# RD Sharma Class 9 Solutions Chapter 5 Factorisation of Algebraic Expressions Ex 5.1

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### RD Sharma Solutions for Class 9 Chapter 5 Factorisation of Algebraic Expressions Ex 5.1 Download PDF

Question 1.
x3 + x – 3x2 – 3
Solution:
x3 + x – 3x2 – 3
x3 – 3a2 + x – 3
⇒  x2(x – 3) + 1(x – 3)
= (x – 3) (x2 + 1)

Question 2.
a(a + b)3 – 3a2b(a + b)
Solution:
a(a + b)3 – 3a2b(a + b)
= a(a + b) {(a + b)2 – 3ab}
= a(a + b) {a2 + b2 + 2ab – 3ab}
= a{a + b) {a2 – ab + b2)

Question 3.
x(x3 – y3) + 3xy(x – y)
Solution:
x(x3 – y3) + 3xy(x – y)
= x(x – y) (x2 + xy + y2) + 3xy(x – y)
= x(x – y) (x2 + xy + y2 + 3y)
= x(x – y) (x2 + xy + y2 + 3y)

Question 4.
a2x2 + (ax2 +1)x + a
Solution:
a2x2 + (ax2 + 1)x + a
= a2x2 + a + (ax2 + 1)x
= a(ax2 + 1) + x(ax2 + 1)
= (ax2 + 1) (a + x)
= (x + a) (ax2 + 1)

Question 5.
x2 + y – xy – x
Solution:
x2 + y – xy – x
= x2-x-xy + y = x(x- l)-y(*- 1)
= (x – 1) (x – y)

Question 6.
X3 – 2x2y + 3xy2 – 6y3
Solution:
x3 – 2x2y + 3xy2 – 6y3
= x2(x – 2y) + 3y2(x – 2y)
= (x – 2y) (x2 + 3y2)

Question 7.
6ab – b2 + 12ac – 2bc
Solution:
6ab – b2 + 12ac – 2bc
= 6ab + 12ac – b2 – 2bc
= 6a(b + 2c) – b(b + 2c)
= (b + 2c) (6a – b)

Question 8.
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) [(x)2 – 2 x x x 2 + (2)2]
= (x – 2) (x – 2)2 = (x – 2)3

Question 9.
(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)      {∵ a2 + b2 + 2ab = (a + b)2}
= [a – b + c + b- c + a]2
= (2a)2 = 4a2

Question 10.
a2 + 2ab + b2 – c2
Solution:
a2 + 2ab + b2 – c2
= (a2 + 2ab + b2) – c2
= (a + b)2 – (c)2         {∵  a2 – b2 = (a + b) (a – b)}
= (a + b + c) (a + b – c)

Question 11.
a2 + 4b2 – 4ab – 4c2
Solution:

Question 12.
x2 – y2 – 4xz + 4z2
Solution:
x2 – y2 – 4xz + 4z2
= x2 – 4xz + 4z2 – y2
= (x)2 – 2 x x x 2z + (2z)2 – (y)2
= (x – 2z)2 – (y)2
= (x – 2z + y) (x – 2z – y)
= (x +y – 2z) (x – y – 2z)

Question 13. Solution: Question 14. Solution: Question 15. Solution: Question 16.
Give possible expression for the length and breadth of the rectangle having 35y2 + 13y – 12 as its area.
Solution:
Area of a rectangle = 35y2 + 13y – 12
= 35y2 + 28y- 15y- 12 (i) If length = 5y + 4, then breadth = 7y – 3
(ii) and if length = 7y-3, then length = 5y+ 4

Question 17.
What are the possible expressions for the dimensions of the cuboid whose volume is 3x2 – 12x.
Solution:
Volume 3x2 – 12x
= 3x(x – 4)
∴ Factors are 3, x, and x – 4
Now, if length = 3, breadth = x and height = x – 4
if length =3, breadth = x – 4, height = x
if length = x, breadth = 3, height = x – 4
if length = x, breadth = x – 4, height = 3
if length = x – 4, breadth = 3, height = x
if length – x – 4, breadth = x, height = 3

Question 18. Solution: Question 19.
(x + 2) (x2 + 25) – 10x2 – 20x
Solution:
(x + 2) (x2 + 25) – 10x2 – 20x
= (x + 2) (x2 + 25) – 10x(x + 2)
= (x + 2) [x2 + 25 – 10x]
= (x + 2) [(x)2 – 2 x x x 5 + (5)2]
= (x + 2) (x – 5)2

Question 20.
2a2 + 26–√ ab +3b2
Solution:
2a2 + 26–√  ab +3 b2
= (2–√ a)2+ 2–√ a x 3–√ b+ (3–√ b)2
= (2–√a + 3–√ b)2

Question 21.
a2 + b2 + 2(ab + bc + ca)
Solution:
a2 + b2 + 2(ab + bc + ca)
= a2 + b2 + 2 ab + 2 bc + 2 ca
= (a + b)2 + 2c(b + a)
= (a + b)2 + 2c(a + b)
= (a + b) (a + b + 2c)

Question 22.
4(x – y)2 – 12(x -y) (x + y) + 9(x + y)2
Solution:
4(x – y)2 – 12(x – y) (x + y) + 9(x + y)2
= [2(x – y)2 + 2 x 2(x – y) x 3(x + y) + [3 (x+y]2        {∵ a2 + b2 + 2 abc = (a + b)2}
= [2(x – y) + 3(x + y)]2
= (2x-2y + 3x + 3y)2
= (5x + y)2

Question 23.
a2 – b2 + 2bc – c2
Solution:
a2 – b2 + 2bc – c2
= a2 – (b2 – 2bc + c2)                                           {∵ a2 + b2 – 2abc = (a – b)2}
= a2 – (b – c)2
= (a)2 – (b – c)2          {∵ a2 – b2 = (a + b) (a – b)}
= (a + b – c) (a – b + c)

Question 24.
xy9 – yx9
Solution:
xy9 – yx9 = xy(y8 – x8)
= -xy(x8 – y8)
= -xy[(x4)2 – (y4)2]
= -xy (x4 + y4) (x4 – y4)                                         {∵ a2-b2 = (a + b) (a – b)}
= -xy (x4 + y4) {(x2)2 – (y2)2}
= -xy(x+ y4) (x2 + y2) (x2 – y2)
= -xy (x4 +y4) (x2 + y2) (x + y) (x -y)
= -xy(x – y) (x + y) (x2 + y2) (x4 + y4)

Question 25.
x4 + x2y2 + y4
Solution:
x4 + x2y2 + y4  = (x2)2 + 2x2y2 + y4 – x2y2           (Adding and subtracting x2y2)
= (x2 + y2)2 – (xy)2                                                        {∵ a2 – b2 = (a + b) (a – b)}
= (x2 + y2 + xy) (x2 + y2 – xy)
= (x2 + xy + y2) (x2 – xy + y2)

Question 26.
x2 + 62–√x + 10
Solution: Question 27.
x2 + 22–√x- 30
Solution: Question 28.
x2 – 3–√x – 6
Solution: Question 29.
x2 + 5 5–√x + 30
Solution: Question 30.
x2 + 2 3–√x – 24
Solution: Question 31.
5 5–√x2 + 20x + 35–√
Solution:  Question 32.
2x2 + 35–√ x + 5
Solution: Question 33.
9(2a – b)2 – 4(2a – b) – 13
Solution: Question 34.
7(x-2y) – 25(x-2y) +12
Solution: Question 35.
2(x+y) – 9(x+y) -5
Solution:
2(x+y) – 9(x+y) -5 All Chapter RD Sharma Solutions For Class 9 Maths

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All Subject NCERT Exemplar Problems Solutions For Class 9

All Subject NCERT Solutions For Class 9

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