1. Finding the area of a rectangle given its length and width. The area formula uses the special product (length)(width), which is a multiplication of two binomials. 2. Finding the diagonal of a square given the length of one side. Using the special product of (a + b)(a - b) = a^2 - b^2, the diagonal can be found as (length)^2 - 0. 3. Finding the area of a triangle given the lengths of two sides and the angle between them. Using the trigonometric identity cos^2(θ) + sin^2(θ

1. EXPLORING
MATH 8
By: John Peter V. Martin

9. Process Questions
 What can you say about the pictures?
 Is there a mathematical view? Why or why not?
 Is there any more examples in our environment? Give 5
more examples.
 How are they related to each other?
 How can you relate those examples to solve real life
problems?
 How can you predict succeeding measures?

10. Advocacy Question
Among the 8 goals of our
school, what are the goals
that are applied in our
activity? Why?

11. Bible Verse
Proverbs 29:14
“ If a king judges the poor
with fairness, his throne
will be established
forever.”

12. Advocacy Questions
What is the meaning of
this Bible verse?
How will you relate this
verse in our activity?

13. If you are Jesus
what will you do if
there are people
who are losing
hope?

14. Transfer Task
 You are part of the RCBC accounting firm as consultant,
your boss assigned your team to help Glisten merchandise
to evaluate the resources and to decide if they are going to
add employees; you are tasked to present varied salary
schemes that could be used in paying the employees
(business plan). Your work will be presented to the board
of trustees in the company and to your costumer. It will
be judged based on the following criteria: explanation,
presentation and organization of the business plan,
accuracy of computations, and demonstration of the
understanding of key concepts of factoring, special
products, rational algebraic expressions and linear
equations in two variables.

15. GRASPS
 GOAL: To present a business plan that will include the varied
salary schemes that could be used in paying the employees.
 ROLE: Consultant, researcher and presenter
 AUDIENCE: The board of trustee in the company and the
costumer
 SITUATION: You are part of the RCBC accounting firm as a
consultant, your boss appointed you to help the aspiring
businessman to decide what business he will pursue; you are
tasked to present a business plan which will include the varied
salary schemes that could be used in paying the employees. Your
work will be presented to the head of the accounting firm and to
your costumer.

16. GRASPS
 PRODUCT: A business plan that will include the
varied salary schemes that could be used in paying the
employees.
 STANDARDS: It will be judged based on the
following criteria: explanation, presentation and
organization of the business plan, accuracy of
computations, and demonstration of the understanding
of key concepts of factoring, special products, rational
algebraic expressions and linear equations in two
variables.

17. MATH IS

18. Process Questions
What is the video all about?
What are the techniques needed to
multiply polynomials?
Is there a technique in solving
special products? What is it?

19. 1. Factors completely different types
of polynomials (polynomials with
common monomial factor,
difference of two squares, sum
and difference of two cubes,
perfect square trinomials, and
general trinomials).

20. Target Goals:
I can identify special products.
I can find special products of certain
polynomials.
I can relate special products with factoring.
I can factor polynomials with common
monomial factor.
I can factor the difference of two squares.

22. Multiplication of polynomials is an extension of the distributive
property. When you multiply two polynomials you distribute each
term of one polynomial to each term of the other polynomial.
We can multiply polynomials in a vertical format like we would
multiply two numbers.
(x – 3)
(x – 2)x_________
+ 6–2x
+ 0–3xx2
_________
x2
–5x + 6

23. Multiplication of polynomials is an application of the distributive
property. When you multiply two polynomials you distribute each
term of one polynomial to each term of the other polynomial.
We can also multiply polynomials by using the FOIL pattern.
(x – 3)(x – 2) = x2
– 5x + 6x(x) + x(–2) + (–3)(x) + (–3)(–2) =

24. Some pairs of binomials have special products.
When multiplied, these pairs of binomials always follow the
same pattern.
By learning to recognize these pairs of binomials, you can use
their multiplication patterns to find the product quicker and
easier.

25. One special pair of binomials is the sum of two numbers times
the difference of the same two numbers.
Let’s look at the numbers x and 4. The sum of x and 4 can be
written (x + 4). The difference of x and 4 can be written (x – 4).
Their product is
(x + 4)(x – 4) =
Multiply using foil, then collect like terms.
x2
– 4x + 4x – 16 = x2
– 16

26. (x + 4)(x – 4) = x2
– 4x + 4x – 16 = x2
– 16
Here are more examples:
(x + 3)(x – 3) = x2
– 3x + 3x – 9 = x2
– 9
(5 – y)(5 + y) = 25 +5y – 5y – y2
= 25 – y2
}What do all of these
have in common?

27. x2
– 16 x2
– 9 25 – y2
What do all of these
have in common?
They are all binomials.
They are all differences.
Both terms are perfect squares.

28. For any two numbers a and b, (a + b)(a – b) = a2
– b2
.
In other words, the sum of two numbers times the difference of
those two numbers will always be the difference of the squares of
the two numbers.
Example: (x + 10)(x – 10) = x2
– 100
(5 – 2)(5 + 2) = 25 – 4 = 21
3 7 = 21

29. The other special products are formed by squaring a binomial.
(x + 4)2
and (x – 6)2
are two example of binomials that have been
squared.
Let’s look at the first example: (x + 4)2
(x + 4)2
= (x + 4)(x + 4) =
Now we FOIL and collect like terms.
x2
+ 4x + 16 =+ 4x x2
+ 8x + 16

30. (x + 4)2
= (x + 4)(x + 4) = x2
+ 4x + 16 =+ 4x x2
+ 8x + 16
Whenever we square a binomial like this, the same pattern always occurs.
See the
first term?
In the final product
it is squared…
…and it appears in the middle term.

31. (x + 4)2
= (x + 4)(x + 4) = x2
+ 4x + 16 =+ 4x x2
+ 8x + 16
Whenever we square a binomial like this, the same pattern always occurs.
What about the
second term?
…and the last term is 4
squared.
The middle number is 2 times 4…

32. (x + 4)2
= (x + 4)(x + 4) = x2
+ 4x + 16 =+ 4x x2
+ 8x + 16
Whenever we square a binomial like this, the same pattern always occurs.
Squaring a binomial will always produce a trinomial whose first
and last terms are perfect squares and whose middle term is 2
times the numbers in the binomial, or…
For two numbers a and b, (a + b)2
= a2
+ 2ab + b2

33. Is it the same pattern if we are subtracting, as in the expression
(y – 6)2
?
(y – 6)2
= (y – 6)(y – 6) = y2
– 6y + 36 =– 6y y2
– 12y + 36
It is almost the same. The y is squared, the 6 is squared and the
middle term is 2 times 6 times y. However, in this product the
middle term is subtracted. This is because we were subtracting in the
original binomial. Therefore our rule has only one small change
when we subtract.
For any two numbers a and b, (a – b)2
= a2
– 2ab + b2

34. Examples:
(x + 3)2
= (x + 3)(x + 3) Remember: (a + b)2
= a2
+ 2ab + b2
= x2
+ 2(3)(x) + 32
= x2
+ 6x + 9
(z – 4)2
= Remember: (a – b)2
= a2
– 2ab + b2
(z – 4)(z – 4)
= z2
– 2(4)(z) + 42
= z2
– 8z + 16

35. You should copy these rules into your notes and try to remember
them. They will help you work faster and make many problems you
solve easier.
For any two numbers a and b, (a – b)2
= a2
– 2ab + b2
For two numbers a and b, (a + b)2
= a2
+ 2ab + b2
For any two numbers a and b, (a + b)(a – b) = a2
– b2
.

36. 1. (2x – 5)(2x + 5)
2. (x + 7)2
3. (x – 2)2
4. (2x + 3y)2

37. 1. (2x – 5)(2x + 5)
(2x – 5)(2x + 5)
22
x2
– 52
4x2
– 25

38. (x + 7)2
x2
+ 2(7)(x) + 72
x2
+ 14x + 49
2. (x + 7)2

39. (x – 2)2
x2
– 2(2)(x) + 22
x2
- 4x + 4
3. (x – 2)2

40. (2x + 3y)2
22
x2
+ 2(2x)(3y) + 32
y2
4x2
+ 12xy + 9y2
4. (2x + 3y)2

41. Finding the common monomial:
What is the common monomial
factor of the equation below?
1.2abc + 2ac – 2a
2.20x2
– 12

42.  Find the greatest common factor (GCF)
 Write each term as the product of the GCF and its
other factor.
1. 2ab = (2)(a)(b)
2ac = (2)(a)(c)
2a = (2)(a)
2abc + 2ac – 2a = 2a(b) + 2a(c) – 2a
= 2a (b + c – 1)2a
Common Monomial Factor
Factoring

43. Common monomial factor
4

44. Try This: ( 15 minutes only )

45. FIST OF
FIVE

46. Muddiest Point
 The part of the lesson that I still find confusing is
________________________________________
________________________________________
_______________________________
because
________________________________________
________________________________________
_______________________________

47. WORK
BY PAIR

48. Individual

49. Assignment
Search geometrical
problems that can be
solved by special product.