Mathematics 8 Worksheet Algebra 1st Quarter Factoring Differences of Squares
1. Mathematics 8 Worksheet Algebra 1st
Quarter Mr. Carlo Justino J. Luna
Republic of the Philippines | DEPARTMENT OF EDUCATION
Region III | Division of City Schools | West District
Tamarind St., Clarkview Subd., Malabanias, Angeles City
School Year 2018-2019
NAME: ________________________________________ SCORE: __________________
GRADE & SECTION: _____________________________ DATE: ____________________
1st Quarter Worksheet #7
Worksheet on factoring the difference of two squares will help us to factorize an algebraic expression
using the following identity: π π
β π π
= (π + π)(π β π)
Find the factors of each expression.
1. π₯2
β 25
2. 4π¦2
β 9
3. 9π2
β 1
4. 36π2
β 1
5. 4π2
β 49
6. π2
β 36
7. 16π2
β 25
8. 25π2
β 4
9. 9π2
β 100
10.49π2
β 81
11.4π₯2
β 9
12.25π¦2
β π§2
13. π2
β 25π2
14.16π2
β 81π2
15.25π₯6
β 16π¦8
16.36π4
β 9π6
17. π10
β 4π8
18.2π₯2
β 18
19.3π2
β 75
20.8π¦2
β 50
2. Mathematics 8 Worksheet Algebra 1st
Quarter Mr. Carlo Justino J. Luna
Republic of the Philippines | DEPARTMENT OF EDUCATION
Region III | Division of City Schools | West District
Tamarind St., Clarkview Subd., Malabanias, Angeles City
School Year 2018-2019
NAME: ________________________________________ SCORE: __________________
GRADE & SECTION: _____________________________ DATE: ____________________
1st Quarter Worksheet #7
Worksheet on factoring the difference of two squares will help us to factorize an algebraic expression
using the following identity: π π
β π π
= (π + π)(π β π)
Find the factors of each expression.
1. π₯2
β 25
= (π₯ + 5)(π₯ β 5)
2. 4π¦2
β 9
= (2π¦ + 3)(2π¦ β 3)
3. 9π2
β 1
= (3π + 1)(3π β 1)
4. 36π2
β 1
= (6π + 1)(6π β 1)
5. 4π2
β 49
= (2π + 7)(2π β 7)
6. π2
β 36
= (π + 6)(π β 6)
7. 16π2
β 25
= (4π + 5)(4π β 5)
8. 25π2
β 4
= (5π + 2)(5π β 2)
9. 9π2
β 100
= (3π + 10)(3π β 10)
10.49π2
β 81
= (7π + 9)(7π β 9)
11.4π₯2
β 9
= (2π₯ + 3)(2π₯ β 3)
12.25π¦2
β π§2
= (5π¦ + π§)(5π¦ β π§)
13. π2
β 25π2
= (π + 5π)(π β 5π)
14.16π2
β 81π2
= (4π + 9π)(4π β 9π)
15.25π₯6
β 16π¦8
= (5π₯3
+ 4π¦4
)(5π₯3
β 4π¦4
)
16.36π4
β 9π6
= (6π2
+ 3π3)(6π2
β 3π3
)
17. π10
β 4π8
= (π5
+ 2π4
)(π5
β 2π4
)
18.2π₯2
β 18
= 2(π₯2
β 9)
= 2(π₯ + 3)(π₯ β 3)
19.3π2
β 75
= 3(π2
β 25)
= 3(π + 5)(π β 5)
20.8π¦2
β 32
= 2(4π¦2
β 16)
= 2(2π¦ + 4)(2π¦ β 4)