© Boardworks Ltd 2005 1 of 73 A1 Algebraic manipulation KS4 Mathematics.

© Boardworks Ltd 2005 1 of 73

A1 Algebraic manipulation

KS4 Mathematics


Contents

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AA1.1 Using index laws


A1.2 Multiplying out brackets

A1.3 Factorization

A1.5 Algebraic fractions

A1.4 Factorizing quadratic expressions


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.


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


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)


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


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


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


Hexagon puzzle


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



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



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


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.


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


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


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


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.


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


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


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A1.3 Factorization

A1.1 Using index laws





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



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


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) =



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) =



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) =


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)


Find the area of the rectangle


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


Expanding two brackets


(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.


Using the grid method to expand brackets


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.


Matching quadratic expressions 1




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



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




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.


The difference between two squares


Matching the difference between two squares


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A1.3 Factorization







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.



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)



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



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


Factorization


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)


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


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A1.3 Factorization






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)



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.


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.


Factorizing quadratic expressions 1




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


Factorizing quadratic expressions 2




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)


Factorizing the difference between two squares


Matching the difference between two squares


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A1.3 Factorization






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)


Equivalent algebraic fractions


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



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)



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


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

+


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)


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

?


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


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


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


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


Addition pyramid – algebraic fractions

© Boardworks Ltd 2005 1 of 73 A1 Algebraic manipulation KS4 Mathematics.

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