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Foundations - Algebra
1. Indices
2. Surds
3. Order of Operations
4. Expanding Brackets
5. Factorisation
6. Algebraic Fractions
7. Completing the Square
8. Manipulating Formuale
Last updated May 29, 2015
1
Indices
The expression am means
. . × a × a}
|a × a × .{z
m times
and we call a the base and m the power or the index. Note that a and m cannot both
equal zero. The rules used to combine these expressions are called the Index Rules.
1.
am × an = am+n
2.
am
= am−n
an
3.
(am )n = amn
4.
(ab)m = am bm
WARNING! It is very common to confuse Rules 1 and 3. Make sure that you understand the difference.
Some consequence of the these rules, which are often encountered, are listed below.
5.
a0 = 1
6.
a−n =
7.
8.
(a 6= 0)
1
(a 6= 0)
an
√
1
an = n a
1 m
m/n
a
= an
√
In Rule 7, if n = 2 then we simply write a. In Rule 8, if n is even then a has to be a
positive number. However, if n is odd then a can be positive or negative.
WARNING! Negative powers of positive numbers do not give negative numbers. See
Example 1.1 below.
1
Example 1.1 Simplify 3−2 .
1
32
1
= .
9
3−2 =
Rule 6
Example 1.2 Simplify 165/4 .
165/4 = (161/4 )5
√
4
= ( 16)5
=2
Rule 8
Rule 7
5
= 32
Example 1.3 Simplify
x2 (xy)3
√
.
xy
x2 (xy)3
x2 × x3 × y 3
√
=
xy
x1/2 × y
x5 × y 3
= 1/2
x ×y
y3
x5
= 1/2 ×
x
y
Rule 4 & Rule 7
Rule 1
= x9/2 × y 2
Rule 2
We normally just write this without the “×”, i.e. x9/2 y.
Exercise 1.1 Without using a calculator, simplify the following, leaving your answers as a
fraction where necessary.
1) 16
1/2
2) 16
−2
4
6)
5
5
11) 3 × 3
−5
−1
3/2
9
7)
4
12) 125
1/3
2
2
3)
3
−3
1
8)
3
13) 2
1/2
×2
2
4)
9)
7/2
9
25
1/2
7
1+
9
55/3
14) 2/3
5
1/2
5) 360
10)
15)
1
27
−2/3
163
259
0
Exercise 1.2 Simplify the following expressions as far as possible.
1) 3x2 × 2x5
2) 7x3 × 3x4
12a7
3a3
42s2
9) 2
7s × 3s3
28t5
7t7
2y 4 14y 6
10) 5 × 3
y
y
5)
2
3) 4x7 × 2x4 × 3x3
14a7 b10
7a5 b5
vm−3 n
11) −4 2
m n
6)
7)
y 12
y4
10x10 × 4x8
8)
5x3
4)
12) 33/2 × 241/2 × 81−1/2 × 2−1/2
Surds
Some square roots can be written as whole numbers or fractions, in other words as rational
numbers. For example
r
9
3
= .
4
2
√
However, other roots cannot. A famous example is 2 - it is an irrational number. When
we have square roots that give irrational numbers we call them surds. They occur often
in trigonometry (and other areas) so it is useful to be able to simplify expressions involving
surds.
√
Example 2.1 Simplify 8.
√
8=
√
4×2 =
√
4×
√
√
2 = 2 2.
Example 2.2 Following the previous example, we can now simplify products of roots.
√
12 ×
√
√
4×3× 3×5
√
√
√
=2 3× 3× 5
√
= 2×3× 5
√
=6 5
15 =
√
Example 2.3 In this example we first expand the brackets (see Section 4) and then simplify.
(1 +
√
2)(3 −
√
√
√
√
2) = 3 + 3 2 − 2 − 2 = 1 + 2 2.
Example 2.4 The following procedure is called rationalising the denominator.
√
√
√
2
2+ 2
2+ 2
√
= √
×√
2
2
2
√
2 2+2
=
√ 2
= 2 + 1.
3
Example 2.5 In this example we rationalise the denominator by multiplying by the
conjugate surd.
√
√
√
1+ 3
1+ 3 2− 5
√ =
√ ×
√
2+ 5
2+ 5 2− 5
√
√
√
2 − 5 + 2 3 − 15
=
4−5
√
√
√
= −2 + 5 − 2 3 + 15.
Exercise 2.1 Simplify the following expressions.
√
√
1) 6 2 × 3 2
√
√
4) 5 6 × 4 3
√
√
7) 72 × 128
√
√
√
√
10) ( 2 + 2 3)(3 2 − 3)
√
√
2) 12 7 × 7
√
√
5) 50 × 20
√
√
8) (3 − 2 5)(1 − 3 5)
√
√
11) (1 − 2)(2 + 2 2)
√
√
3) 2 7 × 3 8
√
√
6) 45 × 80
√
√
9) (2 − 3)(3 + 2 3)
√
√ √
√
12) ( 5 − 2)( 5 + 2)
Exercise 2.2 By rationalising the denominator, simplify the following expressions.
5
1) √
5
√
2
4) √
2−2
√
1− 3
7) √
3+1
1
2) √
7
√
1− 3
5) √
3−2
√
√
5+ 3
√
8) √
5− 3
6
3) √
3
√
2+2 5
√
6)
3− 5
√
√
3 2− 3
√
9) √
3+ 2
WARNING! A very common mistake in simplify square roots is the following:
√
For example,
3
A + B 6=
√
√
A+
√
B.
4x2 + 9 6= 2x + 3.
Order of Operations
If you have worked through the previous exercises then you will appreciate that careful
consideration must be given when evaluating expressions. The order in which operations
are carried our matters. For example
2 × 3 + 4 = 6 + 4 = 10,
4
not
2 × 3 + 4 = 2 × 7 = 14.
The multiplication is carried out before the addition. Also
10 − 3 + 2 = 9,
not
10 − 3 + 2 = 5.
The mnemonic BIDMAS may help you remember the correct order of operations. It stands
for
B
Brackets
I
Indices, i.e. Powers and/or Roots
D/M
Division and/or Multiplication
A/S
Addition and/or Subtraction
Equal care should be used when employing this mnemonic. Working through the following
questions will help boost your skills.
Exercise 3.1 Calculate:
1) 5 × 4 + 2
1
× 36 ÷ 6
3
7) 7 × 4 − 48 ÷ 6
4)
10) 3(7 × 3) + 5
1
13) × 24 ÷ (2 × 2)
3
16) (2 + 4)(1 + 5)2
4(3 + 1)2 − (5 × 3)
1
(2 × 7)
2
(5 + 4)2 p
1
+ (2 × 6)(3 × 4)
21) (9 × 8) −
2
8−5
19)
2) 6 × 8 − 3
1
× 40
4
8) 56 ÷ 7 + 2 × 5
5) 3 ×
11) 4(6 + 3) − 2(12 − 3)
1
14) × 50 ÷ (2 + 3) × (8 − 4)
2
(4 × 4) + 42
17)
11 − 3
√
20) 5 × 4 × 5
5
1
× 24 + 3
2
1
6) 2 + 6 × × 20
2
9) 10 + (8 × 5)
3)
12) 8(12 − 7)(3 + 2)
15) 2 × 7 + (11 − 4)2
18)
(1 + 7)2
(3 − 1)2
4
Expanding Brackets
We have already encountered this procedure in some of the previous exercises. In order to
multiply out brackets all we need is the distributive rule of multiplication over addition:
x × (y + z) = x × y + x × z.
Example 4.1 Expand (x − 3)(2x + 5).
(x − 3)(2x + 5) = x(2x + 5) − 3(2x + 5)
= x × 2x + x × 5 − 3 × 2x − 3 × 5
= 2x2 + 5x − 6x − 15
= 2x2 − x − 15.
Example 4.2 Expand (x − 2)(3x − 5 + 2y).
(x − 2)(3x − 5 + 2y) = x × 3x − x × 5 + x × 2y − 2 × 3x + 2 × 5 − 2 × 2y
= 3x2 − 5x + 2xy − 6x + 10 − 4y
= 3x2 + 2xy − 11x − 4y + 10.
Exercise 4.1 Expand and simplify the following expressions.
1) 3(x2 − 2y)
2) x(2a − xy)
3) (x − 5)(x − 3)
4) (x − 7)(x + 4)
5) (x + 3)2
6) (2x − 1)(3x + 2)
7) (2x − 5)(2 + 3x)
8) (3x + 4)(3x − 4)
9) (3x − 1)2 − (x − 4)2
10) 3(x − 2y)2
11) 3(x − 2y)2
12) (x − 1)3
15) (3x − 7)2 − 7(x − 3)
16) 3(2x − 1)2 − 2(3x − 1)2
13) (3x + 2y)2 + (2y − 3x)2
14) (x − y + z)2 − (x + y − z)2
17) (x − 1)2 (x + 2)
5
18) (x + 1)4
Factorisation
The opposite procedure to multiplying out brackets is that of factorisation. We aim to
take an expression of the form
x2 + 5x + 6
6
and write it as a product of terms:
(x + 2)(x + 3).
You should be familiar with the most common types of factorisation:
• taking out a common factor;
• factorising quadratic polynomials;
• factorising a difference of two squares.
Example 5.1 In the expression x3 + 2x2 − x, x is common to each term, so it can be taken
out as common factor:
x3 + 2x2 − x = x(x2 + 2x − 1).
Example 5.2 In the expression 3(x − 1) + (x − 1)2 the common factor is (x − 1):
3(x − 1) + (x − 1)2 = (3 + (x − 1))(x − 1) = (2 + x)(x − 1).
WARNING! It can be difficult to follow the previous example the first time you see
it. If this is the case, consider:
3A + A2 = (3 + A)A.
Now replace A with x − 1 and you should now see how the factorisation works.
Example 5.3 Consider the quadratic polynomial x2 + 6x + 5. If we wish to factorise it then
we must find numbers a and b that satisfy
x2 + 6x + 5 = (x + a)(x + b).
By expanding the brackets we have
(x + a)(x + b) = x2 + ax + bx + ab = x2 + (a + b)x + ab.
So if
x2 + 6x + 5 = x2 + (a + b)x + ab
then we must have
6 = a + b and 5 = ab,
7
i.e. a and b must add to give 6 and multiply to give 5. An obvious choice is a = 1 and
b = 5 (or vice versa). Hence
x2 + 6x + 5 = (x + 1)(x + 5).
Sometimes finding the correct a and b is less obvious.
Example 5.4 Factorise 2x2 − 5x − 3.
By considering the different ways two numbers can multiply to give −3, we will quickly find
the correct factorisation. We begin by creating the following table, where the numbers in
Rows 1 and 2 are different ways to multiply to get −3, e.g. −1 × 3 = −3, and the terms 2x
and x are the factors of 2x2 .
2x
x
1 −1 −3
3 ←
−3
3
1 −1 ←
x −x −3x
3x ←
−6x 6x
2x −2x ←
−5x 5x −x
x ←
Row 1
Row 2
multiply values in Row 1 by x
multiply values in Row 2 by 2x
add the x-values in the column above
We are looking for −5x so we take the values from Rows 1 and 2 in the first column, namely
+1 and −3. We obtain the factorisation by reading along the appropriate row:
2x2 − 5x − 3 = (2x + 1)(x − 3).
Notice that the table gives the factorisation of similar looking polynomials, e.g.
2x2 + x − 3 = (2x + 3)(x − 1).
WARNING! Always check that you have obtained the correct factorisation. You can
do this by expanding the brackets:
(2x + 1)(x − 3) = 2x2 − 6x + x − 3 = 2x2 − 5x − 3,
as required.
Example 5.5 The expression 4x2 − 9 is an example of a difference of two squares. In
order to factorise it we look to express 4x2 and 9 as two square numbers:
4x2 = (2x)2
and
9 = 32 .
The factorisation is then
4x2 − 9 = (2x + 3)(2x − 3).
8
The general factorisation of any difference of two squares is as follows:
x2 − y 2 = (x − y)(x + y).
Exercise 5.1 Factorise the following expressions as far as possible.
1) x2 + 3x + x
2) x2 − 6x + 9
3) x2 − 4x + 4
4) x2 + 2x − 8
5) x2 − 6x − 16
6) 2x2 − 3x + 1
10) 4x2 − 16y 2
11) 4a2 − 16a
12) 9x2 − 1
7) 6x2 + x − 2
13) 6x2 + 5x − 6
16) (3x − 2)2 − (x − 3)2
8) 9x2 − 6x + 1
14) (x − 1)2 − 4
9) 2d2 + 3d + 1
15) 4x2 − 4x + 3
WARNING! Factorisation is not easy and it takes a lot of practice and experience to
become adept at. However, if you put in the work then you will learn a very useful skill.
6
Algebraic Fractions
The ability to add and subtract algebraic fractions will often help you simplify many different
calculations. The golden rule to follow at all times is:
fractions are added or subtracted by first finding a common denominator.
For example,
8
3
11
2 1
+ =
+
= .
3 4
12 12
12
The same principle applies when handling so called “algebraic fractions.”
Example 6.1
3a 2
3ac 2b 3ac + 2b
+ =
+
+
.
b
c
bc
bc
bc
2
3
5
Example 6.2 Write
+
+ 2
as a single fraction.
y + 5 y − 5 y − 25
3
5
2(y − 5)
3(y + 5)
5
2
+
+ 2
=
+
− 2
y + 5 y − 5 y − 25
(y + 5)(y − 5) (y − 5)(y + 5) y − 25
2y − 10 3y + 15
5
= 2
+ 2
− 2
y − 25 y − 25 y − 25
2y − 10 + 3y + 15 − 5
=
y 2 − 25
5y
= 2
y − 25
9
a 1
+
Example 6.3 Simplify b a .
1
1
+
2
a
b
a 1
a2 + b
+
b a = ab
1
1
b + a2
+
a2 b
a2 b
2
a2 b
a +b
×
=
ab
b + a2
2
ab
=
ab
= a.
Example 6.4 Simplify
x2 + 6x + 8
.
x+2
(x + 2)(x + 4)
x2 + 6x + 8
=
= x + 4.
x+2
x+2
WARNING! Care must be taken when simplifying algebraic fractions. In particular,
the following simplification is incorrect:
a2 + 3
= a + 3.
a
The correct expression is
3
a2 + 3
=a+ .
a
a
10
Exercise 6.1 Write each of the following expressions as a single fraction.
1 1
+
a b
1
4) x +
x
a a2
7) + 2
b
b
2
3d
m
−
10)
n
c
2a
a2
+
13)
2
(a + 2)
a+2
3
a + b3
1
− 4
16)
a − b a − b4
1)
19)
t+1
1
÷
2
t−1
2
3
−
x xy
a2
2
5) +
b
ab
2c 3d
8) 2 −
d
c
x
2y
y
11)
+
+
x+2
x
x+2
2b
a
14)
−
b + 2a b − 2a
2c 3d
17) 2 ×
d
6c
a+1
20) b
a+1
b2
2)
m
n
+
n
m
4
2
6)
−
5a 3a
3y
2
9)
+
x
xy
x
2x
3
12)
+
x+3 x−3x−3
1
2
15)
+
(x + 2)(x − 1) (x − 1)(x − 2)
p+q p−q
18)
×
p
2q
3)
21)
m−3
m2 − 1
×
3
(m + 1)3
2c 3d
−
c
22) d
1
cd
7
Completing the Square
Completing the square is the process that transforms a quadratic polynomial
ax2 + bx + c
into the following expression:
p(x + q)2 + r.
Such an expression allows us to easily sketch the graph of the original polynomial.
Example 7.1 Complete the square in x2 − 10x + 4.
In order to get x2 − 10x we must have (x − 5)2 . Expanding this bracket introduces an extra
term of +25, so we must subtract this from (x − 5)2 . Thus:
x2 − 10x − 4 = (x − 5)2 − 25 + 4
= (x − 5)2 − 21.
Notice what we did in the previous example: the coefficient of x (i.e. the number in front of
x-term) was divided by 2 to give −5. We then subtracted (−5)2 from the square term.
11
Example 7.2 Complete the square in 2x2 + 12x + 5.
2x2 + 12x + 5 = 2[x2 + 6x] + 5
= 2[(x + 3)2 − 9] + 5
= 2(x + 3)2 − 18 + 5
= 2(x + 3)2 − 13
take out a factor of +2,
divide coefficient of x by 2 to obtain +3; subtract 32 ,
multiply out,
add constant terms.
Example 7.3 Complete the square in −3x2 + 4x + 1.
4
2
−3x + 4x + 1 = −3 x − x + 1
3
#
"
2
4
2
−
+1
= −3 x −
3
9
2
2
4
= −3 x −
+ +1
3
3
2
7
2
+
= −3 x −
3
3
2
take out a factor of −3,
2
divide by 2 to obtain − ; subtract
3
2
2
,
−
3
multiply out,
add constant terms.
Exercise 7.1 Complete the square in the following expressions.
8
1) c2 + 4c − 3
2) b2 − 14 − 1
3) y 2 + 5y + 5
5) k 2 − 7k + 2
a
9) a2 + − 1
2
2
13) p + 4p + 1
6) m2 + 11m + 12
7) n2 − n − 10
10) 3y 2 − 12y + 3
11) 2p2 + 4p + 3
4) d2 − 6d + 3
4
8) k 2 − k − 5
3
2
12) 3a + 7a + 1
14) 3y 2 − 15y − 2
15) − 2p2 + 4p + 7
16) 5a2 + 3a − 4
Manipulating Formulae
The following set of questions will test your skills in using and manipulating mathematical
formulae. This is an important skill to have in all areas of science and engineering, and in
other areas like economics and business studies.
12
Exercise 8.1
1. Make x the subject in each of the following equations.
a) x + 2n − m = 0
x
d) = p
m
√
g) x = m + n
j)
n
m
=
n−x
m−x
b) mx + nx = p
x
n
e)
=
m
p
√
h) p x − 1 = r
k)
x+m x−n
x
−
=
n
n
n
m
=n
x
x
s
f) = + 1
r
t
n
m
=
i)
x
x−1
√
√
l) mx + n = m + n
c)
2. Given the formula b = 2a − T
(a) find the value of T when a = 7 and b = 3;
(b) make T the subject of the formula.
3. Given the formula 2a2 = b2 + B
(a) find the value of B when a = 3 and b = −3;
(b) make B the subject of the formula.
4. Given the formula D =
d
3+Q
(a) find the value of Q when d = 1.3 and D = 2.6;
(b) make Q the subject of the formula.
5. The volume V (in cm3 ) of a circular cone of height h cm and base radius r cm is given
by the formula
1
V = πr 2 h.
3
(a) Calculate the radius of a cone of height 5cm and volume 33.5cm. Give your answer
correct to 2 decimal places.
(b) Give r in terms of V , h and π.
an + 1
.
n
7. Make t the subject the formula v = u + at.
6. Make n the subject of the formula C =
8. Make h the subject of the formula S = ar(r + h).
r
b
.
9. Make b the subject of the formulae a =
b+c
10. Rewrite the formula v = at + 16t2 to make a the subject and find its value when
v = 300 and t = 4.
p+x
11. Given the formula y =
, make x its subject.
1 − px
13
9
Answers
Exercise 1.1
1) 4
9)
2)
4
3
1
4
4
9
3)
10) 9
4)
11) 1
3
5
12) 5
25
16
5) 1
6)
13) 16
14) 5
7)
27
8
8) 27
15) 1
Exercises 1.2
1) 6x7
2) 21x7
3) 24x14
4) y 8
5) 4a4
6) 4t−2
7) 2a2 b5
8) 8x15
9) 2s−3
10) 28y 2
11) vmn−1
12) 2
Exercises 2.1
1) 36
2) 84
7) 96
8) 33 − 11 5
√
3) 12 14
√
9) 3
√
√
4) 60 2
√
10) 5 6
√
5) 10 10
6) 60
11) − 2
12) 3
Exercises 2.2
1)
√
5
2)
√
6) 4 + 2 5
√
√
3) 2 3
√
8) 4 + 15
7
7
7) 1
4) − 1 −
√
2
√
9) − 9 + 4 6
5) 1 +
√
3
Exercises 3.1
1) 22
2) 45
3) 15
4) 2
5) 30
6) 62
7) 20
8) 18
9) 50
10) 68
11) 18
12) 200
13) 2
14) 20
15) 63
16) 216
17) 14
18) 8
19) 7
20) 10
21) 21
Exercises 4.1
1) 3x2 − 6y
2) 2ax − x2 y
3) x2 − 8x + 15
7) 6x2 − 11x − 10
8) 9x2 − 16
9) 8x2 − 14x − 15
13) 18x2 + 8y 2
14) − 4xy + 4xz
15 2x2 − 14
4) x2 − 3x − 28
10) 3x2 − 12xy + 12y 2
16) − 6x2 + 1
5) x2 + 6x + 9
11) 9x2 − 36xy + 36y 2
17) x3 − 3x + 2
14
6) 6x2 + x − 2
12) x3 − 3x2 + 3x − 1
18) x4 + 4x3 + 6x2 + 4x + 1
Exercsies 5.1
1) (x + 2)(x + 1)
5) (x − 8)(x + 2)
9) (2d + 1)(d + 1)
13) (3x − 2)(2x + 3)
17) (mn + 7)(mn − 3)
2) (x − 3)2
3) (x − 2)2
4) (x + 4)(x − 2)
10) 4(x − 2y)(x + 2y)
11) 4a(a − 4)
12) (3x − 1)(3x + 1)
6) (2x − 1)(x − 1)
14) (x + 1)(x − 3)
8) (3x − 1)2
7) (3x + 2)(2x − 1)
15) (2x + 1)(2x − 3)
(4x − 5)(2x + 1)
Exercises 6.1
a+b
ab
3
a +2
5)
ab
3y 2 + 2
9)
xy
3a2 + 4a
13)
(a + 2)2
1
17)
d
(m − 1)(m − 3)
21)
3(m + 1)
1)
m2 + n2
mn
2
a + ab
7)
b2
2
x + 3xy + 4y
11)
x(x + 2)
3x + 2
15)
(x − 1)(x2 − 4)
t2 − 1
19)
2
2y − 3
xy
2
6)
15a
nm2 − 3
10)
n2
2
2b − 5ab − 2a2
14)
b2 − 4a2
2
p − q2
18)
2pq
2)
3)
a2 + 1
x
2c − 3d2
8)
d2
x−3
12)
x+3
ab
16)
(a − b)(a2 + b2 )
4)
20) b
22) 2c2 − 3d2
Exercises 7.1
2
1) (c + 2) − 7
2
41
7
−
5) k −
2
4
2
1
17
9) a +
−
4
16
13) (p + 2)2 − 3
2
5
5
3) y +
−
2
4
2
41
1
−
7) n −
2
4
2
2) (b − 7) − 50
2
73
11
−
6) m +
2
4
10) 3(y − 2)2 − 9
2
83
5
−
14) 3 y −
2
4
11) 2(p + 1)2 + 1
15) − 2(p − 1)2 + 9
4) (d − 3)2 − 6
2
49
2
−
8) k −
3
9
2
27
7
−
12) 3 a +
6
12
2
89
3
−
16) 5 a +
10
20
Exercises 8.1
1)
a) x = m − n
mn
p
m
i) x =
m−n
e) x =
p
m+n
r(s + t)
f) x =
t
c) x =
b) x =
m
n
g) x = (m + n)2
j) x = n + m
k) x =
15
2mn
2m − n
d) x = m(p − n)
2
r
+1
h) x =
p
√
l) x = m + 2 2
2) T = 2a − b
1
C−a
v
10) a = − 16t
t
6) n =
2
d − 3D
4) Q =
D
S
8) h =
−r
ar
2
3) B = 2a − b
v−u
a
y−p
11) x =
1 + py
7) t =
16
r
3V
πh
a2 c
9) b =
1 − a2
5) r =