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RBSE Solutions for Class 8 Maths Chapter 10 Factorization Additional Questions

March 21, 2019 by Fazal Leave a Comment

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.

RBSE Solutions

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)

RBSE Solutions

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).

RBSE Solutions

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

RBSE Solutions

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.

RBSE Solutions

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)

RBSE Solutions

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

RBSE Solutions

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

RBSE Solutions

Question 7.
If \({ x }^{ 2 }+\frac { 1 }{ { x }^{ 2 } } =62\) then find the value of \(\left( x+\frac { 1 }{ x } \right) \)
RBSE Solutions for Class 8 Maths Chapter 10 Factorization Additional Questions img-1

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

RBSE Solutions

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)
RBSE Solutions for Class 8 Maths Chapter 10 Factorization Additional Questions img-2
= 3y (5y – 7)

RBSE Solutions

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

RBSE Solutions for Class 8 Maths Chapter 10 Factorization Additional Questions img-3
= 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)

RBSE Solutions

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)

RBSE Solutions

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
RBSE Solutions for Class 8 Maths Chapter 10 Factorization Additional Questions img-4
= 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)
RBSE Solutions for Class 8 Maths Chapter 10 Factorization Additional Questions img-5

RBSE Solutions

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).

RBSE Solutions

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.

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