In this chapter, polynomial division and the factor
and remainder theorems are explained
,,,. However, before this, some essential algebra
revision on basic laws and equations is included
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Chapter 1
Algebra
3x + 2y
1.1 Introduction x−y
In this chapter, polynomial division and the factor
Multiply by x → 3x 2 + 2x y
and remainder theorems are explained (in Sections 1.4
to 1.6). However, before this, some essential algebra Multiply by −y → − 3x y − 2y 2
revision on basic laws and equations is included.
For further Algebra revision, go to website: Adding gives: 3x 2 − xy − 2y 2
http://books.elsevier.com/companions/0750681527
Alternatively,
(3x + 2y)(x − y) = 3x 2 − 3x y + 2x y − 2y 2
1.2 Revision of basic laws
= 3x 2 − xy − 2y 2
(a) Basic operations and laws of indices
The laws of indices are:
a 3 b 2 c4
am Problem 3. Simplify and evaluate when
(i) a m × a n = a m+n (ii) = a m−n abc−2
an a = 3, b = 1
and c = 2.
m √ 8
(iii) (a m )n = a m×n (iv) a n = n a m
a 3 b 2 c4
a −n
1
= n a0 =1 = a 3−1b2−1c4−(−2) = a 2 bc6
(v)
a
(vi) abc−2
When a = 3, b = and c = 2,
1
8
Problem 1. Evaluate 4a 2 bc3−2ac when a = 2,
b = 12 and c = 1 12 a 2 bc6 = (3)2 18 (2)6 = (9) 18 (64) = 72
3 x 2 y3 + x y2
1 3 3 Problem 4. Simplify
4a bc − 2ac = 4(2)
2 3
− 2(2)
2
xy
2 2 2
4 × 2 × 2 × 3 × 3 × 3 12 x 2 y3 + x y2 x 2 y3 x y2
= − = +
2×2×2×2 2 xy xy xy
= 27 − 6 = 21 = x 2−1 y 3−1 + x 1−1 y 2−1
= xy 2 + y or y(xy + 1)
Problem 2. Multiply 3x + 2y by x − y.
, 2 Higher Engineering Mathematics
√ √ (b) Brackets, factorization and precedence
(x 2 y)( x 3 y 2 )
Problem 5. Simplify 1
(x 5 y 3 ) 2 Problem 6. Simplify a 2 − (2a − ab) − a(3b + a).
√ √ 1 1 2
(x 2 y)( x 3 y 2 ) x2 y 2 x 2 y 3 a 2 − (2a − ab) − a(3b + a)
1
= 5 3
(x 5 y 3 ) 2 x y 2 2
= a 2 − 2a + ab − 3ab − a 2
= −2a − 2ab or −2a(1 + b)
= x 2+ 2 − 2 y 2 + 3 − 2
1 5 1 2 3
= x 0 y− 3
1
Problem 7. Remove the brackets and simplify the
expression:
1 1 1
= y − 3 or 1
or √
3 y
2a − [3{2(4a − b) − 5(a + 2b)} + 4a].
y 3
Removing the innermost brackets gives:
Now try the following exercise 2a − [3{8a − 2b − 5a − 10b} + 4a]
Collecting together similar terms gives:
Exercise 1 Revision of basic operations
and laws of indices 2a − [3{3a − 12b} + 4a]
1. Evaluate 2ab + 3bc − abc when a = 2, Removing the ‘curly’ brackets gives:
b = −2 and c = 4. [−16]
2a − [9a − 36b + 4a]
2. Find the value of 5 pq 2r 3 when p = 25 ,
q = −2 and r = −1. [−8] Collecting together similar terms gives:
3. From 4x − 3y + 2z subtract x + 2y − 3z. 2a − [13a − 36b]
[3x − 5y + 5z]
Removing the square brackets gives:
4. Multiply 2a − 5b + c by 3a + b. 2a − 13a + 36b = −11a + 36b or
[6a 2 − 13ab + 3ac − 5b 2 + bc]
36b − 11a
5. Simplify (x y z)(x yz ) and evaluate when
2 3 3 2
x = 12 , y = 2 and z = 3. [x 5 y 4 z 3 , 13 12 ]
Problem 8. Factorize (a) x y − 3x z
(b) 4a 2 + 16ab3 (c) 3a 2 b − 6ab 2 + 15ab.
3 1 − 21
6. Evaluate (a bc−3)(a b2 2 c) when a = 3,
b = 4 and c = 2. [±4 12 ]
(a) x y − 3x z = x( y − 3z)
a2b + a3b 1+a
7. Simplify (b) 4a 2 + 16ab3 = 4a(a + 4b3 )
a 2 b2 b
(c) 3a 2 b − 6ab 2 + 15ab = 3ab(a − 2b + 5)
1 − 12 1
(a 3 b c )(ab)
2 3
8. Simplify √ √
( a 3 b c) Problem 9. Simplify 3c + 2c × 4c + c ÷ 5c − 8c.
√
6 11 3
√
11 1 3 a b
a 6 b 3 c− 2 or √ The order of precedence is division, multiplica-
c 3
tion, addition and subtraction (sometimes remembered
by BODMAS). Hence
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