 2m m )x + (q m2 + 1)
Thus, the remainder is 0 if and only if p + 2m m3 = 0 and q m2 + 1 = 0. Therefore, x4 + px + q is exactly divisible by x2 + mx + 1 if and only if p = m3 2m and q = m2 1. 9. Consider x4 + 2x 4 = 0 in Z7 . Making a table,
Modulo x 0 1 2x 0 2 x4 0 1 4 x + 2x 4 3 6
7 2 4 2 2
3 6 4 6
4 1 4 1
5 3 2 1
6 5 1 2
Therefore, there are no roots in Z7 . 10. By the Rational Roots Theorem, the possible rational roots of f (x) = 2x4 + x3 + x2 + 2x 6 are 1, 2, 3, 6, 1 , 3 . 2 2 Since f (1) = 0, then x 1 is a factor. Dividing out, gives f (x) = (x 1)(2x3 + 3x2 + 4x + 6). Notice that 2x3 + 3x2 + 4x + 6 = x2 (2x + 3) + 2(2x + 3) = (x2 + 2)(2x + 3) .
3 (We could have found that x = 2 is a root of 2x3 + 3x2 + 4x + 6 by trial and error using the above list.) The quadratic factor x2 + 2 has no real (and hence no rational) roots. In Q[x], f (x) = (x 1)(2x + 3)(x2 + 2) and each factor is irreducible (two linear, one irreducible quadratic). In C[x], f (x) = (x 1)(2x + 3)(x 2i)(x + 2i) and each factor is linear and so irreducible.
11. Set w = z 2 . Then the equation becomes w2 + iw + 2 = 0. We solve for w (that is, for z 2 ) using the quadratic formula: w= i i2 4(1)(2) i 9 i 3i = = = 2i or i 2 2 2
1 Therefore, z 2 = i = 1 cis 2 or z 2 = 2i = 2 cis 3 . 2 1 From the rst, z = 1 cis 4 + k for k = 0, 1. 3 From the second, z = 2 cis 4 + m for m = 0, 1.
Therefore, the four complex solutions are z1 = cis
1 4 5 4
= =
3 4 7 4
z2 = cis z3 = 2 cis z4 = 2 cis
1 i 2 1 1 i 2 2
1 2
+
= =
1 2 2 +
1 i 2
= 1 + i =1i
2
1 2
1 i 2
12. Since f (x) Q[x] and 1 + i are 8 1 are roots of f (x), then 1 i is a root (by the Conjugate Roots Theorem since f (x) and R[x]) 8 1 is a root (by Theorem 9.28). Therefore (x (1 + i))(x (1 i)) = x2 2x + 2 is a factor of f (x), as is (x ( 8 1))(x ( 8 1)) = x2 + 2x 7 Note that (x2 2x + 2)(x2 + 2x 7) = x4 9x2 + 18x 14. When divided into f (x), the quotient is 2x2 3x + 1. So f (x) = (2x2 3x + 1)(x2 2x + 2)(x2 + 2x 7) = (2x 1)(x 1)(x2 2x + 2)(x2 + 2x 7) The roots of f (x) are thus 1 , 1, 1 + i, 1 i, 8 1, 8 1. 2 The last factoring consists of irreducibles in Q[x] since each factor is linear, or quadratic with no roots in Q. In R[x], f (x) = (2x 1)(x 1)(x2 2x + 2)(x ( 8 1))(x ( 8 1)) and each factor is irreducible, as it is linear, or quadratic with no roots in R. In C[x], f (x) = (2x 1)(x 1)(x (1 + i))(x (1 i))(x ( 8 1))(x ( 8 i)). 13. Let f (x) = x4 2x3 2x2 + 10x 15. Since f (x) R[x] and c = 1 + 2i is a root of f (x), then c = 1 2i is also a root of f (x) by the Conjugate Roots Theorem. Therefore, (x (1 + 2i))(x (1 2i)) = x2 2x + 3 is a factor of f (x). When divided into f (x), the quotient is x2 5. Therefore, f (x) = (x2 2x + 3)(x2 5). The second quadratic factor has roots 5. Thus, the roots of f (x) are 1 2i and 5. 14. Let f (x) = 2x4 + 5x3 + 6x2 + 4x + 1. By the Rational Roots Theorem, the possible rational roots are 1, 1 . 2 Since f (x) has only positive coecients, it can have only negative real roots. Since f (1) = 0, then x + 1 is a factor, giving f (x) = (x + 1)(2x3 + 3x2 + 3x + 1). 1 Since f ( 2 ) = 0, then 2x + 1 is a factor, giving f (x) = (x + 1)(2x + 1)(x2 + x + 1). The remaining quadratic factor has no real roots, so is irreducible in both Q[x] and R[x]. (This is thus the factorization of f (x) into irreducibles in each of Q[x] and R[x].) 1 i 3 The roots of x2 + x + 1 = 0 are x = , so 2 1 + i 3 1 i 3 f (x) = (x + 1)(2x + 1) x x 2 2
in C[x] and each factor is irreducible. In Z5 [x], we return to the best integer factorization f (x) = (x + 1)(2x + 1)(x2 + x + 1). Let g(x) = x2 + x + 1. Making a table Modulo 5 x 0 1 2 3 4 g(x) 1 3 2 3 1 so g(x) has no roots in Z5 so is irreducible in Z5 [x] by Proposition 9.18. Thus, f (x) = (x + 1)(2x + 1)(x2 + x + 1) and each factor is irreducible in Z5 [x]. 15. Degree 1 The polynomials x and x + 1 are linear and thus irreducible. These are the only linear polynomials in Z[x]. Degree 2 Let g(x) = x2 + ax + b. (The leading coecient has to be 1, since each coecient is either 0 or 1, and the leading coecient cannot be 0.) For g(x) to be irreducible, neither 0 nor 1 is a root, by Proposition 9.18. Thus, g(0) = 0 so g(0) = 1 so g(0) = b = 1. Also, g(x) = 0 so g(1) = 1 so 1 + a + 1 = 1 so a = 1. Thus, g(x) = x2 + x + 1. Degree 3 Let h(x) = x3 + ax2 + bx + c. Similarly to above, h(0) = 1 so c = 1 and h(1) = 1 so a + b = 1 (since g(x) is cubic). Therefore, h(x) = x3 + x2 + 1 or h(x) = x3 + x + 1. Degree 4 Let f (x) = x4 + ax3 + x2 + cx + d. For f (x) to be irreducible, it has no roots in Z2 (so f (0) = 1 and f (1) = 1) and has no irreducible quadratic factors. Since f (0) = 1, then d = 1. Since f (1) = 1, then a + b + c = 1. So f (x) = x4 + x3 + 1, x4 + x2 + 1, x4 + x + 1 or x4 + x3 + x2 + x + 1. But we must also check for any irreducible quadratic factors. Since x2 + x + 1 is the only irreducible quadratic (by above), then the only fourth degree polynomial that we must remove is (x2 + x + 1)(x2 + x + 1) = x4 + 2x3 + 3x2 + 2x + 1 = x4 + x2 + 1. (Every other reducible quadratic has linear factors. This is the only possible quartic with two irreducible quadratic factors.) Therefore, the irreducibles are x1 , x + 1, x2 + x + 1, x3 + x2 + 1, x3 + x + 1, x4 + x3 + 1, x4 + x + 1, x4 + x3 + x2 + x + 1 16. Since p(x) R[x] and 2i is a root, then 2i is a root, by the Conjugate Roots Theorem. Therefore, (x 2i)(x + 2i) = x2 + 4 is a factor of p(x). Thus, p(x) = = = = (x2 + 4)(4x3 + 8x2 x 2) (x2 + 4)(4x2 (x + 2) (x + 2)) (x2 + 4)(4x2 1)(x + 2) (x2 + 4)(2x 1)(2x + 1)(x + 2)
(We could have found these additional factors using the Rational Roots Theorem.) 1 The real roots are 2 , 1 , 2. 2 The last factorization is the factorization of p(x) into irreducible factors in Q[x] and R[x], as each factor is either linear or has no roots in R. In C[x], p(x) = (x 2i)(x + 2i)(2x 1)(2x + 1)(x + 2), and each factor is linear so irreducible. In Z7 [x], we return to the best integer factorization p(x) = (x2 + 4)(2x 1)(2x + 1)(x + 2). Is x2 + 4 irreducible in Z7 [x]? It is enough to check for roots, since x2 + 4 is quadratic. Making a table Modulo 7 x 0 1 2 3 4 5 6 x2 + 4 4 5 1 6 6 1 5 so x2 + 4 has no roots in Z7 , so is irreducible in Z7 [x]. Therefore, p(x) = (x2 + 4)(2x 1)(2x + 1)(x + 2) is the factorization of p(x) into irreducibles in Z7 [x].