Notesale: Turn your study into money

Document Preview

Extracts from the notes are below, to see the PDF you'll receive please use the links above

---

Chapter 1

Algebra
1
...
4
to 1
...
However, before this, some essential algebra
revision on basic laws and equations is included
...
elsevier
...
2

Multiply by x → 3x 2 + 2x y
Multiply by −y →

3x 2 − xy − 2y 2

Adding gives:
Alternatively,

(3x + 2y)(x − y) = 3x 2 − 3x y + 2x y − 2y 2

Revision of basic laws

= 3x 2 − xy − 2y 2

(a) Basic operations and laws of indices
The laws of indices are:
(i) a m × a n = a m+n
(iii)

(a m )n

(v)

a −n

=

a m×n

1
= n
a

am
(ii)
= a m−n
an
√
m
(iv) a n = n a m
(vi)

a0

Problem 3
...


When a = 3, b =
Problem 1
...
Multiply 3x + 2y by x − y
...
Simplify

x 2 y3 + x y2
xy

x 2 y3 + x y2
x 2 y3 x y2
=
+
xy
xy
xy
= x 2−1 y 3−1 + x 1−1 y 2−1
= xy 2 + y or y(xy + 1)

2 Higher Engineering Mathematics

Problem 5
...


2

x2 y 2 x 2 y 3

=

x y

a 2 − (2a − ab) − a(3b + a)
= 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
...


1
y

1
3

1
or √
3 y

Remove the brackets and simplify the

2a − [3{2(4a − b) − 5(a + 2b)} + 4a]
...
Evaluate 2ab + 3bc − abc when a = 2,
b = −2 and c = 4
...
Find the value of 5 pq 2r 3 when p = 25 ,
q = −2 and r = −1
...
From 4x − 3y + 2z subtract x + 2y − 3z
...
Multiply 2a − 5b + c by 3a + b
...
Simplify (x y z)(x yz ) and evaluate when
[x 5 y 4 z 3 , 13 12 ]
x = 12 , y = 2 and z = 3
...
Simplify

− 21

c) when a = 3,
[±4 12 ]


a2b + a3b
a 2 b2
1
2

− 12

(a 3 b c )(ab)
√ √
( a 3 b c)


11

36b − 11a

2

6
...

7
...
Factorize (a) x y − 3x z
(b) 4a 2 + 16ab3 (c) 3a 2 b − 6ab 2 + 15ab
...


Simplify 3c + 2c × 4c + c ÷ 5c − 8c
...
Hence

Algebra
3c + 2c × 4c + c ÷ 5c − 8c
c 
= 3c + 2c × 4c +
− 8c
5c
1
= 3c + 8c2 + − 8c
5
1
1
= 8c2 − 5c +
or c(8c − 5) +
5
5


7
...

8
...


1
...
Simplify
(2a − 3) ÷ 4a + 5 × 6 −3a
...
Solve 4 − 3x = 2x − 11
...
e
...
Solve

3
2a
−
+ 30 − 3a
=
4a 4a

4(2a − 3) − 2(a − 4) = 3(a − 3) − 1
...
Simplify 2( p + 3q − r) − 4(r − q + 2 p) + p
...
Expand and simplify (x + y)(x − 2y)
...
Remove the brackets and simplify:
24 p − [2{3(5 p − q) − 2( p + 2q)} + 3q]
...
e
...

x − 2 3x + 4

By ‘cross-multiplying’:

3(3x + 4) = 4(x − 2)

Removing brackets gives:

9x + 12 = 4x − 8

Rearranging gives:

9x − 4x = −8 − 12

i
...


5
...

[2x y(y + 3x + 4x 2 )]

and

6
...


2
− 3y + 12
3y

−6
= −2
3

Problem 13
...
Factorize 21a 2b2 − 28ab
...
Solve
√
= 2
...
e
...
e
...


ft
m

In problems 1 to 4 solve the equations
1
...


ft
ft
= v from which,
= v−u
m
m
 
ft
= m(v − u)
and
m
m

u+

i
...


f t = m(v − u)

and

m
f = (v − u)
t

X 2 = Z 2 − R 2 and reactance X =

3
...




Z2 − R2



f +p
D
,
Problem 17
...


Rearranging gives:
Squaring both sides gives:

2

[−3]

R 2 + X 2 = Z and squaring both sides gives
R 2 + X 2 = Z 2 , from which,

1

2
...


Problem 16
...
c
...
Make the reactance X the
subject
...


p(d 2 + D 2 ) = f (D 2 − d2 )


D
f +p
=
f −p
d
f +p
D2
= 2
f −p
d

‘Cross-multiplying’ gives:
d2 ( f + p) = D 2 ( f − p)
Removing brackets gives:
d2 f + d2 p = D 2 f − D 2 p

1
1
+
= 0
...

1− t

1
−8
[4]

3(F − f )

...

5
...
Make l the subject of t = 2π

...

7
...
Make r the subject of the formula




x−y
x
1 + r2

...


Solve the simultaneous equations:
7x − 2y = 26

(1)

6x + 5y = 29
...
Remembering that the product of the two inner terms added to the product
of the two outer terms must equal −11x, the only
combination to give this is +1 and −4, i
...
,

(3)

2 × equation (2) gives:
12x + 10y = 58

(4)

equation (3) +equation (4) gives:

3x 2 − 11x − 4 = (3x + 1)(x − 4)

47x + 0 = 188
188
from which,
x=
=4
47
Substituting x = 4 in equation (1) gives:

(3x + 1)(x − 4) = 0 hence

Thus
either

(x − 4) = 0 i
...
x = 4

or

28 − 2y = 26

(b) 4x 2 + 8x + 3 = (2x + 3)(2x + 1)

from which, 28 − 26 = 2y and y = 1

(2x + 3)(2x + 1) = 0 hence

Thus

Problem 19
...

3

(3x + 1) = 0 i
...
x = − 13

(1)
(2)

either

(2x + 3) = 0 i
...
x = − 32

or

(2x + 1) = 0 i
...
x = − 12

Problem 21
...
Determine the equation in x
...
e
...
e
...
e
...
Solve the following equations by
factorization:
(a) 3x 2 − 11x − 4 = 0
(b) 4x 2 + 8x + 3 = 0
...
Solve 4x 2 + 7x + 2 = 0 giving the
answer correct to 2 decimal places
...
123
=
8
−7 + 4
...
123
=
or
8
8
i
...
x = −0
...
39

5

6 Higher Engineering Mathematics
Now try the following exercise

For example,

Exercise 4 Further problems on
simultaneous and quadratic equations
In problems 1 to 3, solve the simultaneous equations

13
——–
16 208
16
48
48
—
··
—

1
...


208
is achieved as follows:
16

[x = 6, y = −1]

(1) 16 divided into 2 won’t go
2
...

3
...

7
2 7

[x = 3, y = 4]

(6) Bring down the 8
(7) 16 divided into 48 goes 3 times

4
...
Determine the quadratic equation in x whose
roots are 2 and −5
...
Solve the following quadratic equations, correct to 3 decimal places:
(a)

−4 = 0

(b) 4t 2 − 11t + 3 = 0
...
637, −3
...
443, 0
...
4

(9) 3 × 16 = 48
(10) 48 − 48 = 0

(b) 8x 2 + 2x − 15 = 0
...

(Note that a polynomial is an expression of the
form

Polynomial division
f (x) = a + bx + cx 2 + d x 3 + · · ·

Before looking at long division in algebra let us revise
long division with numbers (we may have forgotten,
since calculators do the job for us!)

and polynomial division is sometimes required when
resolving into partial fractions—see Chapter 2
...
Divide 2x 2 + x − 3 by x − 1
...
The usual layout is shown below with the dividend
and divisor both arranged in descending powers of the
symbols
...
e
...
The divisor
is then multiplied by 2x, i
...
2x(x − 1) = 2x 2 − 2x,
which is placed under the dividend as shown
...
The process is then repeated, i
...
the
first term of the divisor, x, is divided into 3x, giving
+3, which is placed above the dividend as shown
...
The
remainder, on subtraction, is zero, which completes the
process
...
Put −2x above the
dividend

(5) −2x(x + 1) = −2x 2 − 2x
(6) Subtract
(7)

x into 5x goes 5
...
Simplify

(1) (4) (7)
x 2 − x y + y2

—————————–
x + y x 3 + 0 + 0 + y3
x3 + x2 y
− x2 y
+ y3
− x 2 y − x y2
———————
x y2 + y3
x y2 + y3
———–
· ·
———–

Thus (2x 2 + x − 3) ÷ (x − 1) = (2x + 3)
[A check can be made on this answer by multiplying
(2x + 3) by (x − 1) which equals 2x 2 + x − 3]
Problem 24
...

(1) (4) (7)
3x 2 − 2x + 5

—————————
x + 1 3x 3 + x 2 + 3x + 5
3x 3 + 3x 2
− 2x 2 + 3x + 5
− 2x 2 − 2x
————–
5x + 5
5x + 5
———
· ·
———
(1)

x into 3x 3 goes 3x 2
...

x+y

(1)

x into x 3 goes x 2
...
Put −x y above dividend

(5) −x y(x + y) = −x 2 y − x y 2
(6) Subtract
(7)

x into x y 2 goes y 2
...

Problem 26
...


x +5

——————–
x − 2 x 2 + 3x − 2
x 2 − 2x

14x 2 − 19x − 3

...
Find (5x 2 − x + 4) ÷ (x − 1)
...
Divide (3x 3 + 2x 2 − 5x + 4) by (x + 2)
...
Find

5
...

[x 2 + 2x y + y 2 ]

5x − 2
5x − 10
———
8
———

Problem 27
...


3
...

[5x − 2]

8
...



481
3
2
5x + 18x + 54x + 160 +
x −3

2

1
...

For example, consider the quadratic equation
x 2 + 2x − 8 = 0
...

Hence (x − 2)(x + 4) = 0
...
Therefore,

Thus
4a 3 − 6a 2 b + 5b 3
2a − b
= 2a 2 − 2ab − b2 +

either (x − 2) = 0, from which, x = 2
or
(x + 4) = 0, from which, x = −4

4b3
2a − b

It is clear then that a factor of (x − 2) indicates a root
of +2, while a factor of (x + 4) indicates a root of −4
...
Divide (2x 2 + x y − y 2 ) by (x + y)
...
Divide (3x 2 + 5x − 2) by (x + 2)
...
However, we could
reverse this process
...
We wouldn’t normally solve
quadratic equations this way — but suppose we have
to factorize a cubic expression (i
...
one in which the
highest power of the variable is 3)
...
It is to deal with this kind of case that
we use the factor theorem
...
The factor theorem provides a method of
factorizing any polynomial, f (x), which has simple
factors
...

Problem 28
...

Let

f (x) = x 3 − 7x − 6

If x = 1, then f (1) = 13 − 7(1) − 6 = −12
If x = 2, then f (2) = 23 − 7(2) − 6 = −12
If x = 3, then f (3)

= 33 − 7(3) − 6

=0

If f (3) = 0, then (x − 3) is a factor — from the factor
theorem
...
We can divide x 3 − 7x − 6 by
(x − 3) or we could continue our ‘trial and error’ by substituting further values for x in the given expression —
and hope to arrive at f (x) = 0
...
Firstly, dividing out gives:
x + 3x + 2

—————————
x − 3 x 3 − 0 − 7x − 6
x 3 − 3x 2
2

3x 2 − 7x − 6
3x 2 − 9x
————
2x − 6
2x − 6
———
· ·
———
x 3 − 7x − 6
= x 2 + 3x + 2
Hence
x −3
i
...


x 3 − 7x − 6 = (x − 3)(x 2 + 3x + 2)

x 2 + 3x + 2 factorizes ‘on sight’ as (x + 1)(x + 2)
...

Our expression for f (3) was 33 − 7(3) − 6
...

Therefore let us try some negative values for x
...
Also
f (−2) = (−2)3 − 7(−2) − 6 = 0; hence (x + 2) is
a factor (also as shown above)
...
e
...

Note that the values of x, i
...
3, −1 and −2, are
all factors of the constant term, i
...
the 6
...

Problem 29
...

Let f (x) = x 3 − 2x 2 − 5x + 6 and let us substitute
simple values of x like 1, 2, 3, −1, −2, and so on
...
e
...
6

Dividing a general quadratic expression
(ax 2 + bx + c) by (x − p), where p is any whole
number, by long division (see section 1
...


Now try the following exercise

Exercise 6 Further problems on the factor
theorem
Use the factor theorem to factorize the expressions
given in problems 1 to 4
...


x 2 + 2x − 3

2
...
2x 3 + 5x 2 − 4x − 7

The remainder theorem

[(x + 1)(x + 2)(x − 2)]
[(x + 1)(2x 2 + 3x − 7)]

4
...
Use the factor theorem to factorize
x 3 + 4x 2 + x − 6 and hence solve the cubic
equation x 3 + 4x 2 + x − 6 = 0
...
Solve the equation x 3 − 2x 2 − x + 2 = 0
...
This is, in fact, what the remainder
theorem states, i
...
,
‘if (ax 2 + bx + c) is divided by (x − p),
the remainder will be ap 2 + bp + c’
If, in the dividend (ax 2 + bx + c), we substitute p for
x we get the remainder ap2 + bp + c
...
e
...
e
...

Also, when (x 2 + 3x − 2) is divided by (x − 1), the
remainder is 1(1)2 + 3(1) − 2 = 2
...
However, if the
remainder should be zero then (x − p) is a factor
...

For example, when (2x 2 + x − 3) is divided by
(x − 1), the remainder is 2(1)2 + 1(1) − 3 = 0, which
means that (x − 1) is a factor of (2x 2 + x − 3)
...
e
...
e
...

For example, when (3x 3 + 2x 2 − x + 4) is divided
by (x − 1), the remainder is ap 3 + bp2 + cp + d
(where a = 3, b = 2, c = −1, d = 4 and p = 1),
i
...
the remainder is 3(1)3 + 2(1)2 + (−1)(1) + 4 =
3 + 2 − 1 + 4 = 8
...

Here are some more examples on the remainder
theorem
...
Without dividing out, find the
remainder when 2x 2 − 3x + 4 is divided by (x − 2)
...

Hence the remainder is:
2(2)2 + (−3)(2) + 4 = 8 − 6 + 4 = 6
Problem 31
...

By the remainder theorem, the remainder is given by
ap 3 + bp2 + cp + d, where a = 3, b = −2, c = 1, d =
−5 and p = −2
...
Determine the remainder when
(x 3 − 2x 2 − 5x + 6) is divided by (a) (x − 1) and
(b) (x + 2)
...

(a)

When (x 3 − 2x 2 − 5x + 6) is divided by (x − 1),
the remainder is given by ap 3 + bp2 + cp + d,
where a = 1, b = −2, c = −5, d = 6 and p = 1,

11

= 1−2−5+6 = 0
Hence (x − 1) is a factor of (x 3 − 2x 2 − 5x + 6)
...
Therefore (x − 1)(x + 2)(x ) = x 3 −
2x 2 − 5x + 6
...

or (ii) use the factor theorem where f (x) =
x 3 − 2x 2 − 5x + 6 and hoping to choose
a value of x which makes f (x) = 0
...

(i) Dividing (x 3 − 2x 2 − 5x + 6) by
(x 2 + x − 2) gives:
x −3

————————–
x 2 + x − 2 x 3 − 2x 2 − 5x + 6
x 3 + x 2 − 2x
——————
−3x 2 − 3x + 6
−3x 2 − 3x + 6
——————–
·
·
·
——————–
Thus (x 3 − 2x 2 − 5x + 6)
= (x − 1)(x + 2)(x − 3)
(ii) Using the factor theorem, we let
f (x) = x 3 − 2x 2 − 5x + 6
Then f (3) = 33 − 2(3)2 − 5(3) + 6
= 27 − 18 − 15 + 6 = 0
Hence (x − 3) is a factor
...

Hence the remainder is:
1(3)3 + (−2)(3)2 + (−5)(3) + 6
= 27 − 18 − 15 + 6 = 0
Hence (x − 3) is a factor
...
Find the remainder when 3x 2 − 4x + 2 is
divided by
(a) (x − 2) (b) (x + 1)
...
Determine the remainder when
x 3 − 6x 2 + x − 5 is divided by
(a) (x + 2) (b) (x − 3)
...
Use the remainder theorem to find the factors
of x 3 − 6x 2 + 11x − 6
...
Determine the factors of x 3 + 7x 2 + 14x + 8
and hence solve the cubic equation
x 3 + 7x 2 + 14x + 8 = 0
...
Determine the value of ‘a’ if (x + 2) is a
factor of (x 3 − ax 2 + 7x + 10)
...
Using the remainder theorem, solve the
equation 2x 3 − x 2 − 7x + 6 = 0
...
5]



---