Factoring Polynomials in Modular Approach
7 likes935 views
The document provides lessons on factoring polynomials with common monomial factors and the difference of two squares. It explains that to factor polynomials with common monomial factors, one must find the greatest common factor (GCF) and divide each term by the GCF. It also explains that to factor the difference of two squares, one must get the square root of each term and use those roots to form the sum and difference as factors. Examples are provided to demonstrate these factoring techniques step-by-step. The document aims to teach learners how to completely factor different types of polynomials.
![Factor each expression completely.
1. 𝒂 𝟑
+ 𝟖
2. 𝒎 𝟑
− 𝟐𝟕
Observe the steps in factoring the sum and
difference of two cubes
Step 1
Get the cube root of
each term.
First Term: √𝒂 𝟑𝟑
= a
Second Term: √𝟖
𝟑
= 2
Step 2
Using the two terms a
and 2, form the sum
( a + 2 ) obtain a
binomial.
Step 3
(𝒂 𝟑
+ 𝟖) = (a + 2 ) [𝒂 𝟐
− (2)(a) + (𝟐) 𝟐
]
(𝒂 𝟑
+ 𝟖) = (a + 2 ) (𝒂 𝟐
– 2a + 4 )
1st
term
2nd
term Square the 1st
term
Square
the 2nd
term
Product of
1st
& 2nd
term
Step 1
Get the cube root of
each term.
First Term: √𝒎 𝟑𝟑
= ___
Second Term: √𝟐𝟕
𝟑
= __
Step 2
Using the two terms
___ and ___, form the sum
_________ obtain a
binomial.
Step 3
(𝒎 𝟑
+ 𝟐𝟕) = (m - __ ) [𝒎 𝟐
+ (__)(__) + (__) 𝟐
]
(𝒎 𝟑
+ 𝟐𝟕) = (m - __ ) (𝒎 𝟐
+ ___ + 9 )
1st
term
2nd
term Square the 1st
term Product of
1st
& 2nd
term
Square
the 2nd
term](https://image.slidesharecdn.com/factoringpolynomials-200711051151/85/Factoring-Polynomials-in-Modular-Approach-12-320.jpg)
![3. 𝒑 𝟗
+ 𝟔𝟒
4. 𝒓 𝟔
− 𝒔 𝟏𝟐
5. 𝟖 𝒕 𝟔
+ 𝒗 𝟏𝟐
𝒘 𝟏𝟖
Step 2
Step 3
Step 1
Step 1 Step 2
Step 3
To get the
cube root of a
variable,
divide the
exponent into
3. Example:.
√ 𝒑 𝟐𝟏𝟑
= 𝒑 𝟕
Step 1
Get the cube root of each term.
First Term: √ 𝟖 𝒕 𝟔𝟑
= 2𝒕 𝟐
Second Term: √𝒗 𝟏𝟐 𝒘 𝟏𝟖𝟑
= 𝒗 𝟒
𝒘 𝟔
Step 2
Using the two terms 𝒕 𝟐
and 𝒗 𝟒
𝒘 𝟔
, form the sum
𝒕 𝟐
+ 𝒗 𝟒
𝒘 𝟔
obtain a
binomial.
Step 3
𝟖 𝒕 𝟔
+ 𝒗 𝟏𝟐
𝒘 𝟏𝟖
= (𝟐𝒕 𝟐
+ 𝒗 𝟒
𝒘 𝟔
) [(𝟐𝒕 𝟐
) 𝟐
−( 𝒗 𝟒
𝒘 𝟔
)( 𝟐𝒕 𝟐
) + (𝒗 𝟒
𝒘 𝟔
) 𝟐
]
𝟖 𝒕 𝟔
+ 𝒗 𝟏𝟐
𝒘 𝟏𝟖
= ( 𝟐𝒕 𝟐
+ 𝒗 𝟒
𝒘 𝟔
) (𝟒𝒕 𝟒
- 2 𝒕 𝟐
+ 𝒗 𝟖
𝒘 𝟏𝟐
)
1st
term
2nd
term Square the 1st
term
Squa
re
the
2nd
term
Product of 1st
& 2nd
term](https://image.slidesharecdn.com/factoringpolynomials-200711051151/85/Factoring-Polynomials-in-Modular-Approach-13-320.jpg)
Factoring Polynomials in Modular Approach
- 1. Math Explorer: With Scaffolding Approach toward Independent Learning
- 2. Factoring is the reverse process of multiplication. It is the process of finding the factors of an expression. A polynomial is factored completely if each of its factors can no longer be expressed as product of two other polynomials. Here is the map of the lessons that will be covered in this module. Factoring PolynomialsModule 1 At the end of this module, the learners will: 1. factor completely different types of polynomials 2. find factors of product of polynomials 3. solve problems involving factors of polynomials. Factoring Polynomials with Common Monomial Factor Difference of Two Squares Sum and Difference of Two Cubes Perfect Square Trinomials General Trinomials
- 3. Module 1 The GCF of these three monomials is (3)(x)(x) = 3 𝒙 𝟐 Common monomial factoring is the process of writing a polynomial as a product of two polynomials, one of which is a monomial that factors each term of the polynomial. To ensure that the polynomial is the prime polynomial, use the Greatest Common Factor (GCF) of the terms of the given polynomials. 1. Find the GCF of 4𝒎 𝟐 and 𝟏𝟎𝒎 𝟒 . Solution: Express each as a product of prime factors. 4𝒎 𝟐 = 2 (2) m m 𝟏𝟎𝒎 𝟒 = (5) 2 m m (m) (m) Common Factors = 2 m m The GCF of these two monomials is (2)(m)(m) = 𝟒𝒎 𝟐 2. Find the GCF of 𝟔𝒙 𝟒 , 9𝒙 𝟐 y, and 𝟏𝟓𝒙 𝟓 𝒚. Solution: Express each as a product of prime factors. 𝟔𝒙 𝟒 = (2 ) 3 x x (x) (x) 9𝒙 𝟐 y = (3) 3 x x (y) 𝟏𝟓𝒙 𝟓 𝒚 = (5) 3 x x (x) (x) (x) (y) Common Factors = 3 x x Lesson 1 Polynomial with Common Monomial Factor Observe and analyze below on how to get the GCF and answer number 3 and 4 .
- 4. 3. Find the GCF of 8𝒚 𝟔 and 𝟏𝟎𝒚 𝟒 z. Solution: Express each as a product of prime factors. 8𝒚 𝟔 = 𝟏𝟎𝒚 𝟒 z = Common Factors = 4. Find the GCF of 𝟐𝟎𝒃 𝟒 𝒄 , 5𝒃 𝟐 c, and 𝟏𝟓𝒃 𝟑 𝒄. Solution: Express each as a product of prime factors. 𝟐𝟎𝒃 𝟒 𝒄 = 5𝒃 𝟐 c = 𝟏𝟓𝒃 𝟑 𝒄 = Common Factors = Math Guro GCF – The greatest common factor (GCF) of a set of numbers is the largest factor that all the numbers or variables share. Polynomial – an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Example: 2x +3y – 4xy Monomial – an expression in algebra that contains one term, like 3xy Factors – a number or algebraic expression that divides another number or expression evenly with no remainder. Ex. The factors of 6 is 2 and 3 because (2) (3)= 9.
- 5. The property of exponents stated 𝑎 𝑚 𝑎 𝑛 = 𝑎 𝑚−𝑛 1. ) 𝑥5 𝑥3 = 𝑥5−3 = 𝑥2 2. ) 𝑝7 𝑝4 = 3. ) 105 103 = 105−3 = 102 = 100 4. ) 1525 1523 = 5. ) 45 𝑥10 42 𝑥2 = 45−2 𝑥10−2 = 43 𝑥8 = (4 ∗ 4 ∗ 4 ) 𝑥8 = 64 𝑥8 6.) 36 𝑏17 32 𝑏12 = = = = A fraction is reduced to its lowest terms, when its numerator and denominator have no common factors. Observe the steps: 1. List the prime factors of the numerator and denominator. 2. Find the factors common to both the numerator and denominator. 3. Divide the numerator and denominator by all common factors called CANCELLING. Example 1: To factor polynomials w/ common monomial factor, you should know how to find: 1.GCF 2. Quotient of two monomials 3. Fractions to lowest term
- 6. Now, you are ready to find the factor of polynomials with common monomial. Observe and analyze the steps: 1. Find the greatest common factor (GCF) of the terms in the polynomial. This is the first factor. 2. Divide each term by the GCF to get the other factor. Factor each expression. 1. 𝟏𝟎𝒚 𝟒 + 5𝒚 𝟑 2. 𝟏𝟓𝒎 𝟓 + 5𝒎 𝟐 3. 𝟐𝟓𝒃 𝟑 𝒄 𝟐 − 5𝒃 𝟐 𝒄 Step 1 : Find the GCF. 𝟏𝟎𝒚 𝟒 = (5) (2) (y) (y) (y) (y) 5𝐲 𝟑= (5 ) (y) (y) (y) GCF = 5 y y y GCF = 𝟓𝒚 𝟑 Step 2: Divide out the common factor. 𝟏𝟎𝒚 𝟒 + 5𝒚 𝟑 = 𝟓𝒚 𝟑 ( 𝟏𝟎𝒚 𝟒 𝟓𝒚 𝟑 + 𝟓𝒚 𝟑 𝟓𝒚 𝟑 ) = 𝟓𝒚 𝟑 (𝟐𝒚 𝟒−𝟑 + 𝟏 = 𝟓𝒚 𝟑 (𝟐𝒚 + 1 ) Step 1 Find the GCF. 𝟏𝟓𝒎 𝟓 = 5𝐦 𝟐 = GCF = GCF = Step 2 Divide out the common factor. 𝟏𝟓𝒎 𝟓 + 5𝒎 𝟐 = = = Step 1 Find the GCF. 𝟐𝟓𝒃 𝟑 𝒄 𝟐 = (5)(5)(b)(b)(b)(c)(c) 𝟓𝒃 𝟐 𝒄 = (3) (5)(b)(b) (c) GCF = (5)(b)(b) (c) GCF = 5𝒃 𝟐 𝒄 Step 2 Divide out the common factor. 𝟐𝟓𝒃 𝟑 𝒄 𝟐 − 5𝒃 𝟐 𝒄 = 5𝒃 𝟐 𝒄 ( 𝟐𝟓𝒃 𝟑 𝒄 𝟐 𝟓𝒃 𝟐 𝒄 − 𝟓𝒃 𝟐 𝒄 𝟓𝒃 𝟐 𝒄 ) = 5𝒃 𝟐 𝒄 ( 5𝒃 𝟑−𝟐 𝒄 𝟐−𝟏 – 1) =5𝒃 𝟐 𝒄 ( 5b – 1 ) Any number or variables divided by itself is always equal to 1. All variables always have imaginary exponent one.
- 7. 𝟒. 𝟏𝟐𝒉 𝟒 𝒊 𝟐 − 15𝒉 𝟐 𝒊 Factor each expression. Show your solution with the step-by-step process. 𝟏. 𝟖𝒈 𝟔 + 16𝒈 𝟐 4. 𝟔𝒂 𝟐 𝒃 + 𝟏𝟖𝒂𝒃 𝟐. 𝟒𝟗𝒚 𝟗 + 21𝒚 𝟐 5. 𝟐𝟒𝒑 𝟏𝟗 𝒒 𝟏𝟐 + 𝟒𝒑 𝟏𝟓 𝒒 𝟕 𝟑. 𝟐𝒙 𝟐 𝒚 𝟐 − 14y 6. 𝟑𝟎𝒓 𝟔 + 𝟏𝟎𝒓 𝟐 References Diaz, Z., Mojica M. (2013) . Next Century Mathematics 8; Quezon City ; Phoenix Publishing House , Inc Mathematics 8 Learner’s Module K-12; DepEd K-12 Modified Curriculum Guide and Teacher’s Guide for Mathematics 8 http://www.math.com/school/subject1/lessons/S1U4L2DP.html https://quickmath.com/math-tutorials/common-monomial-factors- factoring-special-products-and.html Skill Booster! Step 1 Find the GCF. 𝟏𝟐𝒉 𝟒 𝒊 𝟐 = 15𝒉 𝟐 𝒊 = GCF = GCF = Step 2 Divide out the common factor. 𝟏𝟐𝒉 𝟒 𝒊 𝟐 − 15𝒉 𝟐 𝒊 = = =
- 8. 𝟐. √𝟒𝟗 = 7 since 7 * 7 = 49 𝟑. √𝟖𝟏 = 𝟒. √𝟏𝟐𝟏 = 𝟓. √𝟔𝟒 = 𝟔. √𝒙 𝟐 = x since 𝑥2÷2 = 𝑥 𝟕. √𝒎 𝟔 = 𝑚3 since 𝑚6÷2 = 𝑚3 𝟖. √𝒓 𝟒 = 𝟗. √𝟒𝒗 𝟐 = 𝟏𝟎. √𝟗𝒗 𝟐 = 3v since √𝟗 * √𝒗 𝟐 = 3 * v = 3v 𝟏𝟏. √𝟏𝟔𝒌 𝟔 = 4𝒌 𝟑 Why? 𝟏𝟐. √𝟒𝒚 𝟖 = Factoring the difference of two squares is one of the common types of factoring a problem that is very easy to identify. The factors of the difference between two squares are the product of the sum and difference of two numbers. To factor the Difference of Two Squares: 1. Get the principal square root of each squares. 2. Using these square roots, form two factors, one a sum and the other a difference. But first, let us have a short review in getting the square root of a number and variable. Observe and analyze the pattern and method, then answer the following. 1. √𝟐𝟓 = 5 since 5 *5 = 25 Lesson 2 Factors of Difference of Two Squares Verify first if the two terms are both perfect squares. 𝒂 𝟐 - 𝒃 𝟐 = ( a + b ) ( a – b ) A variable is perfect square if its exponent is even number.
- 9. You can apply factoring difference of two squares, if and only if, The two terms are both perfect squares The operation used is subtraction Factor each expression carefully. 1. 𝐱 𝟐 - 25 2. 𝐦 𝟐 - 81 3. 𝐛 𝟐 - 4𝐜 𝟐 4. 𝐝 𝟐 - 16𝐞 𝟐 Step 1 Get the square root of each term. First Term: √𝐱 𝟐 = x Second Term: √𝟐𝟓 = 5 Step 2 Using x and 5 , form the sum ( x+ 5) and the difference (x – 5). Thus, 𝐱 𝟐 - 25 = (x+ 5) (x+ 5) Step 1 Step 2 Step 1 Get the square root of each term. First Term: √𝐛 𝟐 = b Second Term: √𝟒𝐜 𝟐 = 2c Step 2 Using b and 2c, form the sum (b + 2c) & the difference (b - 2c). Thus, 𝐛 𝟐 - 4𝐜 𝟐 = (b + 2c) (b – 2c). Step 1 Step 2
- 10. 5. 𝟑𝟔𝐟 𝟐 - 400𝐠 𝟖 𝟔. 𝟏𝟎𝟎𝐫 𝟐 - 143𝐬 𝟒 Factor each expression completely. 𝟏. 𝐚 𝟐 - 1 4. 𝐝 𝟏𝟐 - 4𝐜 𝟐 6. 𝟐. 𝟗𝐩 𝟐 - 81 5. 𝐪 𝟐 - 225𝐫 𝟒 𝟑. 𝐰 𝟒 - 169 6. 𝟏𝟎𝟎𝐫 𝟐 - 289𝐬 𝟏𝟎 References Diaz, Z., Mojica M. (2013) . Next Century Mathematics 8; Quezon City ; Phoenix Publishing House , Inc Mathematics 8 Learner’s Module K-12; DepEd K-12 Modified Curriculum Guide and Teacher’s Guide for Mathematics 8 http://www.themathpage.com/Alg/difference-two- squares.htm Step 1 Get the square root of each term. First Term: √𝟑𝟔𝐟 𝟐 = 36f Second Term: √𝟒𝟎𝟎𝐠 𝟖 = 20𝐠 𝟒 Step 2 Using 36f and 𝟐𝟎𝒈 𝟒 form the sum (36f + 𝟐𝟎𝒈 𝟒 ) and the difference (36f - 𝟐𝟎𝒈 𝟒 ) . Thus, 𝟑𝟔𝐟 𝟐 - 400𝐠 𝟖 = (36f + 𝟐𝟎𝒈 𝟒 ) (36f - 𝟐𝟎𝒈 𝟒 ) Step 1 Step 2 Skill Booster!
- 11. √27 3 = 3 since 3*3*3 = 27 √8 3 = √𝑑153 = 𝑑5 since 𝑑15 ÷3 Type equation here. √64𝑥33 = 8x √𝑛63 = √125𝑦6 𝑧63 = The sum and difference of two cubes is a special case in factoring polynomials. A polynomial in the form a 3 + b 3 is called a sum of cubes. A polynomial in the form a 3 – b 3 is called a difference of cubes. Both of these polynomials have similar factored patterns: But first, let us have a short review in getting the cube root of a number and variable. Observe and analyze the pattern and method, then answer the following. Lesson 3 Factors of Sum and Difference Two Cubes A Sum of Cubes A Difference of Cubes A variable is a perfect cube if its exponent is divisible by 3. To get the cube root, just divide the exponent by 3. i.e. a12 is a perfect cube, its cube root is a4.
- 12. Factor each expression completely. 1. 𝒂 𝟑 + 𝟖 2. 𝒎 𝟑 − 𝟐𝟕 Observe the steps in factoring the sum and difference of two cubes Step 1 Get the cube root of each term. First Term: √𝒂 𝟑𝟑 = a Second Term: √𝟖 𝟑 = 2 Step 2 Using the two terms a and 2, form the sum ( a + 2 ) obtain a binomial. Step 3 (𝒂 𝟑 + 𝟖) = (a + 2 ) [𝒂 𝟐 − (2)(a) + (𝟐) 𝟐 ] (𝒂 𝟑 + 𝟖) = (a + 2 ) (𝒂 𝟐 – 2a + 4 ) 1st term 2nd term Square the 1st term Square the 2nd term Product of 1st & 2nd term Step 1 Get the cube root of each term. First Term: √𝒎 𝟑𝟑 = ___ Second Term: √𝟐𝟕 𝟑 = __ Step 2 Using the two terms ___ and ___, form the sum _________ obtain a binomial. Step 3 (𝒎 𝟑 + 𝟐𝟕) = (m - __ ) [𝒎 𝟐 + (__)(__) + (__) 𝟐 ] (𝒎 𝟑 + 𝟐𝟕) = (m - __ ) (𝒎 𝟐 + ___ + 9 ) 1st term 2nd term Square the 1st term Product of 1st & 2nd term Square the 2nd term
- 13. 3. 𝒑 𝟗 + 𝟔𝟒 4. 𝒓 𝟔 − 𝒔 𝟏𝟐 5. 𝟖 𝒕 𝟔 + 𝒗 𝟏𝟐 𝒘 𝟏𝟖 Step 2 Step 3 Step 1 Step 1 Step 2 Step 3 To get the cube root of a variable, divide the exponent into 3. Example:. √ 𝒑 𝟐𝟏𝟑 = 𝒑 𝟕 Step 1 Get the cube root of each term. First Term: √ 𝟖 𝒕 𝟔𝟑 = 2𝒕 𝟐 Second Term: √𝒗 𝟏𝟐 𝒘 𝟏𝟖𝟑 = 𝒗 𝟒 𝒘 𝟔 Step 2 Using the two terms 𝒕 𝟐 and 𝒗 𝟒 𝒘 𝟔 , form the sum 𝒕 𝟐 + 𝒗 𝟒 𝒘 𝟔 obtain a binomial. Step 3 𝟖 𝒕 𝟔 + 𝒗 𝟏𝟐 𝒘 𝟏𝟖 = (𝟐𝒕 𝟐 + 𝒗 𝟒 𝒘 𝟔 ) [(𝟐𝒕 𝟐 ) 𝟐 −( 𝒗 𝟒 𝒘 𝟔 )( 𝟐𝒕 𝟐 ) + (𝒗 𝟒 𝒘 𝟔 ) 𝟐 ] 𝟖 𝒕 𝟔 + 𝒗 𝟏𝟐 𝒘 𝟏𝟖 = ( 𝟐𝒕 𝟐 + 𝒗 𝟒 𝒘 𝟔 ) (𝟒𝒕 𝟒 - 2 𝒕 𝟐 + 𝒗 𝟖 𝒘 𝟏𝟐 ) 1st term 2nd term Square the 1st term Squa re the 2nd term Product of 1st & 2nd term
- 14. Factor the sum and difference of two cubes. 1. 𝒅 𝟑 + 𝟏 4. 𝒘 𝟔 − 𝟔𝟒 2. 𝒇 𝟗 + 𝟖 5. 𝟖𝒓 𝟑 − 𝟏𝟐𝟓 3. 𝒑 𝟏𝟓 − 𝟐𝟕 6. 𝒄 𝟑 𝒅 𝟑 + 𝟖𝒆 𝟑 References Diaz, Z., Mojica M. (2013) . Next Century Mathematics 8; Quezon City ; Phoenix Publishing House , Inc Mathematics 8 Learner’s Module K-12; DepEd K-12 Modified Curriculum Guide and Teacher’s Guide for Mathematics 8 https://www.cliffsnotes.com/study-guides/algebra/algebra- ii/factoring-polynomials/sum-or-difference-of-cubes ; https://www.varsitytutors.com/hotmath/hotmath_help/topics/sum- and-difference-of-cubes Skill Booster!
- 15. A Perfect Square Trinomial is created when a value is multiplied by itself. It is the sum of two perfect squares and twice the product of the square roots of the squares. Factor each expression. 𝟏. 𝒎 𝟐 + 𝟖𝒎 + 𝟏𝟔 𝟐. 𝒄 𝟐 + 𝟏𝟐𝒄 + 𝟑𝟔 Lesson 4 Factoring Perfect Square Trinomial To factor a perfect square trinomial, you should know how to get the square roots of a number or variable. Examples: √9 = 3 since 3*3= 9 √𝑚2 = 𝑚 since m*m = 𝑚2 𝒂 𝟐 + 2ab + 𝒃 𝟐 = ( a + b ) ( a + b ) = (𝒂 + 𝒃 ) 𝟐 𝒂 𝟐 - 2ab + 𝒃 𝟐 = ( a - b ) ( a - b ) = (𝒂 − 𝒃 ) 𝟐 Step 1 Get the square root of each term. First Term: √𝒎 𝟐 = m Third Term: √𝟏𝟔 = 4 Step 2 Using m and 4, form the sum ( m + 4) and (m + 4) Thus, 𝑚2 + 8𝑚 + 16 = (m + 4 ) (m + 4)= (𝑚 + 4 )2 Step 1 Get the square root of each term. First Term: √𝒄 𝟐 = ___ Third Term: √𝟑𝟔 =___ Step 2 Using ___ and ___, form the sum _______ and _______. Thus, 𝑐2 + 12𝑐 + 36 = _________ ________ = (________ )2
- 16. 𝟑. 𝒓 𝟐 − 𝟏𝟔𝒓 + 𝟔𝟒 𝟒. 𝟒𝟗𝒙 𝟐 − 𝟓𝟔𝒙𝒚 + 𝟏𝟔𝒚 𝟐 𝟓. 𝟑𝟔𝒚 𝟐 − 𝟔𝟎𝒚𝒛 + 𝟐𝟓𝒛 𝟐 What happened to the −𝟓𝟔𝒙𝒚 𝒊𝒏 𝒕𝒉𝒆 𝒕𝒓𝒊𝒏𝒐𝒎𝒊𝒂𝒍 𝟒𝟗𝒙 𝟐 − 𝟓𝟔𝒙𝒚 + 𝟏𝟔𝒚 𝟐 ? Take note that the pattern you have learned is a short cut method in factoring trinomials. Remember when you solve (7x – 4y )(7x – 4y) using the FOIL Method, you will get the product of 𝟒𝟗𝒙 𝟐 − 𝟓𝟔𝒙𝒚 + 𝟏𝟔𝒚 𝟐 . This pattern can only apply if 1st and third term is both perfect square and the middle term is twice the product of square roots of first term and third term. Step 1 Get the square root of each term. First Term: √𝒓 𝟐 = r Third Term: √𝟔𝟒 = 8 You just noticed that the operation used is now subtraction because the 2nd term of the trinomial 𝒓 𝟐 − 𝟏𝟔𝒓 + 𝟔𝟒 is - 16r or in minus sign. Therefore, the operation used in binomial (𝒓 − 𝟖 ) 𝟐 is minus sign. Step 2 Using r and 8, form the difference ( r – 8 ) and (r – 8) Thus, 𝑟2 − 16𝑟 + 16 = (r - 8 ) (r - 8) = (𝑟 − 8 )2 Step 1 Get the square root of each term. First Term: Third Term: Step 2 Answer: (7𝑥 − 4𝑦 )2 Why? Step 1 Get the square root of each term. First Term: Third Term: Step 2
- 17. Factor each expression completely. 𝟏. 𝟒𝒂 𝟐 + 𝟏𝟐𝒂 + 𝟗 𝟒. 𝒖 𝟐 + 𝟐𝟎𝒑 + 𝟏𝟎𝟎 𝟐. 𝒕 𝟐 − 𝟔𝒕 − 𝟗 𝟓. 𝒉 𝟐 𝒊 𝟐 − 𝟖𝒉𝒊 + 𝟏𝟔 𝟑. 𝒌 𝟐 − 𝟏𝟎𝒌 + 𝟐𝟓 𝟔. 𝟏𝟒𝟒𝒔 𝟐 − 𝟕𝟐𝒔 + 𝟗 References Diaz, Z., Mojica M. (2013) . Next Century Mathematics 8; Quezon City ; Phoenix Publishing House , Inc Escaner, J., Sepida, M., Catalla, D. (2013)., Spiral Math 8; Quezon Ciity; Trinita Publishing , Inc Mathematics 8 Learner’s Module K-12; DepEd K-12 Modified Curriculum Guide and Teacher’s Guide for Mathematics 8 https://mathbitsnotebook.com/Algebra1/Factoring/FCPerfSqTri.html https://www.onlinemathlearning.com/perfect-square- trinomial.html Skill Booster! 𝟗𝑿 𝟐 – 42X + 49
- 18. In factoring Non-Perfect Square Trinomial, consider to determine the factors of the leading (square) term; the factors of the last (constant) term may be combined to find the coefficient of the middle term. Trinomials of the form 𝑥2 + bx + c are the products of two binomials. Observe how the last term and the middle term of the resulting trinomial are generated. To factor trinomial, reverse the FOIL process. 1. 𝒙 𝟐 + 5x + 6 = (x + ___) (x + ___) Lesson 5 Factoring Non-Perfect Square Trinomials (x + 2) (x + 3) = 𝒙 𝟐 + 3x + 2x + 6 = 𝒙 𝟐 + 5x + 6 F O I L Sum of 2 & 3 Product of 2 & 3 (x + 4) (x + 5) = ___+___+ ___+__ = ___+___+___ F O I L Sum of 2 & 3 Product of 2 & 3 Find two numbers that when you multiply the result is 6 and when you add, the result is 5. The correct pair is 2 and 3 . Therefore, the factor of 𝑥2 + 2𝑥 + 6 is ( x + 2 ) (x + 3 ).
- 19. 2. 𝒂 𝟐 + 10a + 25 = (a + ___) (a + ___) 3. 𝒎 𝟐 + 10m + 21 = (m + ___) (m + ___) 4. 𝒃 𝟐 + 14b + 45 = (b + ___) (b + ___) 𝟓. 𝒘 𝟐 + 14w+ 48 = ( w + ___ ) (w + ___) Find two numbers that when you multiply the result is 25 and when you add, the result is 10. The correct pair is 5 and 5 . Therefore, the factor of 𝑎2 + 10𝑎 + 25 is ( x + 5 ) (x + 5 ). Find two numbers that when you multiply the result is 21 and when you add, the result is 10. The correct pair is…. Put your solution here….. Put your solution here…..
- 20. 𝟔. 𝒎 𝟐 – 9m + 18 = (m - ___) (m - ___) The correct pair is -3 and -6 . Therefore, the factor of 𝒎 𝟐 – 9m + 18 is (m – 3 ) ( m – 6 ). 𝟕. 𝒒 𝟐 – 6q + 8 = (q - ___) (q - ___ ) 𝟖. 𝒔 𝟐 + 2s - 15 = ( _________ ) ( _________ ) The correct pair is 5 and -3. Therefore, the factor of 𝒔 𝟐 + 2s – 15 is ( s + 5 ) ( s – 3 ). * When you multiply two negative numbers, the result is positive. Example: (-3 )(-6) = 18 *When you add two negative numbers, the result is negative. Example: -3 + -6 = -9 Show your complete answer. * When you multiply positive & negative numbers, the result is negative. Example: (5 )(-3) = -15 *When you add positive & negative numbers, the sign used in the result will be based which absolute num. is greater. Examples: 5 + -3 = 2 We use positive 2 since the sign of the greater value is positive. Find two numbers that when you multiply the result is -15 and when you add, the result is positive 2.
- 21. 𝟗. 𝒛 𝟐 + 5z – 24 = ( _________ ) ( _________ ) 10. 𝒔 𝟐 - 7s - 45 = ( _________ ) ( _________ ) 11. 𝒏 𝟐 - 6n - 16 = ( _________ ) ( _________ ) 12. 𝒃 𝟐 – 5b - 14 = ( _________ ) ( _________ ) Show your complete answer here… Find two numbers that when you multiply the result is -45 and when you add, the result is -7. The correct pair is 7 and -6 . Therefore, the factor of 𝑠2 − 7𝑠 − 45 is ( s + ) (s - ). Find two numbers that when you multiply the result is -6 and when you add, the result is -16. The correct pair is 2 and -8 . Therefore, the factor of 𝑛2 − 6𝑛 − 16 is ( n + 2 ) (n - 8 ). Put your solution here…..
- 22. Factor each expression completely. 1. 𝒏 𝟐 + 3n + 2 𝟒. 𝒈 𝟐 + g - 12 2. 𝒕 𝟐 – 5t + 6 𝟓. 𝒘 𝟐 – 11w + 10 3. 𝒅 𝟐 - d - 6 6. 𝒆 𝟐 – 7e + 10 References Diaz, Z., Mojica M. (2013) . Next Century Mathematics 8; Quezon City ; Phoenix Publishing House , Inc Escaner, J., Sepida, M., Catalla, D. (2013)., Spiral Math 8; Quezon Ciity; Trinita Publishing , Inc Mathematics 8 Learner’s Module K-12; DepEd K-12 Modified Curriculum Guide and Teacher’s Guide for Mathematics 8 http://cortneyhand.weebly.com/teaching-integers-to-students.html http://math.ssmyrl.com/FACT4.HTM Skill Booster!
- 23. Have you wondered how the architect was able to maximize the space of building and was able to place all the stuffs the owners want? Or how a carpenter was able to create a utility box using minimal materials? This lesson will help you solve real-life problems that applied all the factoring techniques you have learned in this module. 1. The area of a rectangular pool is 𝒙 𝟐 − 𝟑𝟓𝒙 + 300. Find the length and the width of the swimming pool. 2. The area of the floor of a house miniature is 𝒙 𝟐 − 𝟏𝟓𝒙 + 50. Find the length and the width of the floor. Lesson 6 Problems Involving Factors of Polynomials To get the value of x in ( x – 20) = 0, transpose -20 to other side into its opposite sign. Like this, x = 20. From negative 20, it becomes positive 20. Given: Area = 𝑥2 − 35𝑥 + 300 Formula: (Length)(width)=Area Solution: Factor 𝑥2 − 35𝑥 + 300 = 0 (x – 20) (x -15) = 0 x – 20 = 0 x – 15 = 0 x = 20 x = 15 Conclusion: The length of the pool is 20 m and its width is 15 m.
- 24. 3. Determine the side of a square garden if its area is 𝟒𝒙 𝟐 − 𝟏𝟔𝒙 + 16. 4. Find the side of a box if its area its area is 𝟏𝟔 𝒙 𝟐 − 𝟔𝟒𝒙 + 64. M Given: Area = 4𝑥2 − 16𝑥 + 16 Formula: (side)(side)=Area Solution: Factor 4𝑥2 − 16𝑥 + 16 = 0 (2x - 4) (2x -4) = 0 2x - 4 = 0 2x - 4 = 0 2x = 4 2 x= 4 𝑥 = 4 2 x = 4 2 X = 2 x = 2 Conclusion: The side of the garden is 2 m and its width is 15 m. Math Guro Area – defined as the space occupied by a flat shape or the surface of an object Trinomial –a polynomial consisting of three terms such as 𝑥2 + 2𝑥 − 5 Binomial –a polynomial consisting of ttwo terms such as 𝑥 + 𝑦 Monomial –a polynomial consisting of one term such ex. 𝑦 Geometry –branch of Mathematics concerned with questions of shape, size, figures and properties of space Polynomial–expressions cosist of variables, coefficients that involves fundamental operations Algebra–branch of Mathematics that deals with letters and symbols
- 25. Points to Ponder Factoring Formulas Points to Ponder! Web Guro https://www.mathsisfun.com/algebra/factoring.html https://www.youtube.com/watch?v=I6TBBzIvgB8 https://www.youtube.com/watch?v=ADj8sGSjewg https://www.youtube.com/watch?v=ADj8sGSjewg https://www.youtube.com/watch?v=8TPofjGXDR4 https://www.youtube.com/watch?v=f6yhfmW41wI
- 26. A. Read and analyze each item carefully and encircle the letter of the best answer. 1. Find the factors of the expression 𝒙 𝟐 – 7x + 1. a. (x – 3)(x – 4) c. (x – 6)(x – 2) b. (x + 3)(x + 4) d. (x + 2)(x + 6) 2. (5a+ 6)2 = a. 25a2+ 36 c. 25a2+ 30a + 36 b. 25a2- 36 d. 25a2 + 60a +36 3. What is the GCF of 54 and 24? a. 3 b. 1 c. 6 d. 2 4.) What is the GCF of 𝒙 𝟒 , 𝒙 𝟐 , 𝒙 𝟕 ? a. 𝑥2 b. 𝑥3 c. 𝑥2 d. x 5.) It is an expression of more than two algebraic terms, especially the sum of several terms that contain different powers of the same variable. a. Polynomials c. binomial b. monomial d. trinomial Summative Test 1.1