Chapter 1
Algebra
1.1 Introduction
In this chapter, polynomial division and the factor
and remainder theorems are explained (in Sections 1.4
to 1.6). However, before this, some essential algebra
revision on basic laws and equations is included.
For further Algebra revision, go to website:
http://books.elsevier.com/companions/0750681527
1.2 Revision of basic laws
(a) Basic operations and laws of indices
The laws of indices are:
(i) a m × a n = a m+n (ii)
a
m
a n = a m−n
(iii) (a m ) n = a m×n
(iv) a
m
n =
n
√
a m
(v) a −n =
1
a n
(vi) a 0 = 1
Problem 1. Evaluate 4a 2 bc 3 −2ac when a = 2,
b =
1
2 and c = 1
1
2
4a
2 bc
3
− 2ac = 4(2)
2
1
2
3
2
3
− 2(2)
3
2
=
4 × 2 × 2 × 3 × 3 × 3
2 × 2 × 2 × 2
−
12
2
= 27 − 6 = 21
Problem 2. Multiply 3x + 2y by x − y.
3x + 2y
x − y
Multiply by x → 3x 2 + 2x y
Multiply by −y →
−3x y − 2y 2
Adding gives:
3x 2 − xy − 2y 2
Alternatively,
(3x + 2y)(x − y) = 3x
2
− 3x y + 2x y − 2y
2
= 3x
2
− xy − 2y
2
Problem 3. Simplify
a
3 b
2 c
4
abc −2 and evaluate when
a = 3, b =
1
8 and c = 2.
a 3 b 2 c 4
abc −2 = a
3−1 b
2−1 c
4−(−2)
= a
2 bc
6
When a = 3, b =
1
8 and c = 2,
a
2 bc
6
= (3)
2
1
8
(2)
6
= (9)
1
8
(64) = 72
Problem 4. Simplify
x 2 y 3 + x y 2
x y
x 2 y 3 + x y 2
x y
=
x 2 y 3
x y
+
x y 2
x y
= x
2−1 y
3−1
+ x
1−1 y
2−1
= xy
2
+ y or y(xy + 1)
Algebra
1.1 Introduction
In this chapter, polynomial division and the factor
and remainder theorems are explained (in Sections 1.4
to 1.6). However, before this, some essential algebra
revision on basic laws and equations is included.
For further Algebra revision, go to website:
http://books.elsevier.com/companions/0750681527
1.2 Revision of basic laws
(a) Basic operations and laws of indices
The laws of indices are:
(i) a m × a n = a m+n (ii)
a
m
a n = a m−n
(iii) (a m ) n = a m×n
(iv) a
m
n =
n
√
a m
(v) a −n =
1
a n
(vi) a 0 = 1
Problem 1. Evaluate 4a 2 bc 3 −2ac when a = 2,
b =
1
2 and c = 1
1
2
4a
2 bc
3
− 2ac = 4(2)
2
1
2
3
2
3
− 2(2)
3
2
=
4 × 2 × 2 × 3 × 3 × 3
2 × 2 × 2 × 2
−
12
2
= 27 − 6 = 21
Problem 2. Multiply 3x + 2y by x − y.
3x + 2y
x − y
Multiply by x → 3x 2 + 2x y
Multiply by −y →
−3x y − 2y 2
Adding gives:
3x 2 − xy − 2y 2
Alternatively,
(3x + 2y)(x − y) = 3x
2
− 3x y + 2x y − 2y
2
= 3x
2
− xy − 2y
2
Problem 3. Simplify
a
3 b
2 c
4
abc −2 and evaluate when
a = 3, b =
1
8 and c = 2.
a 3 b 2 c 4
abc −2 = a
3−1 b
2−1 c
4−(−2)
= a
2 bc
6
When a = 3, b =
1
8 and c = 2,
a
2 bc
6
= (3)
2
1
8
(2)
6
= (9)
1
8
(64) = 72
Problem 4. Simplify
x 2 y 3 + x y 2
x y
x 2 y 3 + x y 2
x y
=
x 2 y 3
x y
+
x y 2
x y
= x
2−1 y
3−1
+ x
1−1 y
2−1
= xy
2
+ y or y(xy + 1)
