1)-Algebraic-Expressions.pptx IAL P1 CHAPTER 1 PPT

Algebraic Expressions
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Prior Knowledge Check
1) Simplify:
a)
b)
2) Write as a single power of 2
a) b)
c)
3) Expand:
a) b)
c)
4) Write down the highest common
factor of:
a) 24 and 16 b) and
c) and
5) Simplify:
a) b)
c)
2𝑚2
𝑛+3𝑚𝑛2
6 𝑥2
−12𝑥−10
28
24
26
3 𝑥+12
10−15 𝑥
12𝑥−30 𝑦
8
2 𝑥
𝑥𝑦
2 𝑥 10 𝑥
5 𝑥
3

You can use the laws of indices to
simplify powers of the same base
1A
a) ¿ 𝑥7
b) ¿ 6 𝑟 5
c) ¿ 𝑏3
d) ¿ 2 𝑥2
e)
¿𝑎6
×2 𝑎2
¿ 2 𝑎8
f)
¿27 𝑥6
÷ 𝑥4
¿2 7 𝑥2

1A
Expand and simplify if possible
¿−21𝑥2
+12𝑥
a)
¿3 𝑦2
−2 𝑦5
b)
¿12 𝑥2
−8 𝑥3
+20 𝑥4
c)
¿10 𝑥2
+6 𝑥−10 𝑥−15
d)
¿10 𝑥2
−4 𝑥−15
LO: To be able to apply laws of indices to simplify expressions:

1A
Simplify
¿ 𝑥4
+𝑥
a)
¿
3 𝑥
2
b)
¿
𝑥7
𝑥
3
+
𝑥4
𝑥
3
¿
3 𝑥2
2 𝑥
−
6 𝑥5
2 𝑥
−3 𝑥4
¿ 4 𝑥5
c) ¿
20 𝑥7
5 𝑥
2
+
15 𝑥3
5 𝑥
2
+3 𝑥
If you have a single
term as the
denominator, you can
simplify the numerator
terms separately…

To find the product of two
expressions you multiply each term
in one expression by each term in
the other expression
1B
(x + 4)(x + 7)
 x2
+ 4x + 7x + 28
 x2
+ 11x + 28
(2x + 3)(x – 8)
 2x2
+ 3x – 16x – 24
 2x2
– 13x - 24
+ 28
+ 7x
+ 7
+ 4x
x2
x
+ 4
x
- 24
- 16x
- 8
+ 3x
2x2
x
+ 3
2x
There are various
methods for doing this,
all are ok!

To find the product of two
expressions you multiply each term
in one expression by each term in
the other expression
If you have more than two brackets,
just multiply any 2 first, and then
multiply the answer by the next one
1B
Expand
( 𝑥+4)(2𝑥−1)(𝑥+3)
¿(2𝑥2
+7 𝑥− 4)
(𝑥+3)
¿2 𝑥3
+6 𝑥2
+7 𝑥2
+21𝑥
−4 𝑥−12
¿2 𝑥3
+13𝑥2
+17 𝑥−12
Multiply the first
pair of brackets
Multiply this
new pair
Simplify

You can write expressions as
products of their factors. This is
known as factorising.
If the terms have a common factor
(or several), then the expression can
be factorized into a single bracket
1C
3 9
x 
a) 3( 3)
x
 
Common
Factor
3
2
5
x x

b) ( 5)
x x
 
x
2
8 20
x x

c) 4 (2 5)
x x
 
4x
2 2
9 15
x y xy

d) 3 (3 5 )
xy x y
 
3xy
2
3 9
x xy

e) 3 ( 3 )
x x y
 
3x

x2
+ 3x 2
+
You get the last number
in a Quadratic Equation
by multiplying the 2
numbers in the brackets
You get the middle
number by adding the 2
(x + 2)(x + 1)

x2
- 2x 15
-
You get the last number
in a Quadratic Equation
by multiplying the 2
You get the middle
number by adding the 2
(x - 5)(x + 3)

x2
- 7x + 12
Numbers that
multiply to give + 12
+3 +4
-3 -4
+12 +1
-12 -1
+6 +2
-6 -2
Which pair adds to
give -7?
(x - 3)(x - 4)
So the brackets were
originally…

x2
+ 10x + 16
Numbers that
+1 +16
-1 -16
+2 +8
-2 -8
+4 +4
-4 -4
Which pair adds to
give +10?
(x + 2)(x + 8)
originally…

x2
- x - 20
Numbers that
multiply to give - 20
+1 -20
-1 +20
+2 -10
-2 +10
+4 -5
-4 +5
Which pair adds to
give - 1?
(x + 4)(x - 5)
originally…

Factorising Quadratics
A Quadratic Equation has the form;
ax2
+ bx + c
Where a, b and c are constants and a ≠
0.
You can also Factorise these equations.
REMEMBER
 An equation with an ‘x2
’ in does not
necessarily go into 2 brackets. You use
2 brackets when there are NO
‘Common Factors’
1E
Examples
a)
2
6 8
x x
 
The 2 numbers in brackets must:
 Multiply to give ‘c’
 Add to give ‘b’
( 2)( 4)
x x
  

ax2
+ bx + c
Where a, b and c are constants and
a ≠ 0.
You can also Factorise these
equations.
1E
Examples
b)
2
4 5
x x
 
( 5)( 1)
x x
  

ax2
+ bx + c
a ≠ 0.
equations.
1E
Examples
c)
2
25
x 
( 5)( 5)
x x
  
(In this case, b = 0)
This is known as ‘the
difference of two squares’
 x2
– y2
= (x + y)(x – y)

ax2
+ bx + c
a ≠ 0.
equations.
1E
Examples
d)
2 2
4 9
x y

(2 3 )(2 3 )
x y x y
  

ax2
+ bx + c
a ≠ 0.
equations.
1E
Examples
d)
2
5 45
x 
 Sometimes, you need to remove
a ‘common factor’ first…
2
5( 9)
x
 
5( 3)( 3)
x x
  

• Expand the following pairs of
brackets
(x + 3)(x + 4)
 x2
+ 3x + 4x + 12
 x2
+ 7x + 12
(2x + 3)(x + 4)
 2x2
+ 3x + 8x + 12
 2x2
+ 11x + 12
+ 12
+ 4x
+ 4
+ 3x
x2
x
+ 3
x
+ 12
+ 8x
+ 4
+ 3x
2x2
x
+ 3
2x
When an x term has a ‘2’
coefficient, the rules are
different…
2 of the terms are
doubled
 So, the numbers in
the brackets add to
give the x term, WHEN
ONE HAS BEEN
DOUBLED FIRST

2x2
- 5x - 3
Numbers that
-3 +1
+3 -1
One of the values to the left
will be doubled when the
brackets are expanded
(2x + 1)(x - 3)
originally…
-6 +1
-3 +2
+6 -1
+3 -2 The -3 doubles so it must
be on the opposite side
to the ‘2x’

2x2
+ 13x + 11
Numbers that
+11 +1
-11 -1
will be doubled when the
(2x + 11)(x + 1)
originally…
+22 +1
+11 +2
-22 -1
-11 -2 The +1 doubles so it must
to the ‘2x’

3x2
- 11x - 4
Numbers that
+2 -2
-4 +1
+4 -1
will be tripled when the
(3x + 1)(x - 4)
originally…
+6 -2
+2 -6
-12 +1
-4 +3
The -4 triples so it must
to the ‘3x’
+12 -1
+4 -3

Indices can be negative numbers or
fractions
1D
Simplify
a) ¿ 𝑥6
b)
¿ 𝑥2
c)
¿ 𝑥2
d) ¿ (125 𝑥
6
)
1
3
¿ (125)
1
3
(𝑥
6
)
1
3
¿ 5 𝑥2

fractions
1D
Simplify
e) ¿
2 𝑥2
𝑥5
−
𝑥
𝑥5
¿
2
𝑥3
−
1
𝑥4
¿2 𝑥− 3
− 𝑥− 4
Either of these forms is correct – check
if the question asks for a specific one!
Simplify
separately
Rewrite

fractions
1D
Evaluate (work out the value of)
a)
¿ √9
¿ 3
b)
¿
3
√64
¿ 4
c) ¿(√49)
3
¿ 343
d) ¿
1
25
3
2
¿
1
(√25 )
3
¿
1
125
You can use a calculator
for these, but you still
need to be able to show
the process, especially
for algebraic versions

fractions
1D
Given that , express in the form where
and are constants
𝑦 =
1
16
𝑥
2
𝑦
1
2
=( 1
16
𝑥2
)
1
2
𝑦
1
2
=( 1
16 )
1
2
(𝑥2
)
1
2
𝑦
1
2
=
1
4
𝑥
Rewrite based
on the question
Each part is raised
to a power ½
Simplify

fractions
1D
Given that , express in the form where
and are constants
𝑦 =
1
16
𝑥
2
4 𝑦− 1
=4( 1
16
𝑥2
)
− 1
4 𝑦
− 1
=4( 1
16 )
− 1
(𝑥
2
)
−1
4 𝑦− 1
=4 (16)(𝑥− 2
)
Rewrite based
on the question
Each part is raised to a
power -1, and will then
be multiplied by 4
Simplify
4 𝑦− 1
=64 𝑥−2
Simplify more

In is an integer that is not a
square number, then is a surd. It
is an example of an irrational
number.
Surds can be used to leave answers
exact without rounding errors, and can
be manipulated by using the following
rules:
1E
Simplify
a) ¿√9×√2
¿3 √2
Make sure that what
you write is clear…
 and are different!
b) ¿ √4 ×√5
2
¿
2 √5
2
¿ √5
Find a factor which is a
square number, which you
can then square root
Simplify the numerator
Simplify the whole
fraction

number.
rules:
1E
Simplify
c)
Try to find a
common factor
¿5√6−2√4 √6+√49 √6
¿5√6−4 √6+7 √6
¿8 √6
Square roots can
be worked out
Simplify

number.
rules:
1E
Expand and simplify if possible
a)
Multiply out
¿5√2−√6
b)
Multiply out
¿10+2√3−5√3−√9
¿10−3√3−3
¿7−3√3
Group together like
terms. Calculate root 9
Simplify

If a fraction has a surd in the
denominator, then it can be useful to
rearrange it so that the denominator
is a rational number. This is called
rationalising the denominator.
For fractions of the form , multiply the
numerator and denominator by
1F
Rationalise
a) 1
√3
× √3
√3
¿
√3
3
b) 1
3+√2
×
3 − √2
3 − √2
¿
3− √2
(3+√2)(3− √2)
¿
3−√2
9+3√2− 3√2− 2
¿
3 −√2
7
Multiply so that the
surd is removed from
the denominator
Multiply both
numerator and
denominator
Multiply out the
brackets
Simplify

1F
Rationalise
c) √5+√2
√5− √2
×
√5+√2
√5+√2
¿
(√5+√2)(√5+√2)
(√5− √2)(√5+√2)
¿
5+√10+√10 +2
5+√10− √10 − 2
¿
7+2√10
3
Multiply both
numerator and
denominator
Multiply out the
brackets
Simplify

1F
Rationalise
d)
Multiply out the
brackets first
¿
1
(1−√3 )(1− √3)
¿
1
4 −2 √3
¿
1
4 −2 √3
×
4+2√3
4+2√3
¿
4+2√3
(4−2√3)(4+2√3)
¿
4+2√3
16+8√3−8 √3−4 √9
¿
4 +2 √3
4
¿
2+√3
2
Multiply to
cancel the surds
Multiply out the
brackets
Simplify
Divide all by 2

1)-Algebraic-Expressions.pptx IAL P1 CHAPTER 1 PPT

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