MATH 135
Fall 2005
Assignment 10 Solutions
Problem List
Chapter 9: 5, 23, 40, 44, 45, 48, 58, 63, 100, 159
Recommended Problems
Exercise 95:
Find the sum, difference and product of each of the following pairs of polynomials with coefficients in the indicated
field.
x
3
+ 2
x
+ 2 and
x
4
+
x
2
+
x
+ 1 in
Z
3
[
x
]
Solution:
If
f
(
x
) =
x
3
+ 2
x
+ 2 and
g
(
x
) =
x
4
+
x
2
+
x
+ 1
∈
Z
3
[
x
] then,
f
(
x
) +
g
(
x
)
=
x
4
+
x
3
+
x
2
f
(
x
)

g
(
x
)
=

x
4
+
x
3

x
2
+
x
+ 1 = 2
x
4
+
x
3
+ 2
x
2
+
x
+ 1
f
(
x
)
g
(
x
)
=
(
x
3
+ 2
x
+ 2)(
x
4
+
x
2
+
x
+ 1)
=
x
7
+
x
5
+
x
4
+
x
3
+ 2
x
5
+ 2
x
3
+ 2
x
2
+ 2
x
+ 2
x
4
+ 2
x
2
+ 2
x
+ 2
=
x
7
+ 3
x
5
+ 3
x
4
+ 3
x
3
+ 4
x
2
+ 4
x
+ 2
=
x
7
+
x
2
+
x
+ 2
.
Exercise 923:
Find the quotient and remainder when
f
(
x
) is divided by
g
(
x
).
f
(
x
) =
x
3
+ 2
x
2
+ 2 and
g
(
x
) = 2
x
2
+
x
+ 1 in
Z
3
[
x
]
Solution:
Using long division (since 2(2) = 1 in
Z
3
),
2
x
2
x
2
+
x
+ 1
x
3
+ 2
x
2
+ 0
x
+ 2
x
3
+ 2
x
2
+ 2
x
x
+ 2
Hence when
f
(
x
) is divided by
g
(
x
), the quotient is 2
x
with remainder
x
+ 2.
Therefore,
f
(
x
) = (2
x
)(
g
(
x
)) + (
x
+ 2).
Exercise 940:
Solve the following polynomial equations over the indicated finite field, stating the multiplicity of any repeated solutions.
x
5
+
x
4
+
x
3
+ 2
x
2
+ 2
x
+ 2 = 0 in
Z
3
Solution:
One approach would be to immediately make a table to find a root in
Z
3
, factor out a linear factor and continue.
A second approach would be to notice a possible factorization. We rewrite the equation as
x
3
(
x
2
+
x
+ 1) + 2(
x
2
+
x
+ 1)
=
0
(
x
3
+ 2)(
x
2
+
x
+ 1)
=
0
Since we are looking over
Z
3
, then either
x
3
+ 2 = 0 or
x
2
+
x
+ 1 = 0.
Now,
x
= 1 is a root to both these equations in
Z
3
so (
x

1) = (
x
+ 2) is a factor. In fact,
x
2
+
x
+ 1 = (
x
+ 2)
2
and
x
3
+ 2 = (
x
+ 2)(
x
2
+
x
+ 1) = (
x
+ 2)
3
(after a bit of work), so the original equation becomes (
x
+ 2)
5
= 0.
Thus,
x
= 1 is the only root and has multiplicity 5.
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Exercise 944:
Find all the rational roots of each of the following rational polynomials.
x
4
+
5
6
x
3

5
18
x
2
+
1
9
Solution:
The first step to solving the equation
x
4
+
5
6
x
3

5
18
x
2
+
1
9
= 0
is to multiply through by the least common multiple of the denominators (18, in this case) to obtain a polynomial
equation with integer coefficients, namely
18
x
4
+ 15
x
3

5
x
2
+ 2 = 0
So by the Rational Roots Theorem, the possibilities for rational roots to this equation are rational numbers of the
form
p
q
where
q

18 and
p

2, i.e. the possible rational roots are
±
1
,
±
1
2
,
±
1
3
,
±
1
6
,
±
1
9
,
±
1
18
,
±
2
,
±
2
3
,
±
2
9
Checking for integer roots first, we see that
x
=

1 is a root (since it satisfies the equation), so we obtain
(
x
+ 1)(18
x
3

3
x
2

2
x
+ 2) = 0
and this new cubic polynomial has the same possibilities for rational roots as the previous quartic polynomial.
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 Spring '08
 Vanderburgh
 Polynomials, Quadratic equation, Complex number, Conjugate Roots Theorem and Theorem

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