Download Lesson 3 Expanding binomial products

Mathematics Stage 5
PAS5.3.1
Algebraic techniques
Lesson 3 Expanding binomial products
Contents
Expanding binomial products ............................................3
Activity answers...............................................................11
Exercise
.......................................................................15
Exercise answers ............................................................19
Lesson 3 Expanding binomial products
1
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PAS5.3.1 Algebraic techniques
Expanding binomial products
You are familiar with the technique of expanding products such as
5 ( x + 3) to find the equivalent expression 5x + 15 .
This process of expanding can be demonstrated using area. The product
5 ( x + 3) represents the area of a rectangle that is 5 units wide and
x + 3 units long.
x
x+3
3
5
The area of the large rectangle is 5 ( x + 3) square units.
The areas of the two smaller interior rectangles are 5x square units and
15 square units.
Therefore 5 ( x + 3) = 5x +15 .
But what if you need to expand an expression such as ( x + 5 ) ( x + 3) ?
This type of expression is called a binomial product, where binomial
means ‘two numbers’ and refers to the two terms in each bracket.
To explore how to expand such expressions, complete the following
activity.
Lesson 3 Expanding binomial products
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Activity – Expanding binomial products
Try these.
1
The rectangle below is given the dimensions x + 5 and x + 3 . The
dotted lines divide the shape into four quadrilaterals.
x
x+5
x
x+3
Find the area of each of the four smaller shapes and write the areas
in each shape.
Complete the following algebraic statement:
Area of whole rectangle = sum of the four inner areas
∴ ( x + 5 ) ( x + 3) = ________________________________________
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PAS5.3.1 Algebraic techniques
2
Draw your own diagram to expand ( y + 2 ) ( y + 4 ) .
∴ ( y + 2 ) ( y + 4 ) = ________________________________________
Check your response by going to the suggested answers section.
By examining the results of the two examples above, a pattern can be
found that will enable you to expand other binomial products without
drawing pictures.
Referring back to the rectangle, each of the two terms on one side ends
up getting multiplied by each of the terms on the other side.
x
5
x
3
This systematic approach can be applied directly to the algebraic
expression. Multiply each term in the first bracket by each term in the
second.
Lesson 3 Expanding binomial products
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( x + 5 ) ( x + 3) = x 2 + 3x + 5x + 15
= x 2 + 8x + 15
The arrows can be drawn in your own solution if you feel they help you
to be systematic.
Some people prefer to look at it in terms of two expansions:
( x + 5 ) ( x + 3) = x ( x + 3) + 5 ( x + 3)
= x 2 + 3x + 5x + 15
= x 2 + 8x + 15
Either way, your method needs to be systematic to avoid confusion.
In the following activity, use a systematic approach to expand the
binomial products without the need to draw rectangles.
Activity – Expanding binomial products
Try these.
3
Expand each of these expressions.
a
( k + 6 ) ( k + 1)
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b
( k + 6 ) ( k − 1)
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c
( k − 6 ) ( k + 1)
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PAS5.3.1 Algebraic techniques
d
( k − 6 ) ( k − 1)
________________________________________
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e
( 2y + 3) ( y − 7 )
______________________________________
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f
( 3x
2
)
− 8 ( x − 2 ) _____________________________________
___________________________________________________
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g
( 4m + 7n ) ( 2n − 5m )
__________________________________
___________________________________________________
___________________________________________________
Check your response by going to the suggested answers section.
When expanding, beware of the signs as this is the most likely error you
will make. Also note that it is not always possible to simplify the
expanded expression by collecting like terms. By looking at the
examples above, can you determine if any terms will be like before your
expand?
You may encounter a binomial product when solving a problem, and it is
often necessary to expand the product to determine a solution or reach a
conclusion.
Lesson 3 Expanding binomial products
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Activity – Expanding binomial products
Try these.
4
The product of two odd numbers will always be odd.
a
Prove that this statement is true.
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b
Use the result above to explain how you would take this further
to show that the product of any number of odd integers will be
odd.
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PAS5.3.1 Algebraic techniques
Check your response by going to the suggested answers section.
Expanding expressions is a valuable skill often needed as a component of
larger solutions. By developing a variety of these skills, you can apply
them when needed in more complex tasks.
Go to the exercises section and complete Exercise – Expanding binomial
products.
Lesson 3 Expanding binomial products
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PAS5.3.1 Algebraic techniques
Activity answers
This section provides answers to the activities found throughout the lesson.
Your answers should be similar to these. If your answers are very
different or if you do not understand an answer, ask your teacher.
Activity – Expanding binomial products
1
The areas are shown on the diagram below.
This square is
x long and x wide.
x
5
This rectangle is
x high and 5 wide.
x
Area = x2
Area = 5x
Area = 3x
Area = 15
This rectangle is x long
and 3 wide.
3
This rectangle is 5 long
and 3 wide.
Area of whole rectangle = sum of the four inner areas
∴ ( x + 5 ) ( x + 3) = x 2 + 3x + 5x + 15
You may have simplified this to be x 2 + 8x + 15 .
Lesson 3 Expanding binomial products
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2
You may have made your y section longer or shorter than that shown
below. However, it is important that the length you assigned to y on
one side is the same as the length you made it on the other side.
y
4
y Area = y2
Area = 4y
2 Area = 2y
Area = 8
∴ ( y + 2 ) ( y + 4 ) = y 2 + 4y + 2y + 8 or y 2 + 6y + 8
3
You may have included arrows or your first line of working may
have shown the two separate expansions.
a
( k + 6 ) ( k + 1) = k 2 + k + 6k + 6
= k 2 + 7k + 6
b
( k + 6 ) ( k − 1) = k 2 − k + 6k − 6
= k 2 + 5k − 6
c
( k − 6 ) ( k + 1) = k 2 + k − 6k − 6
= k 2 − 5k − 6
d
( k − 6 ) ( k − 1) = k 2 − k − 6k + 6
= k 2 − 7k + 6
e
( 2y + 3) ( y − 7 ) = 2y 2 − 14y + 3y − 21
= 2y 2 − 11y − 21
f
( 3x
g
( 4m + 7n ) ( 2n − 5m ) = 8mn − 20m 2 + 14n 2 − 35mn
2
)
− 8 ( x − 2 ) = 3x 3 − 6x 2 − 8x + 16
= 14n 2 − 20m 2 − 27mn
12
(any order)
PAS5.3.1 Algebraic techniques
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a
One way to prove that a number is odd is to show that it is one
more or one less than a number that is even.
Let the two odd numbers be 2n + 1 and 2k + 1 where n and k
are integers.
Product of two odds = ( 2n +1) ( 2k +1)
= 4kn + 2n + 2k +1
= 2 ( 2kn + n + k ) +1
Since k and n are integers, then 2kn + n + k is also an integer.
This means that 2 ( 2kn + n + k ) must be even and hence
2 ( 2kn + n + k ) + 1, the product of the odd numbers, must be
odd.
b
You can explain this in many ways.
Method 1
One method is to start with 3 odd numbers.
The first two multiply to give an odd. You then multiply this
odd answer by another odd which in turn give an odd answer.
Continuing this argument, when multiplying four odd numbers,
the first three give an odd answer which is then multiplied by an
odd giving another odd answer.
And so on.
Method 2
You might combine algebra and words as in the following
explanation.
A product of an unknown number of odd integers can be written
as:
( 2k1 + 1) ( 2k2 + 1) ( 2k3 + 1) ( 2k4 + 1) ...( 2kn −1 + 1) ( 2kn + 1)
where each k is an integer, and n represents the number of odd
integers in the product (for example if there are 10 odd integers
then n = 10).
Lesson 3 Expanding binomial products
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Calculating the first product gives an odd answer.
( 2k1 + 1)( 2k2 + 1)( 2k3 + 1)( 2k4 + 1)...( 2kn −1 + 1) ( 2kn + 1)
= odd × ( 2k3 + 1) ( 2k4 + 1) ...( 2kn −1 + 1) ( 2kn + 1)
Product of two odds gives an odd so:
odd × ( 2k3 + 1) ( 2k4 + 1) ...( 2kn −1 + 1) ( 2kn + 1)
= odd × ( 2k4 + 1) ...( 2kn −1 + 1) ( 2kn + 1)
Continuing this logic, each successive product will result in an
odd answer and hence
( 2k1 + 1) ( 2k2 + 1) ( 2k3 + 1) ( 2k4 + 1) ...( 2kn −1 + 1) ( 2kn + 1) will
always be odd.
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PAS5.3.1 Algebraic techniques
Exercise
Name
___________________________
Teacher
___________________________
Exercise – Expanding binomial expressions
1
By dividing the rectangle below into smaller regions, show that
( n + 7 ) ( n + 6 ) = n 2 + 13n + 42 .
n +7
n+6
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Lesson 3 Expanding binomial products
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2
a
Expand ( x + 5 ) ( x − 3) .
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b
Verify your answer by substituting a value for x.
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3
Expand each of these expressions. (Check your answers by mentally
substituting a value for the pronumeral.)
a
(t − 6 )( y + 2 )
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b
( m + 12 ) ( m + 5 )
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c
( 3k − 8 ) ( 2k − 7 )
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d
(y
2
)(
)
+ 10 y 2 − 2 _____________________________________
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e
( 3p + 9 ) ( 9 − p )
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PAS5.3.1 Algebraic techniques
4
Work through the following task to prove a famous theorem.
The square ABCD below has side length a + b .
a+b
A
B
D
C
The diagram below shows the same square with marks dividing each
side into lengths a and b. The marks are then joined to show a new
shape, EFGH.
a
A
E
b
B
a
F
b
b
H
a
D
b
a
G
a
C
Prove that all four small triangles are congruent.
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b
Prove that EFGH is a square using the proof above and by
proving that ∠HEF = 90° .
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c
The area of the square ABCD is ( a + b ) = ( a + b ) ( a + b )
2
Write this expression in expanded form.
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d
Write an equation for the area of ∆EBF in terms of a and b.
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e
The area of the small square EFGH equals the area of the large
square ABCD minus the area of the four triangles.
Let the length of EF = c.
Write an equation for the area of EFGH in terms of a, b and c.
Simplify the expression.
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f
Which famous theorem have you just proven?
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PAS5.3.1 Algebraic techniques
Exercise answers
This section provides answers to questions found in the ‘Exercise’ section.
Your answers should be similar to these. If your answers are very
different or if you do not understand an answer, ask your teacher.
Exercise – Expanding binomial products
1
Your diagram should be similar to the one below.
n
7
6xn
= 6n
7 x 6 = 42
6
nxn
= n2
7 x n = 7n
n
Area of large rectangle = sum of areas of small inner shapes
Therefore ( n + 7 ) ( n + 6 ) = n 2 + 6n + 7n + 42
= n 2 + 13n + 42
2
a
( x + 5 ) ( x − 3) = x 2 + 2x − 15 .
b
You may have chosen any value for x. Here, x = 5 was chosen.
( x + 5 ) ( x − 3) = 10 × 2
= 20
x 2 + 2x − 15 = 5 2 + 2 × 5 − 15
= 25 + 10 − 15
= 20
Both sides of the equation are equal when x = 5 .
Lesson 3 Expanding binomial products
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3
4
a
(t − 6 ) ( y + 2 ) = ty + 2t − 6y − 12
b
( m + 12 ) ( m + 5 ) = m 2 + 17m + 60
c
( 3k − 8 ) ( 2k − 7 ) = 6k 2 − 37k + 56
d
(y
e
( 3p + 9 ) ( 9 − p ) = −3p 2 + 18 p + 81
a
One proof is shown below.
2
)(
)
+ 10 y 2 − 2 = y 4 + 8y 2 − 20
In ∆HAE and ∆EBF
∠A = ∠B = 90° (vertices of square ABCD)
AE = BF = a (by construction)
HA = EB = b (by construction)
∴ ∆HAE ≡ ∆EBF(SAS)
(Remember, ≡ means 'is congruent to'
and SAS means 'side, angle, side'.)
Similarly, ∆HAE ≡ ∆EBF ≡ ∆FCF ≡ ∆GDH.
(The word ‘similarly’ as used here means ‘and using the same
method of proof’.)
b
Since ∆HAE ≡ ∆EBF ≡ ∆FCF ≡ ∆GDH
Then EF = FG = GH = HA (matching sides of congruent
triangles)
Hence EFGH is a rhombus.
∠AEH + ∠AHE = 90 (sum of small angles in right ∆AHE)
But ∠AHE = ∠BEF
(matching angles in congruent triangles AEH and BFE)
∴ ∠AEH + ∠BEF = 90
Also ∠AEH + ∠BEF + ∠HEF = 180
(angles forming a straight angle)
∴ 90 + ∠HEF = 180
Hence ∠HEF = 90
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PAS5.3.1 Algebraic techniques
Since EFGH is a rhombus and ∠HEF = 90 then EFGH is a
square.
c
Area of large square ABCD = a 2 + 2ab + b 2
d
Area EBF =
e
Area EFGH = Area ABCD − 4 × Area EBF
1
c 2 = a 2 + 2ab + b 2 − 4 × ab
2
2
2
= a + 2ab + b − 2ab
1
ab
2
Hence c 2 = a 2 + b 2
f
a, b and c represent the sides of a right triangle and so you have
proven Pythagoras’s theorem.
Lesson 3 Expanding binomial products
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