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(x⁴ + 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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