Class 9 RD Sharma Solutions Chapter 5 Factorisation of Algebraic Expressions Exercise 5.1 (original) (raw)
Last Updated : 16 Sep, 2024
Chapter 5 of RD Sharma's Class 9 Mathematics textbook, specifically Exercise 5.1, delves into the critical topic of Factorisation of Algebraic Expressions. This exercise introduces students to fundamental factorisation techniques, including the common factor method, grouping method, factorisation of perfect square trinomials, difference of squares, and general trinomials. These skills are essential in algebra, forming the foundation for solving equations, simplifying complex expressions, and understanding advanced mathematical concepts. By mastering these techniques, students develop their ability to recognize patterns, manipulate algebraic expressions efficiently, and enhance their overall mathematical reasoning. The problems in this exercise are designed to progressively build students' confidence and competence, challenging them to apply the appropriate factorisation method for each given expression and achieve complete factorisation.
**Question 1: Factorize x 3 + x – 3x 2 – 3
**Solution:
x3 + x – 3x2 – 3
Here x is common factor in x3 + x and – 3 is common factor in – 3x2 – 3
x3 – 3x2 + x – 3
x2 (x – 3) + 1(x – 3)
Taking (x – 3) common
(x – 3) (x2 + 1)
Therefore, x3 + x – 3x2 – 3 = (x – 3) (x2 + 1)
**Question 2: Factorize a(a + b) 3 – 3a 2 b(a + b)
**Solution:
a(a + b)3 – 3a2b(a + b)
Taking (a + b) as common factor
= a(a + b) {(a + b)2 – 3ab}
= a(a + b) {a2 + b2 + 2ab – 3ab}
= a(a + b) (a2 + b2 – ab)
Question 3: Factorize x(x 3 – y3**) + 3xy(x – y)**
**Solution:
x(x3 – y3) + 3xy(x – y)
= x(x – y) (x2 + xy + y2) + 3xy(x – y)
Taking x(x – y) as a common factor
= x(x – y) (x2 + xy + y2 + 3y)
= x(x – y) (x2 + xy + y2 + 3y)
**Question 4: Factorize a 2 x 2 + (ax 2 + 1)x + a
**Solution:
a2x2 + (ax2 + 1)x + a
= a2x2 + a + (ax2 + 1)x
= a(ax2 + 1) + x(ax2 + 1)
= (ax2 + 1) (a + x)
**Question 5: Factorize x2 + y – xy – x
**Solution:
x2 + y – xy – x
= x2 – x – xy + y
= x(x - 1) – y(x – 1)
= (x – 1) (x – y)
**Question 6: Factorize x 3 – 2x2y + 3xy2 – 6y 3
**Solution:
x3 – 2x2y + 3xy2 – 6y3
= x2(x – 2y) + 3y2(x – 2y)
= (x – 2y) (x2 + 3y2)
**Question 7: Factorize 6ab – b2 + 12ac – 2bc
**Solution:
6ab – b2 + 12ac – 2bc
= 6ab + 12ac – b2 – 2bc
Taking 6a common from first two terms and –b from last two terms
= 6a(b + 2c) – b(b + 2c)
Taking (b + 2c) common factor
Question 8: Factorize (x2 + 1/x2****) – 4(x + 1/x) + 6
**Solution:
(x2 + 1/x2) – 4(x + 1/x) + 6
= x2 + 1/x2 – 4x – 4/x + 4 + 2
= x2 + 1/x2 + 4 + 2 – 4/x – 4x
= (x2) + (1/x)2+ (-2)2 + 2x(1/x) + 2(1/x)(-2) + 2(-2)x
As we know, x2 + y2 + z2 + 2xy + 2yz + 2zx = (x + y + z)2
So, we can write;
= (x + 1/x + (-2))2
or (x + 1/x – 2)2
Therefore, x2 + 1/x2) – 4(x + 1/x) + 6 = (x + 1/x – 2)2
**Question 9: Factorize x(x – 2) (x – 4) + 4x – 8
**Solution:
x(x – 2) (x – 4) + 4x – 8
= x(x – 2) (x – 4) + 4(x – 2)
= (x – 2) [x(x – 4) + 4]
= (x – 2) (x2 – 4x + 4)
= (x – 2) [x2 – 2 (x)(2) + (2)2]
= (x – 2) (x – 2)2
= (x – 2)3
**Question 10: Factorize (x + 2) (x2 + 25) – 10x2 – 20x
**Solution:
(x + 2) (x2 + 25) – 10x(x + 2)
Take (x + 2) as common factor;
= (x + 2)(x2 + 25 – 10x)
= (x + 2) (x2 – 10x + 25)
Expanding the middle term of (x2 – 10x + 25)
= (x + 2) (x2 – 5x – 5x + 25)
= (x + 2){x (x – 5) – 5 (x – 5)}
= (x + 2)(x – 5)(x – 5)
= (x + 2)(x – 5)2
Therefore, (x + 2) (x2 + 25) – 10x (x + 2) = (x + 2)(x – 5)2
**Question 11: Factorize 2a2 + 2√6 ab + 3b2
**Solution:
2a2 + 2√6 ab + 3b2
Above expression can be written as (√2a)2 + 2 × √2a × √3b + (√3b)2
As we know, (p + q)2 = p2 + q2 + 2pq
Here p = √2a and q = √3b
= (√2a + √3b)2
Therefore, 2a2 + 2√6 ab + 3b2 = (√2a + √3b)2
****Question 12: Factorize (a – b + c)**2 **+ (b – c + a)**2 + 2(a – b + c) (b – c + a)
**Solution:
(a – b + c)2 + (b – c + a)2 + 2(a – b + c) (b – c + a)
{Because p2 + q2 + 2pq = (p + q)2}
Here p = a – b + c and q = b – c + a
= [a – b + c + b - c + a]2
= (2a)2
= 4a2
**Question 13: Factorize a2 + b2 + 2(ab + bc + ca)
**Solution:
a2 + b2 + 2ab + 2bc + 2ca
As we know, p2 + q2 + 2pq = (p + q)2
We get,
= (a + b)2 + 2bc + 2ca
= (a + b)2 + 2c(b + a)
Or (a + b)2 + 2c(a + b)
Take (a + b) as common factor;
= (a + b)(a + b + 2c)
Therefore, a2 + b2 + 2ab + 2bc + 2ca = (a + b)(a + b + 2c)
****Question 14: Factorize 4(x - y)**2 **– 12(x – y)(x + y) + 9(x + y)**2
**Solution:
Consider (x – y) = p, (x + y) = q
= 4p2 – 12pq + 9q2
Expanding the middle term, -12 = -6 -6 also 4 × 9 = -6 × -6
= 4p2 – 6pq – 6pq + 9q2
= 2p(2p – 3q) - 3q(2p – 3q)
= (2p – 3q) (2p – 3q)
= (2p – 3q)2
Substituting back p = x – y and q = x + y;
= [2(x - y) – 3(x + y)]2 = [2x – 2y – 3x – 3y ]2
= (2x - 3x - 2y - 3y)2
= [-x – 5y]2
= [(-1)(x + 5y)]2
= (x + 5y)2
Therefore, 4(x - y)2 – 12(x – y)(x + y) + 9(x + y)2 = (x + 5y)2
**Question 15: Factorize a2 – b2 + 2bc – c2
**Solution :
a2 – b2 + 2bc – c2
As we know, (a - b)2 = a2 + b2 – 2ab
= a2 – (b – c)2
Also, we know, a2 – b2 = (a + b)(a - b)
= (a + b – c)(a – (b – c))
= (a + b – c)(a – b + c)
Therefore, a2 – b2 + 2bc – c2 =(a + b – c)(a – b + c)
**Question 16: Factorize a2 **+ 2ab + b2 – c2
**Solution:
a2 + 2ab + b2 – c2
= (a2 + 2ab + b2) – c2
= (a + b)2 – (c)2
We know, a2 – b2 = (a + b) (a – b)
= (a + b + c) (a + b – c)
Therefore, a2 + 2ab + b2 – c2 = (a + b + c) (a + b – c)
Summary
Chapter 5, Exercise 5.1 of RD Sharma's Class 9 Mathematics textbook focuses on the Factorisation of Algebraic Expressions, a fundamental skill in algebra. This exercise introduces students to various factorisation techniques, including the common factor method, grouping method, factorisation of perfect square trinomials, difference of squares, and general trinomials. By mastering these techniques, students develop their ability to simplify complex expressions, solve equations more efficiently, and lay the groundwork for understanding advanced mathematical concepts. The problems in this exercise are carefully designed to progressively build students' confidence and competence, challenging them to recognize patterns, apply appropriate factorisation methods, and enhance their overall mathematical reasoning skills. This foundational knowledge is crucial not only for immediate problem-solving but also for future studies in areas such as calculus, trigonometry, and advanced algebra.