Teach secondary school algebra

1. Algebra
Dennis Almeida
University of Sheffield

2. Introduction
The claim:
The genesis of mathematical knowledge can often
help us teach mathematics in an efficient and
interesting way.
This presentation will examine the historical and,
largely, geometric roots for expanding brackets
involving one unknown, completing the square and
its applications and the binomial expansion.
With respect to school algebra this is certainly the
case.

3. First - a puzzle:
What has bone setting got to do with algebra?

4. What has bone setting got to do with algebra?
•Abu Ja'far Muhammad ibn Musa Al-Khwarizmi wrote the first
treatise on algebra: Hisab al-jabr w’al-muqabala (Calculation by
Restoring and Balancing) in 820 AD. The word algebra is a
corruption of al-jabr which means restoration.
•In Spain, where the Arabs held sway for a long period, there
arose a profession of algebrista’s who dealt in bone setting.
•álgebra. Del lat. tardío algebra, y este abrev. del ár. clás. algabru
walmuqabalah, reducción y cotejo.
•1. f. Parte de las matemáticas en la cual las operaciones
aritméticas son generalizadas empleando números, letras y
signos.
•2. f. desus. Arte de restituir a su lugar los huesos dislocados
Translation: the art of restoring broken bones to their
correct positions

5. 1. Hisab al-jabr w’al-muqabala provided a
compendium of methods to solve a range of equations
involving an unknown.
The implications of Al Khwarizmi work:
2. His name was used to describe any general method
to solve mathematical equations:
Al-Khwarizmi also wrote a treatise on Hindu-Arabic
numerals. The Arabic text is lost but a Latin
translation, Algoritmi de numero Indorum, exists and
this gave rise to the word algorithm deriving from
his name in the title: http://www-history.mcs.st-
and.ac.uk/Biographies/Al-Khwarizmi.html

6. The first known European translation of Khwarizmi’s Algebra:

7. THE BOOK
OF ALGEBRA
IN ARITHMETIC AND GEOMETRY
COMPOSED BY DOCTOR PEDRO NUNES,
THE FOREMOST COSMOGRAPHER IN THE KINGDOM OF PORTUGAL
AND
EMERITUS PROFESSOR IN MATHEMATICS AT THE UNIVERSITY OF
COIMBRA
Published in Anvers
1567

8. Arithmetic and geometry leads to algebra
The (distributive) rules in arithmetic.
Working out (4 + 1)  (4 + 2) geometrically:
4
4
2
1
4 + 1
4
+
2
4  4
4  2
1  4
1  2
(4 + 1)(4 + 2) = 44 + 42 + 14 + 12
x
x
x x
x
x
x
x
(x + 1)(x + 2) = xx + x2 + 1x + 12
(x + 1)(x + 2) = x2 + 2x + x + 2
(x + 1)(x + 2) = x2 + 3x + 2

9. Arithmetic and geometry leads to algebra
Next complete the rectangle.
Reversing the geometry of the last slide:
?
?
x + 1
x2
2x
1x
2
x
x
x2 + 3x + 2 = (x + 1)(x + 2)
x2 + 3x + 2
What should the widths of the added
rectangles be to simultaneously give
an area sum of 3x and area 2 for the
blue corner rectangle?
x
+
2
2
1

10. Arithmetic and geometry leads to algebra
The (distributive) rules in arithmetic.
Working out (4 + 1)  (4 + 1) geometrically:
4
4
1
1
4 + 1
4
+
1
4  4
4  1
1  4
1  1
(4 + 1)(4 + 1) = 44 + 41 + 14 + 11
a a  a
a + b
ba
(a + b)(a + b) = aa + ab + ba + bb
(a + b)(a + b) = a2 + ab + ab + b2
(a + b)2 = a2 + 2ab + b2
a
a
+
b
bb
a  b
b
b

11. Geometry: the mother of algebra
Expansions with negative integers geometrically.
Working out (x + 3)(x – 2) geometrically:
x
x 2
x – 2
3
x
+
3
2x
32
3x – 32
= x2 – 2x + 3x – 32
(x + 3)(x – 2)
(x + 3)(x – 2) = x2 + x – 6
x2
x2 – 2x

12. Geometry: the mother of algebra
Expansions with negative integers geometrically: continued.
Exercise: How could you expand out (x – 3)(x – 2) geometrically?
x
x 2
x – 2
3
x
–
3
2x
x2
= x2 – 2x – 3x + 32
(x – 3)(x – 2)
(x – 3)(x – 2) = x2 – 5x + 6
x2 – 2x
3x – 32
x2 – 2x – (3x – 32)
Area =

13. The difference of two squares geometrically.
a =
b
a
b
a2
a
b
a
–
b2
a – b
b
b
a
a
a – b
Area = a2 – b2
a – b
b
b a
a
a – b

a – b
b
a
Area = (a+b)(a – b)

Therefore a2 – b2 = (a+b)(a – b)

14. Geometry: the mother of algebra
‘Completing’ the square geometrically.
Consider (x2 + 6x) geometrically.
x
x
3
3
Now complete the square for
the diagram by adding the small
square in the lower right.
3
3
x + 3
x
+
3
So (x2 + 6x) = (x + 3)2 – 32
½ of 6
x2 3x
3x

15. Geometry: the mother of algebra
‘Completing’ the square geometrically.
Consider (x2 – 6x) geometrically.
x
x
3
3
3
x – 3
x
–
3
(x2 – 6x) = (x – 3)2 – 32
½ of 6
x2
3x
3x
From the square with area x2 we subtract
a rectangle of area 3x.
Next, to complete the subtraction of 6x,
we subtract another rectangle of area 3x.
x – 3
However the square of side 3 (shown in
blue) needs to be subtracted from within
the remnant square of side (x – 3) .
3
3

16. Geometry: the mother of algebra
‘Completing’ the square rules.
(x2 + 2ax) = (x + a)2 – a2
½ of 2a
(x2 – 2ax) = (x – a)2 – a2
½ of 2a

17. Geometry: the mother of algebra
Completing the square and difference of two squares.
(x2 + 2ax) = x(x + 2a)
= (x + a – a) (x + a + a)
A B
A2
A
– B2
A – B
A + B
=
(A+B)(A– B) = A2 – B2
= (x + a)2 – a2
(x2 + 2ax) = (x + a)2 – a2
Quick exercise: Use this
method to complete the
square for the expression
(x2 – 2ax)
(A – B) (A + B)
A2 – B2

18. Geometry: the mother of algebra
An illustration from Nunes’ book:
The problem is to solve x2 + 10x = 56.
The method is essentially geometric:
x
x2
x
5
5
5x
5x
Area of the L shape = x2 + 10x = 56
52 = 25 Area of the outer square = (x + 5)2
Also area of the outer square =
Area of the L shape + 52 = 56 + 25 = 81
So (x + 5)2 = 81
Thus x + 5 = 9
Negative solution not
considered.
Hence x = 4
The key point in Nunes work is that numbers and the variable x
represents length and the product of any two area.

19. Completing the square application.
‘Completing’ the square to solve quadratic equations.
Example 1. Solve x2 + 10x + 9 = 0
Bracket x terms: (x2 + 10x) + 9 = 0
Complete the square:
{(x + 5)2 – 52} + 9 = 0
Re-arrange to have x terms on LHS:
(x + 5)2 = 25 – 9
(x + 5)2 = 16
(x + 5) = 4
x = +4 – 5 = –1
Example 1a. Solve x2 + 2bx + c = 0
Bracket x terms: (x2 + 2bx) + c = 0
Complete the square:
{(x + b)2 – b2} + c = 0
Re-arrange to have x terms on LHS:
(x + b)2 = b2 – c
(x + b)2 = b2 – c
(x + b) = √(b2 – c)
x = – b +√(b2 – c)
Numbers to letters: Generalising to
algebra
or x = –4 – 5 = –9 or x = – b +√(b2 – c)

20. ‘Completing’ the square to solve quadratic equations.
Example 3. Solve ax2 + bx + c = 0
 
a
ac
b
b
x
a
ac
b
a
b
x
a
ac
b
a
b
x
a
ac
b
a
b
x
a
c
a
b
a
b
x
a
c
a
b
a
b
x
x
a
c
a
b
a
b
x
a
c
x
a
b
x
a
c
bx
ax
x
2
4
2
4
2
Or
4
4
2
4
4
2
Or
.
4
2
Or
2
2
:
LHS
on
terms
have
to
Rearrange
0
2
2
:
square
the
Complete
0
:
by
Divide
0
:
terms
Bracket
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2

















































































Current algorithm for solving quadratic equations

21. The binomial expansion
Binomial means two names or consisting of two terms.
The binomial expansion relates to the removal of brackets in
the expression (a + b)n = (a + b)(a + b)....... (a + b)
We have seen how to do this geometrically for (a + b)2.
To expand (a + b)3 geometrically will, of course, require us to
consider volume.
Al-Sijzi (10th century) proved geometrically that
(a + b)3 = a3 + 3ab(a + b) + b3.
He did this by decomposing a cube of side a + b into the sum
of two cubes of sides a and b and a number of parallelepipeds
of total volume 3ab(a + b).
http://www-history.mcs.st-and.ac.uk/Biographies/Al-Sijzi.html
This is a complex demonstration ….

22. http://www.youtube.com/watch?v=pK5NBIkLXa4
The binomial expansion: al-Sijzi’s demonstration on youtube

23. The binomial expansion
To expand (a + b)n geometrically for n ≥ 4 is not
possible and so the treatment becomes solely thought
experimental.
For this reason the binomial expansion presents some
challenges for teaching.
The next slides offer two treatments for explaining the
binomial expansion. Both derive from history.

24. The binomial expansion: treatment 1
a + b (1a + 1b) 1 1
shorthand
(a + b)2
= (1a + 1b) (1a + 1b)
= (a + b) (a + b)
= 1a2 + 1ba + 1ab + 1b2
= 1a2 + 2ab + 1b2
+
1 1
shorthand
1 2 1
(a + b)3 = (a + b)2(a + b)
= (1a2 + 2ab + 1b2)(1a + 1b)
= 1a3 + 2a2b + 1ab2 + 1a2b + 2ab2 +1b3
= 1a3 + 1a2b + 2a2b + 2ab2 + 1ab2 +1b3
= 1a3 + 3a2b + 3ab2 + 1b3 1 1
shorthand
1 2 1
1 3 3 1

25. The binomial expansion: treatment 1
a + b = (1a + 1b) 1 1
(a + b)2 = 1a2 + 2ab + 1b2 1 1
1 2 1
(a + b)3 = 1a3 + 3a2b + 3ab2 + 1b3
1 1
1 2 1
1 3 3 1
This pattern for the coefficients of the expansion continues:
1 1
1 2 1
1 3 3 1
1 4 6 4 1
(a + b)4 = 1a4 + 4a3b + 6a2b2 + 4ab3 + 1b4

26. The binomial expansion: treatment 1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
The (symmetric) triangular pattern
for the binomial coefficients has
been known since about the 11th
century AD through the work of the
Chinese mathematician Jia Xian.
In the 13th century, Yang Hui
presented the triangle and hence in
China it is still called Yang Hui's
triangle.
In much of the Western world it is
called Pascal’s triangle after the 17th
century French mathematician
Blaise Pascal.

27. The binomial expansion: treatment 2
L G
2 sounds L(aghu) and
G(uru)
How many single
sounds are there ?
L G
1 1
How many types of
different double
sounds are there ?
L G
1 2
L L G L G G
1
How many types of different treble sounds are there ?
L G
1
3
L L G L G G
1
L G
L
G
L L
G L L
G
G
L
G
G
L G
3

28. The binomial expansion: treatment 2
L G
+
(L+ G)2 = L G
+ 2 L G
+
2
2
(L+ G)3 = L G
+ 3 L G
3
3 2
+
G
+ 3 L
2
(L+ G)1 =
Algebraic summary of last slide:
This linguistic derivation of the binomial expansion was formulated
by the Vedic mathematician Pingala in the 2nd century BC.
Evidently one needs to know how many ways sounds can be
composed of 4 , 6, .... L and G’s to be able to determine (L + G)n for
n = 4, 5, 6, ... Appendix 1 shows how this is done.

29. 1 = ½ + ½
1 = ½ + ½1
1 = ½ + ½(½ + ½)
1 = ½ + ½2 + ½2
1 = ½ + ½2 + ½2 (½ + ½)
1 = ½ + ½2 + ½3 + ½3
1 = ½ + ½2 + ½3 + ½3(½ + ½)
1 = ½ + ½2 + ½3 + ½4 + ½4
.
.
.
1 = ½ + ½2 + ½3 + ½4 + ½5 + …………………..
1
1
1
1
½
¼ = ½2
½3
½4
1 = ½ + ½2 + ½3 + ½4 + ½5 + …
Geometry and series

30. 1.5 = 1 + ½
1.5 = 1 + ⅓1.5
1.5 = 1 + ⅓ (1 + ½)
1.5 = 1 + ⅓ + ⅓½
1.5 = 1 + ⅓ + ⅓⅓1.5
1.5 = 1 + ⅓ + ⅓2(1 + ½)
1.5 = 1 + ⅓ + ⅓2 + ⅓2 ½
1.5 = 1 + ⅓ + ⅓2 + ⅓2⅓1.5
1.5 = 1 + ⅓ + ⅓2 + ⅓3(1 + ½)
.
.
1.5 = 1 + ⅓ + ⅓2 + ⅓3 + ⅓4 + ………………
Geometry and series: extending the last result.

31. ...
..........
..........
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
4
3
2
3
2
2
2
2





















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



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





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


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









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

















x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
Geometry and series: Generalising (useful for KS5)
Appendix 2 shows how to use this
method to factorise difference of cubes,
fourth powers, etc.

32. Appendix1:Generalisingtreatment2ofThebinomialexpansion
A B
Suppose we want to arrange three different letters:
C
For the letter in the first position there are 3 choices: A, B or C
Having chosen the first letter, there are 2 choices for the letter in the
second position. A total of 3  2 choices.
Having chosen the first and second letter, there is just 1 choice for the
letter in the third position. A total of 3  2  1 choices.
So there are 3! = 3  2  1 different ways to arrange A, B and C.

33. A B
Now suppose we want to arrange three letters but two are the same:
For the letter in the first position there are 3 choices: A, B or B
Having chosen the first letter, there are 2 choices for the letter in the second position.
A total of 3  2 choices.
Having chosen the first and second letter, there is just 1 choice for the letter in the
third position. A total of 3  2  1 = 3! choices.
B
However there will be some duplications. To see this we annotate the second B:
A B Bꞌ
A Bꞌ B
The arrangement above is the same as:
There are 2 = 2  1 = 2! duplications.
So the number of arrangements of 3 letters two of which are the same =
!
2
!
1
!
3
!
2
!
3

2
3
!
2
!
1
!
3
:
Notation C

Appendix1:Generalisingtreatment2ofThebinomialexpansion

34. So we can use the subscript to keep a track of the number of B’s.
.
2!1!
3!
1!2!
3!
However 1
3
2
3
C
C 


The result of the last slide :
2
2 C
3
A
1
and
s
B’
with
letters
3
arranging
of
ways
of
number
The 
2
3
also
B
1
and
s
A’
2
with
letters
3
arranging
of
ways
of
number
The C

1
1 C
3
B
and
s
A’
2
with
letters
3
arranging
of
ways
of
number
The 
1
s
B'
and
s
A’
3
with
letters
3
arranging
of
ways
of
number
The 3

 0
0 C
2
2 C
3
s
B'
and
s
1A’
with
letters
3
arranging
of
ways
of
number
The 
1
s
B'
and
s
A’
0
with
letters
3
arranging
of
ways
of
number
The 3

 3
3 C
Appendix1:Generalisingtreatment2ofThebinomialexpansion

35. Binomial expansion for (L + G)3 using the new notation:
So for power 4:
(L+ G)3 = L G
+ 3C1 L G
3
3 2
G
+ 3C2 L
2
3C0 + 3C3
(L+ G)4 = L G
+ 4C1
L G
4
4 3
G
+ 4C2
L
2
4C0
+ 4C4
2
G
+ 4C3
L
3
And so on for higher powers.
(L+ G)n = L G
+ nC1
L G
n
n n–1
G
+ nC2
L
2
nC0 +..………….+ nCn
n –2
In general:
  !
!
!
Where
r
r
n
n
Cr
n


Appendix1:Generalisingtreatment2ofThebinomialexpansion

36. Appendix 2: Difference of cubes, fourth powers, etc
The appendix looks at a type of expression involving
two terms and so they are binomial in that sense.
The expressions we look at are generalisations of the
difference of two squares (x2 – y2) and how they might
be factorised.
i.e. We see how (x3 – y3), (x4 – y4), etc can be factorised.
The method used dates from the 14th century and
involves repeated substitution and some algebra. So it is
likely only to be suitable for very able students.

37. Appendix 2: Difference of cubes, fourth powers, etc
 
 
 
    
2
2
3
3
2
2
3
3
2
2
3
3
3
2
2
2
2
3
2
2
3
2
2
2
gives
(0)
Using
or
gives
(5)
arranging
-
Re
)
5
.........(
1
get
We
(1).
by
given
for
expression
by the
(4)
in
the
replace
Now
)
4
........(
:
get
we
by
(3)
multiply
now
we
If
)
3
.........(
1
get
We
(1).
by
given
for
expression
by the
(2)
in
the
replace
Now
)
2
........(
:
get
we
by
(1)
multiply
we
If
)
1
........(
1
or
1
Then
(0)
.....
1
Write
y
xy
x
y
x
y
x
y
xy
x
y
x
t
y
xy
x
ty
tx
ty
y
xy
x
y
ty
xy
x
tx
y
xy
x
tx
x
ty
y
x
y
ty
x
tx
y
x
tx
x
ty
ty
tx
y
x
t




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








tx
tx
tx
tx
tx
tx
tx

38. Appendix 2: Difference of cubes, fourth powers, etc
 
 
    
5
,
for
on
so
And
gives
(0)
Using
gives
arranging
-
Re
1
gives
before
done
had
what we
Repeating
)
5
.........(
derived
had
we
slide
previous
In the
)
1
........(
1
Then
(0)
.....
1
Write
3
2
2
3
4
4
3
2
2
3
4
4
4
3
2
2
3
3
2
2
3
3
2
2
3
4
3
2
2
3



































n
y
x
y
xy
y
x
x
y
x
y
x
y
xy
y
x
x
y
x
t
ty
y
xy
y
x
x
y
ty
xy
y
x
x
y
xy
y
x
x
tx
ty
y
xy
x
tx
ty
y
x
t
n
n
tx
tx
The method appears
to be contrived but
its real power lies in
being able to
factorise xn – yn for
any natural number
n.
The calculation that
follows shows how:
v
v

39. Appendix 2: Difference of cubes, fourth powers, etc
 
 
 
    
2
2
3
3
2
2
3
3
2
2
3
3
3
2
2
2
2
3
2
2
3
2
2
2
gives
(0)
Using
or
gives
(5)
arranging
-
Re
)
5
.........(
1
:
get
We
(1).
by
given
for
expression
by the
(4)
in
the
replace
Now
)
4
........(
:
get
we
by
(3)
multiply
we
If
)
3
.........(
1
:
get
We
(1).
by
given
for
expression
by the
(2)
in
the
replace
Now
)
2
........(
:
get
we
by
(1)
multiply
we
If
)
1
........(
1
or
1
Then
(0)
.....
1
Write
y
xy
x
y
x
y
x
y
xy
x
y
x
t
y
xy
x
ty
tx
ty
y
xy
x
y
ty
xy
x
tx
y
xy
x
tx
x
ty
y
x
y
ty
x
tx
y
x
tx
x
ty
ty
tx
y
x
t






































tx
tx
tx
tx
tx
tx
tx
c
Multiply (1) by x and then substitute for tx.
Repeat twice to see how (x3 + y3) factorises.
c
c
n.b. If we generalise then this factorisation for sum of powers of two variables only works for odd powers.