Class 8 RD Sharma Solutions - Chapter 7 Factorization - Exercise 7.3

Exercise 7.3 in Chapter 7 of RD Sharma's Class 8 Mathematics textbook focuses on advanced factorization techniques. This exercise builds upon the foundational factoring methods introduced earlier and challenges students with more complex algebraic expressions. It typically covers factoring cubic expressions, sum and difference of cubes, expressions with multiple variables, and special patterns. These problems are designed to enhance students' algebraic manipulation skills and prepare them for more advanced mathematical concepts.

Factorize each of the following:

Question 1. 6x(2x - y) + 7y(2x - y)

Solution: 

As we know that (2x - y) is a common factor in the above expression.

So we will take out it and the expression can be written like,

(2x - y)(6x + 7y) 

Question 2. 2r(y - x) + s(x - y)

Solution:

We know that 

(y - x) = -(x - y) or (x - y) = -(y - x)

So we will write the above expression like

2r(y - x) - s(y - x)

Taking (y - x) as common from this expression

(2r - s)(y - x)

Question 3. 7a(2x - 3) + 3b(2x - 3)

Solution:

We know that in this expression (2x - 3) is common factor, 

So we take common from it.

(2x - 3)(7a + 3b)

Question 4. 9a(6a - 5b) - 12a² (6a - 5b)

Solution:

Taking (6a - 5b) common from the above expression

(9a - 12a²)(6a - 5b)

Now taking a common from it

9a - 12a²

=> 3 * 3a - 4 * 3a

=> 3a(3 - 4a)

Now writing 3a(3 - 4a) instead of (9a - 12a²)

3a(3 - 4a)(6a - 5b)

Question 5. 5(x - 2y)² + 3(x - 2y)

Solution:

Taking (x - 2y) common from the above expression

(x - 2y)[5(x - 2y) + 3]

=>(x - 2y)(5x - 10y + 3)

Question 6. 16(2l - 3m)² - 12(3m - 2l)

Solution: 

Taking  2l - 3m = -(3m - 2l) or (3m - 2l) = -(2l - 3m)

So above expression can be written as

 => 16(2l - 3m)² + 12(2l - 3m)

Taking common (2l - 3m)

=> (2l - 3m)[16(2l - 3m) + 12]

=> (2l - 3m){4[4(2l - 3m) + 3}

=> 4(2l - 3m)(8l - 12m + 3)

Question 7. 3a(x - 2y) - b(x - 2y)

Solution:

Taking (x - 2y) common from above expression

=> (x - 2y)(3a - b)

Question 8. a²(x + y) + b²(x + y) + c²(x + y)

Solution:

Taking (x + y) common from above expression

=>(x + y)[ a² + b² + c²]

Question 9. (x - y)² + (x - y)

Solution: 

This can be written as

=>(x - y)(x - y) + (x - y)

Taking (x - y) common from the above expression

=> (x - y)[x - y + 1]

Question 10. 6(a + 2b) - 4(a + 2b)²

Solution:

Taking (a + 2b) from the above expression

=> (a + 2b)[6 - 4(a + 2b)]

=> Again taking 2 as a common factor

=> 2(a + 2b)[3 - 2(a + 2b)]

=> 2(a + 2b)(3 - 2a - 4b)

Question 11. a(x - y) + 2b(y - x) + c(x - y)²

Solution:

We can write

y - x = -(x - y)

then

=> a(x - y) - 2b(x - y) + c(x - y)²

Taking (x - y) from above expression

=> (x - y)(a - 2b + c(x - y))

=> (x - y)(a - 2b + cx - cy)

Question 12. -4(x - 2y)² + 8(x - 2y)

Solution: 

Taking 4(x - 2y) as a common factor from expression

=> 4(x - 2y)[-(x - 2y) + 2]

=> 4(x - 2y)[2 - x + 2y]

Question 13. x³(a - 2b) + x²(a - 2b)

Solution: 

Taking x²(a - 2b) as common factor from above expression

=> x²(a - 2b)(x + 1)

Question 14. (2x - 3y)(a + b) + (3x - 2y)(a + b)

Solution: 

Taking (a + b) as a common factor from expression

(a + b)(2x - 3y + 3x - 2y)

(a + b)(5x - 5y)

Taking 5 as common factor from (5x - 5y)

=> 5(a + b)(x - y)

Question 15. 4(x + y)(3a - b) + 6(x + y)(2b - 3a)

Solution: 

Taking 2(x + y) as a common factor

=> 2(x + y)[2(3a - b) + 3(2b - 3a)]

=> 2(x + y)[6a - 2b + 6b - 9a]

=> 2(x + y)(4b - 3a)

Summary

Exercise 7.3 in Chapter 7 of RD Sharma's Class 8 Mathematics textbook delves into advanced factorization techniques, building upon earlier concepts to challenge students with more complex algebraic expressions. This exercise typically covers a range of topics including factoring cubic expressions (x³ ± y³), applying the formulas for sum and difference of cubes, handling expressions with multiple variables, and recognizing special factorization patterns. Students encounter problems that require them to factor higher degree polynomials, such as fourth-degree expressions, and to work with more intricate algebraic structures. The primary goal of this exercise is to enhance students' algebraic manipulation skills, deepen their understanding of polynomial structures, and prepare them for more advanced mathematical concepts they will encounter in higher grades. By mastering these advanced factorization techniques, students develop crucial problem-solving skills and gain a stronger foundation for future studies in algebra, calculus, and other areas of mathematics where factorization plays a key role.