© boardworks ltd 2005 1 of 73 a1 algebraic manipulation ks4 mathematics

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© Boardworks Ltd 2005 of 73 A1 Algebraic manipulation KS4 Mathematics

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© Boardworks Ltd 2005 1 of 73

A1 Algebraic manipulation

KS4 Mathematics

© Boardworks Ltd 2005 2 of 73

Contents

A

A

A

A

AA1.1 Using index laws

A1 Algebraic manipulation

A1.2 Multiplying out brackets

A1.3 Factorization

A1.5 Algebraic fractions

A1.4 Factorizing quadratic expressions

© Boardworks Ltd 2005 3 of 73

Multiplying terms

Simplify:

x + x + x + x + x = 5x

Simplify:

x × x × x × x × x = x5

x to the power of 5

x5 as been written using index notation.

xn

The number x is called the base.

The number n is called the index or power.

© Boardworks Ltd 2005 4 of 73

We can use index notation to simplify expressions.

For example,

3p × 2p = 3 × p × 2 × p = 6p2

q2 × q3 = q × q × q × q × q = q5

3r × r2 = 3 × r × r × r = 3r3

3t × 3t = (3t)2 or 9t2

Multiplying terms involving indices

© Boardworks Ltd 2005 5 of 73

Multiplying terms with the same base

For example,

a4 × a2 = (a × a × a × a) × (a × a)

= a × a × a × a × a × a

= a6

When we multiply two terms with the same base the indices are added.When we multiply two terms with the same base the indices are added.

= a (4 + 2)

In general,

xm × xn = x(m + n)xm × xn = x(m + n)

© Boardworks Ltd 2005 6 of 73

Dividing terms

Remember, in algebra we do not usually use the division sign, ÷.

Instead, we write the number or term we are dividing by underneath like a fraction.

For example,

(a + b) ÷ c is written as a + bc

© Boardworks Ltd 2005 7 of 73

Like a fraction, we can often simplify expressions by cancelling.

For example,

n3 ÷ n2 =n3

n2

=n × n × n

n × n

= n

6p2 ÷ 3p =6p2

3p

=6 × p × p

3 × p

2

= 2p

Dividing terms

© Boardworks Ltd 2005 8 of 73

Dividing terms with the same base

For example,

a5 ÷ a2 =a × a × a × a × a

a × a= a × a × a = a3

4p6 ÷ 2p4 =4 × p × p × p × p × p × p

2 × p × p × p × p= 2 × p × p = 2p2

= a (5 – 2)

= 2p(6 – 4)

When we divide two terms with the same base the indices are subtracted.When we divide two terms with the same base the indices are subtracted.

In general,

xm ÷ xn = x(m – n)xm ÷ xn = x(m – n)

2

© Boardworks Ltd 2005 9 of 73

Hexagon puzzle

© Boardworks Ltd 2005 10 of 73

Sometimes terms can be raised to a power and the result raised to another power.

For example,

(y3)2 = (pq2)4 =

Expressions of the form (xm)n

y3 × y3

= (y × y × y) × (y × y × y)

= y6

pq2 × pq2 × pq2 × pq2

= p4 × q (2 + 2 + 2 + 2)

= p4 × q8

= p4q8

© Boardworks Ltd 2005 11 of 73

Expressions of the form (xm)n

For example,

(a5)3 = a5 × a5 × a5

= a(5 + 5 + 5)

= a15

When a term is raised to a power and the result raised to another power, the powers are multiplied.When a term is raised to a power and the result raised to another power, the powers are multiplied.

= a(3 × 5)

In general,

(xm)n = xmn(xm)n = xmn

© Boardworks Ltd 2005 12 of 73

Expressions of the form (xm)n

Rewrite the following without brackets.

1) (2a2)3 = 8a6 2) (m3n)4 = m12n4

3) (t–4)2 = t–8 4) (3g5)3 = 27g15

5) (ab–2)–2 = a–2b4 6) (p2q–5)–1 = p–2q5

7) (h½)2 = h 8) (7a4b–3)0 = 1

© Boardworks Ltd 2005 13 of 73

The zero index

Look at the following division:

y4 ÷ y4 = 1

But using the rule that xm ÷ xn = x(m – n)

y4 ÷ y4 = y(4 – 4) = y0

That means that

y0 = 1

In general, for all x 0,

x0 = 1x0 = 1

Any number or term divided by itself is equal to 1.

© Boardworks Ltd 2005 14 of 73

Negative indices

Look at the following division:

b2 ÷ b4 =b × b

b × b × b × b=

1b × b

=1b2

But using the rule that xm ÷ xn = x(m – n)

b2 ÷ b4 = b(2 – 4) = b–2

That means that

b–2 = 1b2

In general,

x–n = 1xn

© Boardworks Ltd 2005 15 of 73

Negative indices

Write the following using fraction notation:

u–1 = 1u

2b–4 = 2b4

x2y–3 = x2

y3

This is the reciprocal of u.

2a(3 – b)–2 = 2a

(3 – b)2

© Boardworks Ltd 2005 16 of 73

Negative indices

Write the following using negative indices:

2a

=

x3

y4=

p2

q + 2=

3m(n2 + 2)3

=

2a–1

x3y–4

p2(q + 2)–1

3m(n2 + 2)–3

© Boardworks Ltd 2005 17 of 73

Indices can also be fractional.

Fractional indices

x × x =12

12 x + =

12

12 x1 = x

But, x × x = x

x1 = x

So, x = x x = x 12

Similarly, x × x × x =13

13

13 x + + =

13

13

13

But, x × x × x = x3 3 3

So, x = x x = x 13 3

The square root of x.

The cube root of x.

© Boardworks Ltd 2005 18 of 73

x = x x = x

In general,

Fractional indices

Also, we can write x as x . mn

1n × m

Using the rule that (xm)n = xmn, we can write

1n n

We can also write x as xm × . mn

1n

x × m = (x )m = (x)m1n

1n n

In general,

x = xmx = xm x = (x)mx = (x)mmn n or

mn n

x = (xm) = xm1nm×

n1n

© Boardworks Ltd 2005 19 of 73

Here is a summary of the index laws.

xm × xn = x(m + n)

xm ÷ xn = x(m – n)

Index laws

(xm)n = xmn

x1 = x

x0 = 1 (for x = 0)

x = x 1n

n

x = x 12

x = xm or (x)mnmn n

x–1 = 1x

x–n = 1xn

© Boardworks Ltd 2005 20 of 73

Contents

A

A

A

A

A

A1.2 Multiplying out brackets

A1.3 Factorization

A1.1 Using index laws

A1 Algebraic manipulation

A1.5 Algebraic fractions

A1.4 Factorizing quadratic expressions

© Boardworks Ltd 2005 21 of 73

Look at this algebraic expression:

Expanding expressions with brackets

3y(4 – 2y)

This means 3y × (4 – 2y), but we do not usually write × in algebra.

To expand or multiply out this expression we multiply every term inside the bracket by the term outside the bracket.

3y(4 – 2y) = 12y – 6y2

© Boardworks Ltd 2005 22 of 73

Look at this algebraic expression:

Expanding expressions with brackets

–a(2a2 – 2a + 3)

When there is a negative term outside the bracket, the signs of the multiplied terms change.

–a(2a2 – 3a + 1) = –2a3 + 3a2 – a

In general, –x(y + z) = –xy – xz

–x(y – z) = –xy + xz

–(y + z) = –y – z

–(y – z) = –y + z

© Boardworks Ltd 2005 23 of 73

Expanding brackets and simplifying

Sometimes we need to multiply out brackets and then simplify.

For example, 3x + 2x(5 – x)

We need to multiply the bracket by 2x and collect together like terms.

3x + 10x – 2x2

= 13x – 2x2

3x + 2x(5 – x) =

© Boardworks Ltd 2005 24 of 73

Expanding brackets and simplifying

Expand and simplify: 4 – (5n – 3)

We need to multiply the bracket by –1 and collect together like terms.

4 – 5n + 3

= 4 + 3 – 5n

= 7 – 5n

4 – (5n – 3) =

© Boardworks Ltd 2005 25 of 73

Expanding brackets and simplifying

Expand and simplify: 2(3n – 4) + 3(3n + 5)

We need to multiply out both brackets and collect together like terms.

6n – 8 + 9n + 15

= 6n + 9n – 8 + 15

= 15n + 7

2(3n – 4) + 3(3n + 5) =

© Boardworks Ltd 2005 26 of 73

We need to multiply out both brackets and collect together like terms.

15a + 10b – 2a – 5ab

= 15a – 2a + 10b – 5ab

= 13a + 10b – 5ab

Expanding brackets then simplifying

5(3a + 2b) – a(2 + 5b) =

Expand and simplify: 5(3a + 2b) – a(2 + 5b)

© Boardworks Ltd 2005 27 of 73

Find the area of the rectangle

© Boardworks Ltd 2005 28 of 73

Find the area of the rectangle

What is the area of a rectangle of length (a + b) and width (c + d)?

a b

c

d

ac bc

ad bd

In general,(a + b)(c + d) = ac + ad + bc + bd

© Boardworks Ltd 2005 29 of 73

Expanding two brackets

Look at this algebraic expression:

(3 + t)(4 – 2t)

This means (3 + t) × (4 – 2t), but we do not usually write × in algebra.

To expand or multiply out this expression we multiply every term in the second bracket by every term in the first bracket.

(3 + t)(4 – 2t) = 3(4 – 2t) + t(4 – 2t)

= 12 – 6t + 4t – 2t2

= 12 – 2t – 2t2

This is a quadratic

expression.

© Boardworks Ltd 2005 30 of 73

Using the grid method to expand brackets

© Boardworks Ltd 2005 31 of 73

Expanding two brackets

With practice we can expand the product of two linear expressions in fewer steps. For example,

(x – 5)(x + 2) = x2 + 2x – 5x – 10

= x2 – 3x – 10

Notice that –3 is the sum of –5 and 2 …

… and that –10 is the product of –5 and 2.

© Boardworks Ltd 2005 32 of 73

Matching quadratic expressions 1

© Boardworks Ltd 2005 33 of 73

Matching quadratic expressions 2

© Boardworks Ltd 2005 34 of 73

Squaring expressions

Expand and simplify: (2 – 3a)2

We can write this as,

(2 – 3a)2 = (2 – 3a)(2 – 3a)

Expanding,

(2 – 3a)(2 – 3a) = 2(2 – 3a) – 3a(2 – 3a)

= 4 – 6a – 6a + 9a2

= 4 – 12a + 9a2

© Boardworks Ltd 2005 35 of 73

Squaring expressions

In general,

(a + b)2 = a2 + 2ab + b2

The first term squared …

… plus 2 × the product of the two terms …

… plus the second term squared.

For example,

(3m + 2n)2 = 9m2 + 12mn + 4n2

© Boardworks Ltd 2005 36 of 73

Squaring expressions

© Boardworks Ltd 2005 37 of 73

The difference between two squares

Expand and simplify (2a + 7)(2a – 7)

Expanding,

(2a + 7)(2a – 7) = 2a(2a – 7) + 7(2a – 7)

= 4a2 – 14a + 14a – 49

= 4a2 – 49

When we simplify, the two middle terms cancel out.

In general,

(a + b)(a – b) = a2 – b2 (a + b)(a – b) = a2 – b2

This is the difference between two squares.

© Boardworks Ltd 2005 38 of 73

The difference between two squares

© Boardworks Ltd 2005 39 of 73

Matching the difference between two squares

© Boardworks Ltd 2005 40 of 73

Contents

A

A

A

A

A

A1.3 Factorization

A1.2 Multiplying out brackets

A1.1 Using index laws

A1 Algebraic manipulation

A1.5 Algebraic fractions

A1.4 Factorizing quadratic expressions

© Boardworks Ltd 2005 41 of 73

Factorizing expressions

Factorizing an expression is the opposite of expanding it.

a(b + c) ab + ac

Expanding or multiplying out

FactorizingOften:When we expand an expression we remove the brackets.When we factorize an expression we write it with brackets.

© Boardworks Ltd 2005 42 of 73

Factorizing expressions

Expressions can be factorized by dividing each term by a common factor and writing this outside of a pair of brackets.

For example, in the expression

5x + 10

the terms 5x and 10 have a common factor, 5.

We can write the 5 outside of a set of brackets

5(x + 2)

We can write the 5 outside of a set of brackets and mentally divide 5x + 10 by 5.

(5x + 10) ÷ 5 = x + 2

This is written inside the bracket.

5(x + 2)

© Boardworks Ltd 2005 43 of 73

Factorizing expressions

Writing 5x + 10 as 5(x + 2) is called factorizing the expression.

Factorize 6a + 8

6a + 8 = 2(3a + 4)

Factorize 12n – 9n2

12n – 9n2 = 3n(4 – 3n)

The highest common factor of 6a and 8 is 2.

(6a + 8) ÷ 2 = 3a + 4

The highest common factor of 12n and 9n2 is 3n.

(12n – 9n2) ÷ 3n = 4 – 3n

© Boardworks Ltd 2005 44 of 73

Factorizing expressions

Writing 5x + 10 as 5(x + 2) is called factorizing the expression.

3x + x2 = x(3 + x)2p + 6p2 – 4p3

= 2p(1 + 3p – 2p2)

The highest common factor of 3x and x2 is x.

(3x + x2) ÷ x = 3 + x

The highest common factor of 2p, 6p2 and 4p3 is 2p.

(2p + 6p2 – 4p3) ÷ 2p

= 1 + 3p – 2p2

Factorize 3x + x2 Factorize 2p + 6p2 – 4p3

© Boardworks Ltd 2005 45 of 73

Factorization

© Boardworks Ltd 2005 46 of 73

Factorization by pairing

Some expressions containing four terms can be factorized by regrouping the terms into pairs that share a common factor. For example,

Factorize 4a + ab + 4 + b

Two terms share a common factor of 4 and the remaining two terms share a common factor of b.

4a + ab + 4 + b = 4a + 4 + ab + b

= 4(a + 1) + b(a + 1)

4(a + 1) and + b(a + 1) share a common factor of (a + 1) so we can write this as

(a + 1)(4 + b)

© Boardworks Ltd 2005 47 of 73

Factorization by pairing

Factorize xy – 6 + 2y – 3x

We can regroup the terms in this expression into two pairs of terms that share a common factor.

xy – 6 + 2y – 3x = xy + 2y – 3x – 6

= y(x + 2) – 3(x + 2)

y(x + 2) and – 3(x + 2) share a common factor of (x + 2) so we can write this as

(x + 2)(y – 3)

When we take out a factor of

–3, – 6 becomes + 2

© Boardworks Ltd 2005 48 of 73

Contents

A

A

A

A

A

A1.4 Factorizing quadratic expressions

A1.3 Factorization

A1.2 Multiplying out brackets

A1.1 Using index laws

A1 Algebraic manipulation

A1.5 Algebraic fractions

© Boardworks Ltd 2005 49 of 73

Quadratic expressions

A quadratic expression is an expression in which the highest power of the variable is 2. For example,

x2 – 2, w2 + 3w + 1, 4 – 5g2 ,t2

2The general form of a quadratic expression in x is:

x is a variable.

a is a fixed number and is the coefficient of x2.

b is a fixed number and is the coefficient of x.

c is a fixed number and is a constant term.

ax2 + bx + c (where a = 0)

© Boardworks Ltd 2005 50 of 73

Factorizing expressions

Remember: factorizing an expression is the opposite of expanding it.

Expanding or multiplying out

FactorizingOften:When we expand an expression we remove the brackets.

(a + 1)(a + 2) a2 + 3a + 2

When we factorize an expression we write it with brackets.

© Boardworks Ltd 2005 51 of 73

Factorizing quadratic expressions

Quadratic expressions of the form x2 + bx + c can be factorized if they can be written using brackets as

(x + d)(x + e)

where d and e are integers.

If we expand (x + d)(x + e) we have,

(x + d)(x + e) = x2 + dx + ex + de

= x2 + (d + e)x + de

Comparing this to x2 + bx + c we can see that:

The sum of d and e must be equal to b, the coefficient of x.

The product of d and e must be equal to c, the constant term.

© Boardworks Ltd 2005 52 of 73

Factorizing quadratic expressions 1

© Boardworks Ltd 2005 53 of 73

Matching quadratic expressions 1

© Boardworks Ltd 2005 54 of 73

Factorizing quadratic expressions

Quadratic expressions of the form ax2 + bx + c can be factorized if they can be written using brackets as

(dx + e)(fx + g)

where d, e, f and g are integers.

If we expand (dx + e)(fx + g)we have,

(dx + e)(fx + g)= dfx2 + dgx + efx + eg

= dfx2 + (dg + ef)x + eg

Comparing this to ax2 + bx + c we can see that we must choose d, e, f and g such that: a = df,

b = (dg + ef)

c = eg

© Boardworks Ltd 2005 55 of 73

Factorizing quadratic expressions 2

© Boardworks Ltd 2005 56 of 73

Matching quadratic expressions 2

© Boardworks Ltd 2005 57 of 73

Factorizing the difference between two squares

A quadratic expression in the form

x2 – a2

is called the difference between two squares.

The difference between two squares can be factorized as follows:

x2 – a2 = (x + a)(x – a)

For example,

9x2 – 16 = (3x + 4)(3x – 4)

25a2 – 1 = (5a + 1)(5a – 1)

m4 – 49n2 = (m2 + 7n)(m2 – 7n)

© Boardworks Ltd 2005 58 of 73

Factorizing the difference between two squares

© Boardworks Ltd 2005 59 of 73

Matching the difference between two squares

© Boardworks Ltd 2005 60 of 73

Contents

A

A

A

A

A

A1.5 Algebraic fractions

A1.3 Factorization

A1.2 Multiplying out brackets

A1.1 Using index laws

A1 Algebraic manipulation

A1.4 Factorizing quadratic expressions

© Boardworks Ltd 2005 61 of 73

Algebraic fractions

The rules that apply to numerical fractions also apply to algebraic fractions.

For example, if we multiply or divide the numerator and the denominator of a fraction by the same number or term we produce an equivalent fraction.

3x4x2

and are examples of algebraic fractions.2a

3a + 2

For example,

3x4x2

=34x

=68x

=3y4xy

=3(a + 2)4x(a + 2)

© Boardworks Ltd 2005 62 of 73

Equivalent algebraic fractions

© Boardworks Ltd 2005 63 of 73

Simplifying algebraic fractions

We simplify or cancel algebraic fractions in the same way as numerical fractions, by dividing the numerator and the denominator by common factors. For example,

Simplify 6ab3ab2

6ab3ab2

=6 × a × b

3 × a × b × b

2

=2b

© Boardworks Ltd 2005 64 of 73

Simplifying algebraic fractions

Sometimes we need to factorize the numerator and the denominator before we can simplify an algebraic fraction. For example,

Simplify 2a + a2

8 + 4a

=a4

2a + a2

8 + 4a=

a (2 + a)4(2 + a)

© Boardworks Ltd 2005 65 of 73

Simplifying algebraic fractions

Simplify b2 – 363b – 18

b2 – 36 is the difference

between two squares.

b2 – 363b – 18

=(b + 6)(b – 6)

3(b – 6)

b + 63

=

If required, we can write this as

63

=b3

+b3

+ 2

© Boardworks Ltd 2005 66 of 73

Manipulating algebraic fractions

Remember, a fraction written in the form

a + bc

can be written asbc

ac

+

However, a fraction written in the form

ca + b

cannot be written ascb

ca

+

For example,

1 + 23

=23

13

+ but3

1 + 2=

32

31

+

© Boardworks Ltd 2005 67 of 73

Multiplying and dividing algebraic fractions

We can multiply and divide algebraic fractions using the same rules that we use for numerical fractions.

In general, ab

× =cd

acbd

ab

÷ =cd

ab

× =dc

adbc

and,

For example,3p4

× =2

(1 – p)6p

4(1 – p)=

3

2

3p2(1 – p)

© Boardworks Ltd 2005 68 of 73

23y – 6

÷ =4

y – 2

This is the reciprocal

of4

y – 2

23y – 6

×4

y – 2

23(y – 2)

×=4

y – 2

16

=

Multiplying and dividing algebraic fractions

2

What is2

3y – 6 ÷

4y – 2

?

© Boardworks Ltd 2005 69 of 73

Adding algebraic fractions

We can add algebraic fractions using the same method that we use for numerical fractions. For example,

What is1a

+2b

?

We need to write the fractions over a common denominator before we can add them.

1a

+2b

=b + 2a

abbab

+2aab

=

In general,

+ =ab

cd

ad + bcbd

© Boardworks Ltd 2005 70 of 73

Adding algebraic fractions

What is3x

+y2

?

We need to write the fractions over a common denominator before we can add them.

3x

+y2

=

=6 + xy

2x

+62x

xy2x

=

+3 × 2x × 2

y × x2 × x

© Boardworks Ltd 2005 71 of 73

Subtracting algebraic fractions

We can also subtract algebraic fractions using the same method as we use for numerical fractions. For example,

We need to write the fractions over a common denominator before we can subtract them.

In general,

What is – ?p3

q2

– =p3

q2

– =2p6

3q6

2p – 3q6

– =ab

cd

ad – bcbd

© Boardworks Ltd 2005 72 of 73

Subtracting algebraic fractions

What is – ?

–(2 + p) × 2q

4 × 2q3 × 42q × 4

2 + p4

32q

=–2 + p

432q

= –2q(2 + p)

8q128q

=2q(2 + p) – 12

8q4

6

=q(2 + p) – 6

4q

© Boardworks Ltd 2005 73 of 73

Addition pyramid – algebraic fractions