#### Transcript A1 Algebraic manipulation

```KS4 Mathematics
A1 Algebraic manipulation
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Contents
A1 Algebraic manipulation
A A1.1 Using index laws
A A1.2 Multiplying out brackets
A A1.3 Factorization
A A1.5 Algebraic fractions
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Multiplying terms
Simplify:
x + x + x + x + x = 5x
x to the power of 5
Simplify:
x × x × x × x × x = x5
x5 as been written using index notation.
xn
The number n is called
the index or power.
The number x is called the base.
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Multiplying terms involving indices
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
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or
9t2
Multiplying terms with the same base
When we multiply two terms with the same base the indices
For example,
a4 × a2 = (a × a × a × a) × (a × a)
=a×a×a×a×a×a
= a6 = a (4 + 2)
In general,
xm × xn = x(m + n)
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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
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is written as
a+b
c
Dividing terms
Like a fraction, we can often simplify expressions by
cancelling.
For example,
3
n
n3 ÷ n2 = 2
n
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2
6p
6p2 ÷ 3p =
3p
2
n×n×n
=
n×n
6×p×p
=
3×p
=n
= 2p
Dividing terms with the same base
When we divide two terms with the same base the indices
are subtracted.
For example,
a5
÷
a2
a×a×a×a×a
=
a×a
= a × a × a = a3 = a (5 – 2)
2
4p6 ÷ 2p4 =
4×p×p×p×p×p×p
= 2 × p × p = 2p2 = 2p(6 – 4)
2×p×p×p×p
In general,
xm ÷ xn = x(m – n)
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Hexagon puzzle
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Expressions of the form (xm)n
Sometimes terms can be raised to a power and the result
raised to another power.
For example,
(y3)2 = y3 × y3
(pq2)4 = pq2 × pq2 × pq2 × pq2
= (y × y × y) × (y × y × y)
= p4 × q (2 + 2 + 2 + 2)
= y6
= p4 × q8
= p4q8
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Expressions of the form (xm)n
When a term is raised to a power and the result raised to
another power, the powers are multiplied.
For example,
(a5)3 = a5 × a5 × a5
= a(5 + 5 + 5)
= a15 = a(3 × 5)
In general,
(xm)n = xmn
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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
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The zero index
Look at the following division:
Any number or term divided
by itself is equal to 1.
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 = 1
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Negative indices
Look at the following division:
b2 ÷ b4 =
b×b
b×b×b×b
=
1
1
= 2
b×b
b
But using the rule that xm ÷ xn = x(m – n)
b2 ÷ b4 = b(2 – 4) = b–2
That means that
b–2 =
1
b2
x–n =
1
xn
In general,
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Negative indices
Write the following using fraction notation:
u–1 =
2b–4 =
2
b4
x2y–3 =
x2
y3
2a(3 –
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1
u
b)–2 =
This is the
reciprocal of u.
2a
(3 – b)2
Negative indices
Write the following using negative indices:
2
= 2a–1
a
x3
= x3y–4
4
y
p2
= p2(q + 2)–1
q+2
3m
2 + 2)–3
3m(n
=
(n2 + 2)3
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Fractional indices
Indices can also be fractional.
1
2
1
2
x ×x = x
1
2
1
+2
= x1 = x
But,
x × x = x
So,
x = x
Similarly,
But,
So,
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The square
root of x.
1
2
1
3
1
3
1
3
x ×x ×x = x
1
3
+
1
3
+
1
3
= x1 = x
x × x × x = x
3
3
3
x = x
1
3
3
The cube
root of x.
Fractional indices
In general,
x = x
1
n
n
1
m
Also, we can write x n as x n × m .
Using the rule that (xm)n = xmn, we can write
1
n
x
m
n
We can also write x as
xm ×
1
n
×m
n
1
n
= (x )m = (x)m
.
1
x m× n
=
1
(xm)n
n
= xm
In general,
m
n
x =
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n m
x
or
m
n
n
x = (x)m
Index laws
Here is a summary of the index laws.
x–1 = 1
x
x–n = 1n
x
xm × xn = x(m + n)
xm ÷ xn = x(m – n)
x0
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(xm)n = xmn
x = x
x1 = x
x = x
= 1 (for x = 0)
1
2
1
n
m
n
x =
n
xm
n
n
or (x)m
Contents
A1 Algebraic manipulation
A A1.1 Using index laws
A A1.2 Multiplying out brackets
A A1.3 Factorization
A A1.5 Algebraic fractions
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Expanding expressions with brackets
Look at this algebraic expression:
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
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Expanding expressions with brackets
Look at this algebraic expression:
–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
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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 + 2x(5 – x) = 3x + 10x – 2x2
= 13x – 2x2
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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 – 5n + 3
= 4 + 3 – 5n
= 7 – 5n
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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.
2(3n – 4) + 3(3n + 5) = 6n – 8 + 9n + 15
= 6n + 9n – 8 + 15
= 15n + 7
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Expanding brackets then simplifying
Expand and simplify: 5(3a + 2b) – a(2 + 5b)
We need to multiply out both brackets and collect together
like terms.
5(3a + 2b) – a(2 + 5b) = 15a + 10b – 2a – 5ab
= 15a – 2a + 10b – 5ab
= 13a + 10b – 5ab
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Find the area of the rectangle
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Find the area of the rectangle
What is the area of a rectangle of
length (a + b) and width (c + d)?
a
b
c
ac
bc
d
bd
In general,
(a + b)(c + d) = ac + ad + bc + bd
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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
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This is a
expression.
Using the grid method to expand brackets
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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 …
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… and that
–10 is the
product of
–5 and 2.
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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
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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
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Squaring expressions
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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.
This is the difference
In general,
between two squares.
(a + b)(a – b) = a2 – b2
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The difference between two squares
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Matching the difference between two squares
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Contents
A1 Algebraic manipulation
A A1.1 Using index laws
A A1.2 Multiplying out brackets
A A1.3 Factorization
A A1.5 Algebraic fractions
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Factorizing expressions
Factorizing an expression is the opposite of expanding it.
Expanding or multiplying out
a(b + c)
ab + ac
Factorizing
Often:
When we expand an expression we remove the brackets.
When we factorize an expression we write it with brackets.
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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 and mentally
divide 5x + 10 by 5.
(5x + 10) ÷ 5 = x + 2
This is written inside the bracket.
5(x + 2)
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Factorizing expressions
Writing 5x + 10 as 5(x + 2) is called factorizing the
expression.
Factorize 6a + 8
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Factorize 12n – 9n2
The highest common
factor of 6a and 8 is 2.
The highest common
factor of 12n and 9n2 is 3n.
(6a + 8) ÷ 2 = 3a + 4
(12n – 9n2) ÷ 3n = 4 – 3n
6a + 8 = 2(3a + 4)
12n – 9n2 = 3n(4 – 3n)
Factorizing expressions
Writing 5x + 10 as 5(x + 2) is called factorizing the
expression.
Factorize 3x + x2
The highest common
factor of 3x and x2 is x.
(3x +
x 2)
÷x=3+x
Factorize 2p + 6p2 – 4p3
The highest common factor
of 2p, 6p2 and 4p3 is 2p.
(2p + 6p2 – 4p3) ÷ 2p
= 1 + 3p – 2p2
3x + x2 = x(3 + x)
2p + 6p2 – 4p3
= 2p(1 + 3p – 2p2)
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Factorization
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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)
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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.
When we take
xy – 6 + 2y – 3x = xy + 2y – 3x – 6
out a factor of
–3, – 6
= y(x + 2) – 3(x + 2)
becomes + 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)
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Contents
A1 Algebraic manipulation
A A1.1 Using index laws
A A1.2 Multiplying out brackets
A A1.3 Factorization
A A1.5 Algebraic fractions
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A quadratic expression is an expression in which the
highest power of the variable is 2. For example,
2
t
x2 – 2,
w2 + 3w + 1,
4 – 5g2 ,
2
The general form of a quadratic expression in x is:
ax2 + bx + c
(where a = 0)
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.
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Factorizing expressions
Remember: factorizing an expression is the opposite of
expanding it.
Expanding or multiplying out
a2 + 3a + 2
(a + 1)(a + 2)
Factorizing
Often:
When we expand an expression we remove the brackets.
When we factorize an expression we write it with brackets.
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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.
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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
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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)
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Factorizing the difference between two squares
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Matching the difference between two squares
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Contents
A1 Algebraic manipulation
A A1.1 Using index laws
A A1.2 Multiplying out brackets
A A1.3 Factorization
A A1.5 Algebraic fractions
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Algebraic fractions
3x
2a
and
are examples of algebraic fractions.
2
4x
3a + 2
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.
For example,
3x
3
6
3y
3(a + 2)
=
=
=
=
2
4x
4x
8x
4xy
4x(a + 2)
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Equivalent algebraic fractions
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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,
6ab
Simplify
3ab2
2
6ab
6×a×b
=
2
3ab
3×a×b×b
2
=
b
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Simplifying algebraic fractions
Sometimes we need to factorize the numerator and the
denominator before we can simplify an algebraic fraction. For
example,
2a + a2
Simplify
8 + 4a
2a + a2
8 + 4a
a (2 + a)
=
4(2 + a)
a
=
4
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Simplifying algebraic fractions
b2 – 36 is the
difference
between two
squares.
b2 – 36
Simplify
3b – 18
b2 – 36
(b + 6)(b – 6)
=
3b – 18
3(b – 6)
b+6
3
If required, we can write this as
=
b
b
6
+
=
+ 2
3
3
3
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Manipulating algebraic fractions
Remember, a fraction written in the form
a+b
c
can be written as
a
b
+
c
c
However, a fraction written in the form
c
a+b
cannot be written as
c
c
+
a
b
For example,
1+2
3
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1
2
=
+
3
3
but
3
1+2
3
3
=
+
1
2
Multiplying and dividing algebraic fractions
We can multiply and divide algebraic fractions using the same
rules that we use for numerical fractions.
In general,
and,
For example,
a
c
ac
×
=
b
d
bd
a
c
a
d
÷
=
×
=
b
d
b
c
bc
3
3p
6p
3p
2
×
=
=
(1 – p)
4(1 – p)
2(1 – p)
4
2
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Multiplying and dividing algebraic fractions
2
4
?
What is
÷
3y – 6
y–2
2
3y – 6
4
2
÷
=
y–2
3y – 6
This is the
reciprocal
4
of
y–2
y–2
×
4
y–2
2
=
×
3(y – 2)
4 2
1
=
6
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We can add algebraic fractions using the same method that
we use for numerical fractions. For example,
1
2
What is
+
?
a
b
We need to write the fractions over a common denominator
b
2a
b + 2a
1
2
+
=
+
=
a
b
ab
ab
ab
In general,
a
c
+
=
b
d
bd
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y
3
What is
+
?
x
2
We need to write the fractions over a common denominator
y
y×x
3
3×2
+
=
+
x
x×2
2×x
2
xy
6
+
=
2x
2x
6 + xy
=
2x
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Subtracting algebraic fractions
We can also subtract algebraic fractions using the same
method as we use for numerical fractions. For example,
p
q
What is
–
?
3
2
We need to write the fractions over a common denominator
before we can subtract them.
p
q
2p
3q
2p – 3q
–
=
–
=
6
6
6
3
2
In general,
a
c
–
=
b
d
bd
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Subtracting algebraic fractions
2+p
3
What is
–
?
2q
4
2+p
3
(2 + p) × 2q
3×4
–
=
–
2q
4
4 × 2q
2q × 4
2q(2 + p)
12
=
–
8q
8q
2q(2 + p) – 12
=
8q
4
6
q(2 + p) – 6
=
4q
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