RBSE Solutions for Class 8 Maths Chapter 10 Factorization Additional Questions

RBSE Solutions for Class 8 Maths Chapter 10 Factorization Additional Questions is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Additional Questions.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 8
Subject	Maths
Chapter	Chapter 10
Chapter Name	Factorization
Exercise	Additional Questions
Number of Questions	39
Category	RBSE Solutions

Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Additional Questions

I.Objective Type Questions

Question 1.
Factors of expression x² + (a + b) x + ab are
(a) (x + a)(x – b)
(b) (x – a) (x + b)
(c) (x + a)(x + b)
(d) (x – a) (x – b)

Question 2.
Square of (2x + 3) is
(a) 4x² + 6x + 9
(b) 4x² + 2x + 9
(c) 4x² + 12x + 9
(d) 4x² + 9

Question 3.
Square of (6x + 1) is
(a) 36x² + 1
(b) 36x² + 6x + 1
(c) 36x² + 6
(d) 36x² + 12x + 1.

Question 4.
The product(RBSESolutions.com)of expression (2a – 3) (2a + 3) is
(a) 4a² + 2a + 9
(b) 4a² – 9
(c) 4a² – 6
(d) 4a² – 4a + 9

Question 5.
Common factor of 6x + 18xy is
(a) y
(b) 6y
(c) 6x
(d) xy

Question 6.
Factor of 2x³ + x² + 2x + 1 is
(a) (2x + 1)(x² + 1)
(b) (x² + 2)(x + 1)
(c) (x + 2)(x² + 1)
(d) (x² + 1)(x + 1)

Question 7.
Factor of 4x² + 8xy + 4y² is
(a) (2x + 2y)²
(b) (2x – 2y)²
(c) (2x + y)²
(d) (x + 2y)²

Question 8.
The factors(RBSESolutions.com)of expression a² + 2ab + b² are
(a) (a + b)(a – b)
(b) (a + b)²
(c) (a – b)²
(d) (a² + b²)²

Question 9.
The factors of expression a² – b² are
(a) (a² – b²)
(b) (a² – b²) (a + b)
(c) (a + b) (a + b)
(d) (a – b) (a + b)

Answers
1. (c)
2. (c)
3. (d)
4. (b)
5. (c)
6. (a)
7. (a)
8. (b)
9. (d).

II. Fill in the blanks

Question 1.
The equation which is true for all value of variables is called ___

Question 2.
(x + a) (x + b) = x² + (….) x + (ab)

Question 3.
a² – b² = (…) x (a – b)

Question 4.
(x – 1) (x + 1) is equal to ___

Question 5.
The value(RBSESolutions.com)of 3.5 x 3.5 – 2.5 x 2.5 is ___

Answers
1. Identity
2. (a + b)
3. (a + b)
4. x² – 1
5. 6

III. True/False Type Questions

Question 1.
Factors of a² – a + ab – a are (a – 1) (a + b).

Question 2.
Expansion of \({ \left( x+\frac { 1 }{ x } \right) }^{ 2 }\) is \({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } +2\)

Question 3.
Factors of x² – 7x + 12 are (x + 3) (x + 4).

Question 4.
If x = 2 and y = – 1 then value of x² + 4xy + 4y² is 0.

Answers
1. True
2. True
3. False
4. True.

IV. Matching Type Questions

Part 1	Part 2
1. 103 × 107	(a) 11021
2. 22y – 33z	(b) 2y
3. common factor of 2y, 22xy	(c) – 4y
4. – 36y³ ÷ 9y²	(d) 11 (2y – 3z)

Answers
1. ⇔ (a)
2. ⇔ (d)
3. ⇔ (b)
4. ⇔ (c)

V. Very Short Answer Type Questions

Question 1.
Factorize 3x³ + 3x² + x + 1.
Solution
3x³ + 3x² + x + 1
= 3x² (x + 1) + 1 (x + 1)
= (x + 1) (3x² + 1)

Question 2.
Find common factor of 2(x + y) + 3(x + y) + 5(x + y)
Solution
Common factor is (x + y)

Question 3.
Simplify 2x(2x² + 2x – 9).
Solution
2x (2x² + 2x – 9)
= 2x × 2x² + 2x × 2x – 2x × 9
= 4x³ + 4x² – 18x

Question 4.
Simplify (x + 2) (x + 3).
Solution
(x + 2) (x + 3)
= x² + (2 + 3)x + 2 × 3
= x² + 5x + 6

Question 5.
Find the factor of 4x² – a².
Solution
4x² – a²
= (2x)² – a²
= (2x + a) (2x – a)

Question 6.
If x + y = 20, xy = 34 then find the value of x² + y².
Solution
(x + y)² = x² + y² + 2xy
⇒ (20)² = x² + y² + 2(34)
⇒ 400 = x² + y² + 68
⇒ x² + y² = 400 – 68
⇒ x² + y² = 332

Question 7.
If \({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =62\) then find the value of \(\left( x+\frac { 1 }{ x } \right) \)

Question 8.
If x² – x – 42 = (x + k) (x + 6) then find the value of k.
Solution
x² – x – 42 = (x + k) (x + 6)
⇒ x² – 7x + 6x – 42 = (x + k) (x + 6)
⇒x (x – 7) + 6 (x – 7) = (x + k) (x + 6)
⇒ (x – 7) (x + 6) = (x + k) (x + 6)
On comparison
k = – 7

VI. Short Answer Type Questions

Question 1.
Factorize the following expression
x² + 8x + 16
Solution
x² + 8x + 16
= x² + 2 × x + 4 + (4)²
= (x + 4)²

Question 2.
Simplify :
39y³ (50y² – 98) ÷ 26y² (5y + 7)
Solution
39y³ (50y² – 98) ÷ 26y² (5y + 7)

= 3y (5y – 7)

Question 3.
Divide 8(x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z².
Solution
8(x³y²z² + x²y³z² + x²y²z³) ÷ 4x²y²z².

= 2x + 2y + 2z
= 2(x + y + z)

Question 4.
Factorize x⁴ – (x – z)⁴.
Solution
x⁴ – (x – z)⁴
= (x²)² – {(x – z)²}²
= {x² – (x – z)²} {x² + (x – z)²}
= {x – (x – z)} {x + (x – z)} {x² + x² – 2xz + z²}
= z (2x – z) (2x² – 2xz + z²)

Question 5.
Factorize 8x²y³ (a + b)² + 24x³y² (a + b)² – 16x³y³ (a + b)².
Solution
8x²y³ (a + b)² + 24x³y² (a + b)² – 16x³y³ (a + b)²
= 8x²y² (a + b)²  [y + 3x – 2xy]

Question 6.
Factorize
(2x + 3y)² – 5(2x + 3y) – 14
Solution
Let 2x + 3y = a
then (2x + 3y)² – 5(2x + 3y) – 14
= a² – 5a – 14
= a² – 7a + 2a – 14
= a(a – 7) + 2(a – 7)
= (a – 7) (a + 2)
= (2x + 3y – 7) (2x + 3y + 2)

Question 7.
Factorize by(RBSESolutions.com)common factor method
p²qr + pq²r + pqr²
(ii) Divide the polynomial
18m³ + 6m² + 12m by 3m
Solution
(i) p²qr + pq²r + pqr²
= pqr (p + q + r))
(ii) 18m³ + 6m² + 12 m
= 6m (3m² + m + 2)
Now (18m³ + 6m² + 12m) ÷ 3m

= 2(3m² + m + 2)

Question 8.
Factorize the expression and divide
(z² – 4z – 12) ÷ (z + 2)
Solution
z² – 4z – 12
= z² – 6z + 2z – 12
= z(z – 6) + 2(z – 6)
= (z – 6) (z + 2)
∴(z² – 4z – 12) + (z + 2)

Question 9.
Factorize the(RBSESolutions.com)following expressions : [Solve any two]
(i) 5pq + 5p + 3q² + 3q
(ii) a² – 5a + 6
(iii) p⁴ – 81
Solution
(i) 5pq + 5p + 3q² + 3q
= (5pq + 3q²) + (5p + 3q)
= q(5p + 3q) + 1(5p + 3 q)
= (5p + 3q)(q + 1)
(ii) a² – 5a + 6
= a² – 3a – 2a + 6
= a(a – 3) – 2(a – 3)
= (a – 3) (a – 2).
(iii) p⁴ – 81
= (p²)² – (9)²
Identity a² – b² = (a + b) (a – b)
= (p² + 9) (p² – 9)
= (p² + 9) [(p)² – (3)²]
= (p² + 9) (p + 3) (p – 3).

We hope the given RBSE Solutions for Class 8 Maths Chapter 10 Factorization Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Additional Questions, drop a comment below and we will get back to you at the earliest.