SD_Patterns and Algebra 1_Calacat_Vergara.pptx

PATTERNS and ALGEBRA 1
CHERRY S. CALACAT (Tagbilaran City Division)

DEPARTMENT OF EDUCATION
ENABLING OBJECTIVES
• Recognize and use the appropriate method in
factoring algebraic expressions;
• Perform operations on rational algebraic
expressions;
• Carefully simplify rational expressions with
polynomials;
2

ENABLING OBJECTIVES
• Set apart misconceptions in factoring and simplifying
rational algebraic expressions when practicing
problem solving skills;
• Create problems/activities/tasks that promote
problem solving and critical thinking skills.
3

BUREAU OF CURRICULUM DEVELOPMENT
Factoring
Operations on Rational Expression
KEY CONTENTS
4

Least Learned COMPETENCIES
5

6

Why is Factoring
Polynomials an
important Skill?
7
It helps us
understand more
about our
equations
We rewrite our
polynomials in
simpler form
We yield a lot of
information when we
apply the principles of
factoring to equations

 (𝑥𝑦2−3𝑧)(𝑥𝑦2+3𝑧) =
(𝑎 − 2𝑏) (𝑎2
+2𝑎𝑏 + 4𝑏2
) =
 (𝑎 + 2𝑏) (𝑎2
−2𝑎𝑏 + 4𝑏2
) =
(2𝑥 − 𝑦 + 3)2
=
PATTERNS and ALGEBRA 1 Special Products
8
𝒙𝟐
𝒚𝟒
− 𝟗𝒛𝟐
𝒂𝟑
− 𝟖𝒃𝟑
𝒂𝟑
+ 𝟖𝒃𝟑
𝟒𝒙𝟐
+ 𝒚𝟐
− 𝟒𝒙𝒚 + 𝟏𝟐𝒙 − 𝟔𝒚 + 𝟗

6𝑥2 + 17𝑥 + 7 =
4𝑥2
− 12 𝑥𝑦 + 9𝑦2
=
 4𝑥2
+ 12 𝑥𝑦 + 9𝑦2
=
𝑥4
− 23𝑥2
+ 49 =
PATTERNS and ALGEBRA 1 Special Products
9
𝟐𝒙 + 𝟏 (𝟑𝒙 + 𝟕)
(𝟐𝒙 − 𝟑𝐲)(𝟐𝒙 − 𝟑𝐲)
(𝟐𝒙 + 𝟑𝐲)(𝟐𝒙 + 𝟑𝐲)
(𝒙𝟐
− 𝟑𝒙 − 𝟕)(𝒙𝟐
+ 𝟑𝒙 − 𝟕)

10
When can
factoring a
polynomial be
performed?
When a polynomial is
expressed as the product
of polynomials, each of
which has rational
coefficients.

11
When do we say
that a polynomial
is in its factored
form?
A polynomial is
completely
factored if none
of its factors can
be factored.
Not all polynomials
can be written in
factored form.
(Prime Polynomials)

What are the
common
factoring
techniques?
12

36𝑐𝑦5 − 56𝑐2𝑦3𝑧
A. Common Monomial Factor:
36𝑐𝑦5
− 56𝑐2
𝑦3
𝑧
36𝑐𝑦5 − 56𝑐2𝑦3𝑧
4𝑐𝑦3(9𝑦2 − 14𝑐𝑧)
4𝑐𝑦3(9𝑦2 − 14𝑐𝑧)
FACTORING
13

B. Difference of Two Squares:
𝑥2 − 𝑦2 = (𝑥 + 𝑦)(𝑥 − 𝑦)
(2𝑥 − 3)2
− 𝑦 + 2 2
= [ 2𝑥 − 3 + 𝑦 + 2 ][ 2𝑥 − 3 − 𝑦 + 2 ]
(2𝑥 − 3)2
− 𝑦 + 2 2
= [ 2𝑥 − 3 + 𝑦 + 2 ][ 2𝑥 − 3 − 𝑦 + 2 ]
(2𝑥 − 3)2
− 𝑦 + 2 2
= [ 2𝑥 − 3 + 𝑦 + 2 ][ 2𝑥 − 3 − 𝑦 + 2 ]
(2𝑥 − 3)2
− 𝑦 + 2 2
= [ 2𝑥 − 3 + 𝑦 + 2 ][ 2𝑥 − 3 − 𝑦 + 2 ]
FACTORING
14

C. Perfect Square Trinomial:
𝑥2 ± 2𝑥𝑦 + 𝑦2 = (𝑥 ± 𝑦)2
𝑥4
− 14𝑥2
𝑦 + 49𝑦2
= (𝑥2
− 7𝑦)2
𝑥4
− 14𝑥2
𝑦 + 49𝑦2
= (𝑥2
− 7𝑦)2
𝑥4
− 14𝑥2
𝑦 + 49𝑦2
= (𝑥2
− 7𝑦)2
𝑥4
− 14𝑥2
𝑦 + 49𝑦2
= (𝑥2
− 7𝑦)2
FACTORING
15

C. Simple Trinomial:
𝑥2 + 2𝑥 − 35 = (𝑥 + 7)(𝑥 − 5)
𝑥2 + 2𝑥 − 35 = (𝑥 + 7)(𝑥 − 5)
𝑥2 + 2𝑥 − 35 = (𝑥 + 7)(𝑥 − 5)
𝑥2 + 2𝑥 − 35 = (𝑥 + 7)(𝑥 − 5)
FACTORING
16

D. General Trinomial by Grouping:
2𝑥2 + 7𝑥 + 5 = 2𝑥2 + 5𝑥 + 2𝑥 + 5
2𝑥2 + 7𝑥 + 5 = 2𝑥2 + 5𝑥 + 2𝑥 + 5
2𝑥2 + 7𝑥 + 5 = 2𝑥2 + 5𝑥 + 2𝑥 + 5
2𝑥2
+ 7𝑥 + 5 = 𝑥(2𝑥 + 5) + 1(2𝑥 + 5)
2𝑥2 + 7𝑥 + 5 = (2𝑥 + 5)(𝑥 + 1)
FACTORING
17

E. Sum and Difference of Cubes:
𝑥3
± 𝑦3
= (𝑥 ± 𝑦)(𝑥2
∓ 𝑥𝑦 + 𝑦2
)
125𝑥3
− 64𝑦3
= (𝑥 ± 𝑦)(𝑥2
∓ 𝑥𝑦 + 𝑦2
)
125𝑥3
− 64𝑦3
= (5𝑥 − 4𝑦)(𝑥2
∓ 𝑥𝑦 + 𝑦2
)
125𝑥3
− 64𝑦3
= (5𝑥 − 4𝑦)(𝑥2
∓ 𝑥𝑦 + 𝑦2
)
125𝑥3
− 64𝑦3
= (5𝑥 − 4𝑦)(25𝑥2
+ 20𝑥𝑦 + 16𝑦2
)
125𝑥3
− 64𝑦3
= (5𝑥 − 4𝑦)(25𝑥2
+ 20𝑥𝑦 + 16𝑦2
)
125𝑥3
− 64𝑦3
= (5𝑥 − 4𝑦)(25𝑥2
+ 20𝑥𝑦 + 16𝑦2
)
FACTORING
18

F. Grouping:
𝑥𝑦 − 𝑦 + 𝑥 − 1
𝑥𝑦 − 𝑦 + 𝑥 − 1
𝑦(𝑥 − 1) + (𝑥 − 1)
𝑦(𝑥 − 1) + (𝑥 − 1)
(𝑥 − 1)(𝑦 + 1)
FACTORING
19

G. Factoring 𝒙𝒏 ± 𝒚𝒏:
If n is an even number, then 𝒙𝒏 − 𝒚𝒏 can be considered as
the difference of two squares.
𝑥8
− 𝑦8
= 𝑥4 2
− 𝑦4 2
𝑥8
− 𝑦8
= 𝑥4 2
− 𝑦4 2
= (𝑥4
−𝑦4
)(𝑥4
+ 𝑦4
)
= (𝑥2
−𝑦2
)(𝑥2
+ 𝑦2
)(𝑥4
+ 𝑦4
)
= (𝑥 − 𝑦)(𝑥 + 𝑦)(𝑥2
+ 𝑦2
)(𝑥4
+ 𝑦4
)
FACTORING
20

H. Factoring 𝒙𝒏 ± 𝒚𝒏:
If n is a multiple of 3, then 𝒙𝒏
± 𝒚𝒏
can be considered as the
sum and difference of two cubes.
𝑥6
+ 𝑦6
= 𝑥2 3
+ 𝑦2 3
= (𝑥2
+ 𝑦2
)(𝑥4
− 𝑥2
𝑦2
+𝑦4
)
FACTORING
21

I. Factoring 𝒙𝒏 ± 𝒚𝒏:
If n is odd and not a multiple of 3, then
𝒙𝒏 + 𝒚𝒏 = (𝒙 + 𝒚)(𝒙𝒏−𝟏 − 𝒙𝒏−𝟐𝒚 + 𝒙𝒏−𝟑𝒚𝟐 − ⋯ + 𝒚𝒏−𝟏)
𝒙𝒏 − 𝒚𝒏 = (𝒙 − 𝒚)(𝒙𝒏−𝟏 + 𝒙𝒏−𝟐𝒚 + 𝒙𝒏−𝟑𝒚𝟐 + ⋯ + 𝒚𝒏−𝟏)
FACTORING
22

Factor 𝑥5
− 32𝑦5
𝑥5
− 32𝑦5
= 𝑥 5
− (2𝑦)5
= (𝑥 − 2𝑦)(𝑥4
+ 𝑥3
(2𝑦) + 𝑥2
2𝑦 2
+ 𝑥(2𝑦)3
+ (2𝑦)4
)
= (𝑥 − 2𝑦)(𝑥4
+ 2𝑥3
𝑦 + 4𝑥2
𝑦2
+ 8𝑥𝑦3
+ 16𝑦4
)
FACTORING
23

98 ∙ 102 =
If 𝑥4
− 16 = 𝐶 𝑥2
+ 𝐴 𝑥 + 𝐵 𝑥 − 2 , what is the
sum of A, B and C ?
Given that 𝑥 + 𝑦 = 10, if 𝑥𝑦 = 5, find the value of
𝑥3
+ 𝑦3
.
24
PATTERNS and ALGEBRA 1 Non-Routine
FactoringTechniques
(100 – 2)(100 + 2) = 10000 – 4 = 9996
Ans. 7
Ans. 850

If 𝑎2 + 3𝑎 + 9 = 0, what is 𝑎3 ?
If 𝑥 = 3 + 1, then what is
𝑥2
7−2𝑥+𝑥2 ?
 Given that 𝑟 +
1
𝑟
2
= 3, find the value of 𝑟3
+
1
𝑟3.
25
FactoringTechniques
Ans. 27
Ans.
𝟑+𝟏
𝟑
Ans. 0

Find m so that 𝑚𝑥4
− 42𝑥2
+ 49 is a perfect
square.
Factor completely: 4𝑝2
− 9𝑏2
+ 6𝑏𝑐 − 𝑐2
26
FactoringTechniques
Ans. 9
Ans. (𝟐𝒑 + 𝟑𝒃 − 𝒄)(𝟐𝒑 − 𝟑𝒃 + 𝒄)

Activity 1
Factoring Techniques
27

Rational Expressions
• Basic Definitions
• Simplification of Rational Expressions
• Multiplication and Division of Rational Expressions
• Addition and Subtraction of Rational Expressions
28

Definition of Rational Expression
If 𝑃 𝑥 and 𝑄 𝑥 are polynomials and 𝑄 𝑥 ≠ 0,
then
𝑃(𝑥)
𝑄(𝑥)
is a rational expression in x where 𝑃 𝑥 and
𝑄 𝑥 are the numerator and denominator, respectively
of the expression.
29

Examples of Rational Expression
7
𝑥−1
2𝑥+1
𝑥2−1
𝑥−2
𝑥2−4
𝑥3+4𝑥2−3𝑥
2𝑥4+5𝑥3−5𝑥−2
𝑥2+7𝑥−6
1
30
Remark:Arationalexpressionisinitssimplestformifthenumeratoranddenominator
havenocommonfactors.

Operations on R.E. Multiplication and Division
31
1. Perform the indicated operations and simplify:
𝑥2
− 𝑦2
2𝑥2 + 𝑥𝑦 − 3𝑦2
∙
6𝑥2
+ 13𝑥𝑦 + 6𝑦2
𝑥 + 𝑦
(𝑥+𝑦)(𝑥−𝑦)
(2𝑥+3𝑦)(𝑥−𝑦)
∙
(3𝑥+2𝑦)(2𝑥+3𝑦)
𝑥+𝑦
= 3𝑥 + 2𝑦

32
𝑎3
− 𝑏3
2𝑎2 + 4𝑎𝑏 + 2𝑏2
÷
𝑎3
+ 𝑎2
𝑏 + 𝑎𝑏2
𝑎2 − 𝑏2
÷
3(2𝑎2
− 3𝑎𝑏 + 𝑏2
)
6𝑎 + 6𝑏
=
(𝑎−𝑏)(𝑎2+𝑎𝑏+𝑏2)
2(𝑎2+2𝑎𝑏+𝑏2)
∙
(𝑎+𝑏)(𝑎−𝑏)
𝑎(𝑎2+𝑎𝑏+𝑏2)
∙
6(𝑎+𝑏)
3(2𝑎−𝑏)(𝑎−𝑏)
=
𝑎 − 𝑏
𝑎(2𝑎 − 𝑏)

33
𝑎5
− 1
𝑎 + 1
∙
𝑎2
+ 1
𝑎4 − 1
÷
𝑎4
+ 𝑎3
+ 𝑎2
+ 𝑎 + 1
𝑎4 − 𝑎3 + 𝑎 − 1
=
(𝑎 − 1)(𝑎4
+ 𝑎3
+ 𝑎2
+ 𝑎 + 1)
𝑎 + 1
∙
𝑎2
+ 1
(𝑎2 − 1)(𝑎2 + 1)
∙
(𝑎2
− 1)(𝑎2
− 𝑎 + 1)
𝑎4 + 𝑎3 + 𝑎2 + 𝑎 + 1
=
(𝑎 − 1)(𝑎2
− 𝑎 + 1)
𝑎 + 1

Operations on R.E. Addition and Subtraction
34
2𝑥
𝑥2 − 5𝑥 − 6
+
1
𝑥2 − 6𝑥
+
6𝑥 + 4
𝑥3 − 5𝑥2 − 6𝑥
=
2𝑥
(𝑥 − 6)(𝑥 + 1)
+
1
𝑥(𝑥 − 6)
+
6𝑥 + 4
𝑥(𝑥 − 6)(𝑥 + 1)
=
2𝑥2+ 𝑥+1 +6𝑥+4
𝑥(𝑥−6)(𝑥+1)
=
2𝑥2+7𝑥+5
𝑥(𝑥−6)(𝑥+1)
=
(2𝑥+5)(𝑥+1)
𝑥(𝑥−6)(𝑥+1)
=
2𝑥+5
𝑥(𝑥−6)

35
1
3𝑎 + 4
−
𝑎 − 7
3𝑎2 + 13𝑎 + 12
−
4
𝑎2 + 4𝑎 + 3
=
1
3𝑎 + 4
−
𝑎 − 7
3𝑎 + 4 𝑎 + 3
−
4
𝑎 + 3 𝑎 + 1
=
𝑎 + 3 𝑎 + 1 − 𝑎 − 7 𝑎 + 1 − 4(3𝑎 + 4)
(3𝑎 + 4)(𝑎 + 3)(𝑎 + 1)
=
𝑎2 + 4𝑎 + 3 − 𝑎2 − 6𝑎 − 7 − 12𝑎 − 16
(3𝑎 + 4)(𝑎 + 3)(𝑎 + 1)
=
−2𝑎 − 6
(3𝑎 + 4)(𝑎 + 3)(𝑎 + 1)
=
−2(𝑎 + 3)
(3𝑎 + 4)(𝑎 + 3)(𝑎 + 1)
=
−2
(3𝑎 + 4)(𝑎 + 1)

36
6. If
𝑎
𝑥+1
+
𝑏
(𝑥+1)2 +
𝑐
(1+𝑥)3 =
𝑥2+3𝑥+3
(𝑥+1)3 , determine the
value of a, b and c.
𝑎
𝑥+1
+
𝑏
𝑥+1 2 +
𝑐
𝑥+1 3 =
𝑥2+3𝑥+3
𝑥+1 3
𝑎 𝑥 + 1 2
+ 𝑏 𝑥 + 1 + 𝑐
𝑥 + 1 3
𝑙𝑒𝑡 (𝑥 + 1) = 𝑥
a=1 b=3 c=3

Activity 2
Operations on Rational Expression
37

MA. LIEYBEEM M. VERGARA (Danao City Division)

TERMINAL OBJECTIVES
At the end of this virtual seminar, the Grade 8
Mathematics teachers will be able to teach
competently the most critical content in Patterns and
Algebra 1.
40

ENABLING OBJECTIVES
• Develop the ability to approach and solve problems
involving linear equations;
• Create problems/activities/tasks that promote
problem solving and critical thinking skills; and
• Appreciate the importance of graphing applications
like “desmos” and “geogebra” in solving the least
learned competencies.
41

42

A Linear Equation is an equation in which the highest
power of the variable is always 1 and is also called as first
degree equation.
The graph of linear equation is always a straight line.
43
Linear Equation

Forms of Linear Equation
a. 𝑨𝒙 + 𝑩𝒚 + 𝑪 = 𝟎 General Form
b. 𝑨𝒙 + 𝑩𝒚 = 𝑪 Standard Form
c. 𝒚 = 𝒎𝒙 + 𝒃 Slope-intercept Form
d. 𝒚 − 𝒚𝟏 = 𝒎(𝒙 − 𝒙𝟏) Point-slope Form
e.
𝒙
𝒂
+
𝒚
𝒃
= 𝟏 Intercept Form
f. 𝒙 = 𝒂 Vertical
g. 𝒚 = 𝒃 Horizontal
44

Linear Equation Miscellaneous Problems
45
1. What is the equation of the line through
(– 2, 6) with x-intercept thrice the y-
intercept.
Solution:
 Let b--- be the y-intercept
 (𝑜, 𝑏) and (3𝑏, 0) are points on the line
 𝑚 =
0−𝑏
3𝑏−0
=
−𝑏
3𝑏
= −
𝟏
𝟑
 Point slope form
𝑦 − 6 = −
1
3
𝑥 + 2
3𝑦 − 18 = −𝑥 + 2
𝒚 = −
𝟏
𝟑
𝒙 +
𝟏𝟔
𝟑

46
2. Give the slope-intercept form of the line
whose y-intercept is twice the x-intercept
and is passing through (2, – 3).
Solution:
 Let a--- be the x-intercept
 (𝑎, 0) and (0,2𝑎) are points on the line
 𝑚 =
2𝑎−0
0−𝑎
=
2𝑎
−𝑎
= −𝟐
𝑦 + 3 = −2 𝑥 − 2
𝑦 = −2𝑥 − 4 + 3
𝒚 = −𝟐𝒙 + 𝟏

47
3. Find the equation of the line intersecting
the line 2𝑦 − 5𝑥 = 11 at its y-intercept
such that these two lines are
perpendicular.
2𝑦 − 5𝑥 = 11 Find the y-intercept
𝑦 =
5
2
𝑥 +
11
2
5𝑥 + 2𝑦 = 𝑐
Slope of a line
perpendicular
to the given line
𝒎 = −
𝟐
𝟓
Use point (0,
11
2
)
5
11
2
+ 2(0) = 𝑐
55
2
= 𝑐
5𝑥 + 2𝑦 =
55
2
2
𝟏𝟎𝒙 + 𝟒𝒚 = 𝟓𝟓

48
4. A right triangle has its right angle at (– 4, 1)
and the equation of one of its legs is
2𝑥 − 3𝑦 + 11 = 0. Find the equation of the
other leg.
3𝑦 = −2𝑥 − 11
𝒚 =
𝟐
𝟑
𝒙 +
𝟏𝟏
𝟑
1 = −
3
2
−4 + 𝑏
𝟏 = 𝟔 + 𝒃
𝒃 = −𝟓
𝒚 = −
𝟑
𝟐
𝒙 − 𝟓
Plot the point (-4,1)

3𝑥 − 𝑘𝑦 = −5
49
5. Find the values of k and m so that the system of
linear equation
3𝑥 − 𝑘𝑦 = −5
7𝑦 − 4𝑥 = 𝑚
has (– 2, 1) as the only solution.
Substitute (−2, 1)
3 −2 − 𝑘 1 = −5
𝑘 = −1
7𝑦 − 4𝑥 = 𝑚
7 1 − 4 −2 = 𝑚
𝑚 = 15

51
3𝑥 + 𝑦 = −2
2𝑥 − 𝑦 = −3
5𝑥 = −5
𝒙 = −𝟏
3𝑥 + 𝑦 = −2
3(1) + 𝑦 = −2
𝒚 = 𝟏
𝑥 + 𝑎𝑦 − 3 = 0
−1 + 𝑎(1) − 3 = 0
𝒂 = 𝟒

52
Find the midpoint of
two points (A,B)
𝐴, 𝐵 =
−2 + 4
2
,
−1 + 3
2
= (1,1)
Solve for the slope
𝑚 𝐴, 𝐵 =
3 − (−1)
4 − (−2)
=
4
6
=
2
3
Substitute using point slope form
𝑦 − 1 = −
3
2
(𝑥 − 1)
𝒎 ⊥= −
𝟑
𝟐
3𝑥 + 2𝑦 − 5 = 0
2𝑦 − 2 = −3𝑥 + 3

54

55
9. Find the equation of the line through
5
2
, 3 that forms the area of 15
sq. units with the axes.

THANKYOU
CHERRY S. CALACAT (Tagbilaran City Division)
MA. LIEYBEEM M. VERGARA (Danao City Division)

SD_Patterns and Algebra 1_Calacat_Vergara.pptx

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