Unit 1 algebra

1 Objectives After completing this chapter you should be able to       

Simplify expressions and collect like terms apply the rules of indices multiply out brackets factorise expressions including quadratics simplify algebraic fractions manipulate surds use Rules of indices

Chapter outline 1.1

Algebraic Expressions

1.2

Addition and subtraction of like terms

1.3

Multiplying algebraic expressions and expanding brackets

1.4

Factorising algebraic expressions

1.5

Algebraic Fractions

1.6

Surds

1.7

Indices

1.1 Algebraic Expressions In order to extend the techniques of arithmetic so that they can be more useful in applications we introduce letters or symbols to represent quantities of interest. Algebra is the use of letters and symbols to express generalizations. For example, we know that 4  4  2  4 and 5  5  2  5 ; these are particular examples of the fact that any number added to itself is the same as twice that number. Using a letter to represent any number, this general fact can be written very concisely as x  x  2 x . This is an example of an algebraic expression.

Letters such as x and y in mathematical expressions are known as variables. Variables are symbols that are used to represent unspecified numbers. Any letter may be used as a variable.

The number in front of the letter is known as coefficient. An algebraic expression consists of one or more numbers and variables along with one or more arithmetic operations. An algebraic expression that is a number, a variable, or a product or quotient of numbers and variables is called a term. Example of terms are: 7

a

 2x

4 2 y 7

0.75a 4b  3

2 2/3 x 3

Arithmetic of algebraic terms is defined as follows; Addition (  ): If the letters x and y stand for two numbers, their sum is written as x  y . Note that x  y is the same as y  x just as 3 + 6 is the same as 6 + 3.

Subtraction (  ): The quantity x  y is called the difference of x and y , and means the number

y subtracted from the number x . Note that x  y is the not the same as y  x .

Multiplication (  ): Five times the number x is written as 5 x . We usually omit the multiplication sign. The quantity xy is called the product of x and y .

Division (  ): x  y or

x is called quotient of the two numbers x and y . This is also known as y

an algebraic fraction.

Frequently we shall need to multiply a number by itself several times, for example a  a  a  a  a  a  a  a  a. To abbreviate such a quantity, we use the notation a 9 , which means

the number a is multiplied itself by 9 times. In the quantity x y , x is called the base and y is the power or index. Note that the plural of index is indices.

Example 1 The following are some example of verbal and their algebraic expressions.

Verbal Expression

Algebraic Expression

2 less than the product of 5 and a number y.

5y  2

the product of 4 and a divided by the product of 3 and b .

4a  3b

nine feet shorter than the height of the tree (T = tree height).

T 9

one-third as costly as a first-class ticket (f = price of first class ticket).

f 3

x  x  x  x  x  x5 a number x is multiplied 5 times by itself.

EXERCISE 1A

1.2 Addition and Subtraction of like terms Like terms contain the same combination of letters. For example, 3 y,72 y and 0.5 y are all 1 multiples of y and so are like terms. Similarly, 5 x 2 ,3x 2 and x 2 are all multiples of x 2 and so 2 are like terms. Like terms can be collected together and added or subtracted in order to simplify them.

Example 2 Simplify 3 x  7 x  2 x .

Solution: 3x  7 x  2 x  8 x

All three terms are multiples of

and so are like terms.

Example 3 Simplify 3x  2 y .

Solution: 3x  2 y  3x  2 y

and are not like terms. One is a multiple of the other is a multiple of . The expression cannot be simplified.

and

Example 4 Simplify x  7 x  x 2

Solution: x  7 x  x 2  8x  x 2

The like terms are and . These can be simplified to Note that and are not like terms and so this expression cannot be simplified.

.

Example 5 Simplify ab  a 2  7b 2  9ab  8b 2

Solution: ab  a 2  7b 2  9ab  8b 2  ab  9ab  a 2  7b 2  8b 2  10ab  a 2  b 2

EXERCISE 1B

The terms and . The terms simplify to .

are like and they simplify to and are also like and they

1.3 Multiplying Algebraic Expressions and Expanding Brackets Multiplication The multiplication sign is usually omitted, so that, for example, 2 x means 2  x and x  y can be simplified to xy

If a string numbers and letters are multiplied, the multiplication can be done in any order, for example 2 x  3 y  2  x  3  y  6 xy

Powers can be used to simplify expressions such as x  x  x  x 3 and x 2  x3  x5

Rules for determining the sign of the answer when multiplying positive and negative algebraic expressions are the same as those used for multiplying numbers.

Example 6 Simplify 4 xy   5 2

Solution: 4 xy 2  5  4 xy  4 xy  5  4 x y4 x y5  80 x 2 y 2

positive

positive = positive

positive

negative = negative

negative

positive = negative

negative

negative = positive

Example 7 Simplify the following algebraic expressions (a) 4 x 2  7 x 5

(b) (a )  (b)

(c)  3ab 2 c  (12c 2 d )

Solution: (a) 4 x 2  7 x 5

 4  7  x2  x5  28 x 7

(b) (a)  (b)

 ab

Note that when multiplying, the order in which we write down the terms does not matter.

Here we have the product of a positive and a negative quantity.

The result will be negative.

(c)  3ab 2 c  (12c 2 d )

 3ab 2 c  (12c 2 d )   3   12   a  b 2  c  c 2  d

Here we have the product of a negative and a negative quantity.

The result will be positive.

 36abc 3 d

Expanding Brackets Expanding an expression means multiplying it out.

To expand brackets from expressions of the form a(b  c) and a(b  c) , multiply a with all bracketed terms:

Example 8 Expand the brackets from (a) 6( x  5)

(b) 8(2 x  4)

(c)

 ( x  y)

Solution: (a) 6( x  5)  6 x  65  6 x  30

In the expansion here, it is intended that the 6 multiplies both terms in the brackets.

(b) 8(2 x  4)  8  2 x  8  (4)

In the expansion here 8 multiplies both terms in the brackets.

 16 x  32

(c)

 ( x  y)

 x  y

The expression means . So -1 multiplies to both the terms in the bracket.

Expanding brackets from expressions of the form (a  b)(c  d ).

In the expansion (a  b)(c  d ) it is intended that the quantity (c  d ) multiplies to both the a and the b in the second bracket. Therefore

Example 9 Expand the brackets (a) (3  x)(2  y )

Solution:

(b) ( x  6)( x  3)

(c) (1  x)(2  x)

(a)

(3  x)(2  y )

(b) ( x  6)( x  3)

 3(2  y )  x(2  y )

 x( x  3)  6( x  3)

 6  3 y  2 x  xy

 x 2  3 x  6 x  18  x 2  3 x  18

(c)

(1  x)(2  x)

 1(2  x)  x(2  x)  2  x  2x  x2  2  3x  x 2

Difference of Two Squares Consider the expansion ( x  4)( x  4)  x( x  4)  4( x  4)

 x 2  4 x  4 x  16  x 2  16 This is known as difference of two squares. In general we have

Squares (2 x  3) 2 means (2 x  3)(2 x  3) (2 x  3) 2  (2 x  3)(2 x  3)  2 x(2 x  3)  3(2 x  3)  4x 2  6x  6x  9  4 x 2  12 x  9

In general

and

(first term)2 + 2(first)(second term) + (second term)2

EXERCISE 1C

1.4 Factorising Algebraic Expressions A number is factorized when it is written as a product. For example, 15 may be factorized into 3 5 . We say that 3 and 5 are factors of 15. Algebraic expressions can also be factorized.

Consider the expression 5 x  20 y . Both 5 x and 20 y have the number 5 common to both terms.

We say that 5 is a common factor of both 5 and 20. Thus 5 x  20 y  5( x  4 y ) . Removal of the brackets will result in the original expression and can always be used to check your answer. We can think of factorizing as the reverse of expanding brackets.

Simple Factorisation Example 10 Factorise these expressions completely. (a) 3 x  12

(b) 8 x 2  12 x

(d) 9 x 2  15 xy 2

(e) 3x 2  9 xy

(c) 6 x  3 x 2  9 xy

Solution: (a)

3x  12  3( x  4)

(b)

8 x 2  12 x  4  2  x  4  3 x  4 x(2 x  3) 2

(c)

6 x  3 x 2  9 xy  3 x (2  x  3 y )

3 is a common factor of

and 12

4 and are common factors of the two terms.

Factor out 4 and

Note that 3 and are both factors of the three terms in the expression. So take the common factor out and write the remaining inside the bracket.

(d)

9 x 2 y  15 xy 2  3 xy (3 x  5 y )

(e)

3,

and

are common factors of

. So take

and

outside the bracket.

3x 2  9 xy  3 x( x  3 y )

EXERCISE 1D Factorising Quadratic Expressions Expression of the form ax 2  bx  c , where a , b and c are numbers , are called quadratic expressions. The numbers b or c may equal to zero but a must not be zero. The number a is called the coefficient of x 2 , b is the coefficient of x , and c is called the constant term.

We see that 2 x 2  3 x  1, x 2  3 x  2, x 2  7 and 2 x 2  x are all quadratic expressions.

To factorise such an expression means to express it as a product of two terms.

Quadratic Expressions where the coefficient of x2 is 1 Consider the expression ( x  m)( x  n) . Removing the brackets we find ( x  m)( x  n)  x( x  n)  m( x  n)  x 2  mx  nx  mn  x 2  (m  n) x  mn Note that the coefficient of the x term is the sum m  n and the constant term is the product mn .

Using this information several quadratic expressions can be factorized by careful inspection. For example, suppose we wish to factorise x 2  5 x  6 . We know that x 2  (m  n) x  mn can be factorised to ( x  m)( x  n) . We seek values of m and n so that x 2  5 x  6  x 2  (m  n) x  mn Comparing the coefficients of x on both sides we require 5 mn

Comparing the constant terms on both sides we require 6  mn

By inspection we see that m  3 and n  2 have this property and so x 2  5 x  6  ( x  3)( x  2)

Example 11 Factorise the quadratic expression x 2  8 x  12.

Solution: The factorization of x 2  8 x  12 will be of the form ( x  m)( x  n) . This means that mn must equal 12 and m  n must equal 8. The two numbers must therefore be 2 and 6. So x 2  8 x  12  x 2  6 x  2 x  12  x( x  6)  2( x  6)  ( x  6)( x  2)

Factorise the first two terms and the second two terms separately and then factorise again.

Example 12 Factorise x 2  10 x  25

Solution: x 2  10 x  25  x 2  5 x  5 x  25  x( x  5)  5( x  5)

Here

 ( x  5)( x  5)

multiply together to give .

1. Work out . Here 2. Work out two factors of which add to give you .

3. Rewrite term using these two factors.

4. Factorise first two terms and last two terms 5. is a common factor, so take that

 ( x  5) 2

and

.

You need to find two brackets

outside the bracket. This now completely factorised.

Difference of Two Squares

that

The expression x 2  y 2 is called the difference of two squares. Factorising such expressions will give the following results.

Note that x 2  y 2  ( x  y )( x  y ) .

Example 13 Factorise x 2  121

Solution: x 2  121  x 2  112 ( x  11)( x  11)

This difference of two squares and so we can use the result above.

Example 14 Factorise 4 x 2  9 y 2

Solution: 4 x 2  9 y 2  2 2 x 2  32 y 2  (2 x) 2  (3 y ) 2  (2 x  3 y )(2 x  3 y )

Example 15 Factorise x 2  5 x  6

Solution: x 2  5 x  6  x 2  3x  2 x  6  x( x  3)  2( x  3)  ( x  3)( x  2)

Quadratic Expressions where the coefficient of x2 is not 1

These expression are little harder to factorise. All possible factors of the first and last terms must be found, and various combinations of these should be attempted until the required answer is found. This involves trial and error along with educated guesswork and practice.

Example 16 Factorise, if possible, the expression 2 x 2  11x  12

Solution: 2 x 2  11x  12  2 x  3 x  8 x  12 2

Here the coefficient of steps below.

is not 1. In order to factorise, follow the

 x(2 x  3)  4(2 x  3)  (2 x  3)( x  4)

Example 17 Factorise 4 x 2  6 x  2

1. Multiply the coefficient of ( which is 2) by the constant term ( which is 12) to get 24.

2. Work out all possible factors of 24 in pairs such as (1, 24) (2, 12) (3, 8) (4, 6) (-1, -24) (-2, -12) (-3, -8) and (-4, 6) 3. Pick the pair which when added gives the coefficient of , which is 11. The two factors are 3 and 8. Now write as .

4. Factorise the first two and the last two 5. Take the common bracket out.

Solution:

x

4x 2  2x  4x  2

4 x 2  2 x  4 x  2  2 x(2 x  1)  2(2 x  1) 2 x(2 x  1)  2(2 x  1)  (2 x  1)(2 x  2)

(2 x  1)(2 x  2)  2(2 x  1)( x  1)

EXERCISE 1E Fdsfdsfdsfsdf Sdfsdfdsfsdf

1.4 Algebraic Fractions The value of a fraction is unaltered when the numerator and denominator are each multiplied or divided by the same number. e.g.

4 2 6 14     ...

10 5 15 35

and

ax x 2 x x p  q      ... ay y 2 y y  p  q 

Cancelling Common Factors

When simplifying algebraic expressions, it is sensible to factorise numerators and denominators where possible; look both for common factors and for factors of quadratic expressions.

Example 18 Simplify

24 xy 18 x

Solution: 24 xy 4 y  18 x 3

Top and bottom are divided by 6 and by

Example 19 Simplify the following. (a)

2x 2  4x 7y2

(b)

3 x 1  x2 x2

Solution: (a)

2x 2 2x 2 1  4x   2 2 4x 7y 7y

take the reciprocal of the second term

2x 2 28y 2 x x  14 y 2 

(b)

x 1 3 3 ( x  2)    x  2 x  2 ( x  2) ( x  1) 3  x 1

Notice that we use brackets to enclose two terms. This makes it easier to see what cancels.

Example 20 Simplify the following algebraic fractions

(a)

2a 2  2ab 6ab  6b 2

(b)

x2  4 x2

(c)

x y x2  y2

Solution: (a)

2a 2  2ab 2a (a  b)  6ab  6b 2 6b(a  b) a  3b

2a is a common factor of the numerator and 6b is a common factor of the denominator. So factorise both numerator and denominator.

a  b is a factor for both numerator and denominator.

So, cancel it out.

(b)

(c)

x2  4 x2  4  x  2 ( x  2) ( x  2)( x  2)  ( x  2)  x2

x y x y  2 2 ( x  y )( x  y ) x y 1  x y

Factorise the numerator and cancel out the common factors.

Factorise the denominator, and cancel out the common factors.

Example 21 Simplify (a)

x 2x  x 2

(d)

3 x  xy x 2  5x

Solution:

(b)

x4 ( x  4) 2

(e)

x 2 1 x 2  3x  2

(c)

x2 x  3x  2 2

(a)

(b)

(c)

x x  2 x(2  x) 2x  x 1  2 x

The denominator factorises to x(2  x) . Also note that

1( x  4) x4  2 ( x  4)( x  4) ( x  4) 1  x4

There is a factor of x  4 in both numerator and denominator.

1( x  2) x2  x  3 x  2 ( x  2)( x  1) 1  x 1

numerator can be written as 1 x .

2

(d)

(e)

3x  xy x(3  y )  x 2  5 x x( x  5) 3 y  5 x ( x  1)( x  1) x 2 1  2 x  3 x  2 ( x  1)( x  2) x 1  x2

Note that the denominator is a quadratic expression which factorises to give ( x  2)( x  1) . Now, ( x  2) is a common factor to both numerator and denominator. So cancel it out.

x is a factor common to both numerator and denominator.

This is cancelled.

Numerator is a difference of two squares which gives factorises to give ( x  1)( x  1) and denominator factorises to give ( x  1)( x  2) . So ( x  1) is a factor common to both numerator and denominator.

Multiplication and Division of Algebraic Fractions To multiply two algebraic fractions together we multiply their numerators together and multiply their denominators together.

To divide two algebraic fractions together, take the reciprocal (invert) of the second fraction and multiply.

Any common factor in the result should be cancelled.

Example 22 Simplify 4 3y  x 16

(a)

4 x  5 y

(b)

(e)

4 x 2 3x 3  y yz

 x 3 (f) 5     25 

(c)

1 x 2

(g)

x 2  4x  3 x  4  2x  8 x 1

(d)

1  ( a  b) 2

Solution: (a)

4 x 4x   5 y 5y

We multiply the numerators together and multiply denominators together.

(b)

4 3y 4 3y   x 16 16 x 12 y  16 x 3y  4x

numerators together and denominators are multiplied together and the common factors are cancelled.

(c)

(d)

(e)

1 1 x x   2 2 1 x  2 1 ab 1  ( a  b)   2 1 2 ab  2 4 x 2 3x 3 4 x 2  3x 3   y yz y  yz 12 x 5  2 y z

(f)

 x 3 5 x 3 5     25  1 25 5( x  3)  25 x 3  5

Write x as

x .

1

Write a  b as

ab .

1

We multiply the numerators together and multiply denominators together. No common factor.

A common factor of 5 can be cancelled from numerator and denominator.

(g)

x 2  4x  3 x  4  x 1 2x  8 ( x  3)( x  1) ( x  4)   2( x  4) ( x  1) ( x  3)( x  1)( x  4)  2( x  4)( x  1) x3  2

Before multiplying the two fractions together we should try to factorise if possible so that common factors can be identified.

Example 23 (a)

10a a 2  b 3b

(b)

x  2 3x  6  x  4 x 2  16

Solution: (a)

(b)

10a a 2 10a 3b    b 3b b a2 30ab  2 a b 30  a x  2 3x  6  x  4 x 2  16 x  2 x 2  16   x  4 3x  6 x  2 ( x  4)( x  4)   3( x  2) x4 ( x  2)( x  4)( x  4)  3( x  4)( x  2) x4  3

The second fraction is inverted and then multiplied by the first.

First take the reciprocal of the second fraction . Second, factorise all expressions if possible. Multiply the numerators together and denominators together. Lastly, cancel the common factors.

Addition and Subtraction of Algebraic Fractions The method is the same as that for adding or subtracting numerical fractions. Note that it is not correct to simply add or subtract the numerator and denominator. The lowest common denominators must first be found. This is the simplest expression that contains all original denominators as its factors. Each fraction is then written with this common denominator. The

fractions then can be added or subtracted by adding or subtracting just the numerators, and dividing the result by the common denominator.

Example 24 Simplify the numerical fractions

1 3  3 4

Solution: Here the LCM of the denominators 3 and 4 is 12. So the common denominator is 12.

1 4 3 3 4 9      3 4 4 3 12 12 Since we have a common denominator now, add the numerators to get 1 3 13   .

3 4 12 Note that this last fraction has no common factor to cancel.

Example 25 Simplify (a)

3 1  4 x

(b)

3 4  x x2

(d)

1 1  x  4 ( x  4) 2

(e)

3 4x  2 x 1 x 1

(c)

2 5  x  3 x 1

Solution: (a)

3 1 3x 4    4 x 4x 4x 3x  4  4x

First rewrite the fractions to ensure they have a common denominator. The common denominator is 4 x here.

Write

3x 1 3 4 as . , and as 4x 4 x 4x

Add the two numerators.

Note that no further simplification is possible.

(b)

(c)

(d)

3 4 3x 4    x x2 x2 x2 3x  4  x2

The expression

3 3x is written as 2 , which makes x x

the denominators of both terms x 2 but leaves the value of the expression unaltered. Note that both the original denominators, x and x 2 , are factors of the new denominator. The fractions are then added by adding just the numerators. No further simplification is possible.

2 5  x  3 x 1 2( x  1) 5( x  3)   ( x  3)( x  1) ( x  3)( x  1) 2( x  1)  5( x  3)  ( x  3)( x  1) 2 x  2  5 x  15  ( x  3)( x  1) 7 x  13  ( x  3)( x  1)

The LCM of the denominators is ( x  3)( x  1) . So write

1 1  x  4 ( x  4) 2

LCM of denominators is ( x  4) 2 .

( x  4) 1  2 ( x  4) ( x  4) 2 x  4 1  ( x  4) 2 x 3  ( x  4) 2

2( x  1) 5( x  3) 2 5 as as and . Now x3 ( x  3)( x  1) x 1 ( x  3)( x  1) add the numerators and simplify.



Write

( x  4) 1 1 and as is unchanged.

2 x4 ( x  4) ( x  4) 2

Add the two numerators and simplify the numerator. See that there is no further simplification possible.

(e)

3 4x  2 x 1 x 1 3( x  1) 4x   ( x  1)( x  1) ( x  1)( x  1) 3( x  1)  4 x  ( x  1)( x  1) 3x  3  4 x  ( x  1)( x  1) 3 x  ( x  1)( x  1)

We can write x 2  1 as ( x  1)( x  1) .

The LCM of the denominators is ( x  1)( x  1) .

Write

3( x  1) 3 4x and 2 as is unchanged. x 1 ( x  1)( x  1) x 1

Now, subtract 4 x from 3( x  1) and simplify the numerator. See if that there is factor common to both numerator and denominator. So no further simplification is possible.

1.6 Surds At first numbers were used only for counting, and 1, 2, 3, . . . . were all was needed. These are natural numbers, or positive integers.

Then it was found that numbers could also be useful for measurement A rational number is a number which can be written in the form

p where p and q are integers q

and q  0 . All proper fractions, improper fractions, mixed numbers and integers are rational numbers. So are terminating and recurring decimals fractions.

An irrational number is one that is not rational. If an irrational number is written as a decimal fraction, the decimal is infinite and has no repeating pattern.

A surd is an expression containing one or more irrational roots of numbers.

The square root of a number may be rational or irrational. The square root of every square number is rational. The square root of very prime number is irrational.

Expressions such as

4,

25 have exact numerical values, but expressions such as

5 , . . . cannot be written as numerically exact quantities.

2,

3,

Expressions such as

2 or

3

9 are called surds. This section is about calculating with surds.

To simplify surds, use these rules:

To simplify other roots use these rules:

When two irrational number are multiplied together, the result may be rational or irrational. Since a  b  ab , ab will rational if and only if ab is a rational square number.

Example 26 Which of these numbers are irrational? (a)

35

(b)

63

(c)

1 32

Solution: (a)

35  5  7  5  7

irrational

(b)

63  9  7  9  7  3 7

irrational

(c)

1 1 1 1 1     32 32 16  2 16  2 4 2

irrational

(d)

5

1 32

(d) 5

rational

5 1 1 1 1    32 5 32 5 2 5 2

Operations with surds If a, b, c , and d are integers then:

Multiplication:

Division:

Addition:

Subtraction:

Example 27 Simplify the following leaving your answer in surd form: (a) 2 3  5 7

(b) 4 15  2 5

(c) 8 2  5 2

(d) 7 3  2 12

Solution: (a)

2 3  5 7  2  5 3 7  10 21

(b)

4 15  2 5 

4 15 2 5

2 3

(c)

8 2  5 2  (8  5) 2

7 3  2 12  7 3  2 4  3

(d)

 7 3  2 4  3

 13 2

 7 34 3  ( 7  4) 3 3 3

Example 28 Expand and simplify (a)

2  2 7 5  7 

4  3 4  3 

(b)

Solution: (a)

2  2 7 5  7   2(5 



7)  2 7 5  7



 10  2 7  10 7  2 7  7 10  (2  10) 7  2  7  10  8 7  14  4  8 7

(b)

4  3 4  3   44  3  34  3   16  4 3  4 3  3  3  16  3  13

Rationalising the denominator A fraction whose denominator contains a surd is more awrkward to deal with than one where a surd occurs in the numerator. There is a technique for transferring the surd expression from the denominator to the numerator. Removing the surd from the denominator is called rationalising the denominator.

To rationalize the denominator of: 

a

the fraction

, multiply by

b



. That is

b

a

the fraction

b

, multiply by

bk c

bk c bk c

. That is

Example 29 Write down each of these fractions with a rational denominator. (a)

2 5

4

(b)

32

7

Solution: (a)

2 5



2 5

7

(b)



7

4 32



7



2 5 7

7

4 2 3



77

4

2 

1

2  3   42  3   8  4 3  8  4 43 3  2 3  2   3  2



1

1  2  1  2  2

Solution:

2 35 7



Example 30 Simplify



2

.

2

3

1

1



1  2  1  2  2

2

  

1

1



1 2 2  2 1 2 2  2 1 3 2 2 1

3  2



1 3 2 2

3  2 2   1  3  2 2  2  3  2 2  3  2 2  3  2 2  

3 2 2 3 2 2  98 98  3 2 2 3 2 2 

4 2

Example 31 Simplify the expression

3 22 3 3 2 2 3

.

Solution:

3 2  2 3  3 2  2 3  3 3 2  2 3  3 2  2 3  3 2   6 6  6 6  2 3   3 2   2 3 

3 22 3 3 2 2



2

2

2

2

18  12 6  12 6 30  12 6  6  5 2 6 

Example 32 Simplify (a)

3

16

(b)

3

12  3 18

Solution (a)

3

16  3 2  8  3 8  3 2  2  3 2

(b)

3

12  3 18  3 12 18  3 216  6

EXERCISE 1F

1.8 Indices In an expression such as 3 4 , the base is 3 and the 4 is called the power or index (the plural is indices). Working with indices involves using some properties which apply to any base, so we express these rules in terms of a general base a (i.e. a stands for any number).

In general, the symbol a m stands for the result of multiplying m as together:

Another way of describing this is ‘ a raised to the m th power’, or more shortly ‘ a to the power m ’. When this notation is used, expressions can often be simplified by using a few simple rules known as Rules of Indices.

Special powers such as zero and one are defined as

The multiplication rule. m of these m  n of these   nofthese    a  a  a  a  a   a  a  a  a   a  a  a  a   a  a m  n . m

n

Closely linked with this is the division rule.

mofthese  a  a  a  a  a   a m

n

provided that m  n .



m  n of these nofthese    a  a  a   a  a  a  a   a  a m  n .

Another rule is the power-on-power rule.

a 

m n

of these brackets n  m of these m of these m of these m n of these         a  a  a   a  a  a  a   a  a  a  a   a  a  a  a   a

 a mn

The factor rule, has two bases but one index:

a  b 

m

of these brackets m of these m of these m        ( a  b )  ( a  b )  ( a  b )    ( a  b)  a  a  a    a  b  b  b    b

 am bm.

Negative Power rule.

Rational Power rule.

Summarising all the rules we have Rules of Indices

1.

2.

3.

4.

5.

6.

7.

8.

Example 33 Simplify (a)

2a b  4a b

(b) 4a 2 b  3ab 1

(d)

x 

(e)

2

2 7

3



4



2

a b   b a

1/ 3

 x 3

3

4

5

Solution: (a)

2a b   4a b   2 a  b  4a b   8a b   4a b   8  4 a  a  b 2

3

4

2 3

3

2 3

3

3

4

6

 2a

64

b

 2a 2 b 2 .

3 1

Factor rule

4

4

3

 b1



Power-on-power Rearranging Division rule

(c)

23  27 43

(b)

(d)



4a 2 b  3ab 1

x 

2 7



2

(c)

1

 4a 2 b 

1   3a   b  1  4a 2 b  1 9a 2  2 b 2 b  4a 2 b  2 9a 4  b 1 2 9 4  b3 9

2

 

2 3 7 2 23 210  6 2  210 6 

 24

(e)

1 x3 1  x 14  3 x 14  3 x

 x  3  x 2 7 

x

23  27 23  27  2 43 22

a b   b a  a b b a   a b  1/ 3

3

4 / 31

 a1/ 3b 2

Example 34 1

(b) 16



3 4



3 4

 25  (c)    9 

3 / 2

Solution: 1

(a)

92  9  3

 25  (c)    9 

3 / 2

 9     25 

(b) 16 3/ 2

5

4/3

11

Simplify (a) 9 2

4

 

 24 3



3 4

1  2 4 4 / 4   2 3  .

8

 9   3  3 27        5 25 125  

5/3

1/ 3

5 / 31 / 3

1