RBSE Solutions for Class 8 Maths Chapter 10 Factorization Ex 10.2 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Exercise 10.2.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 8 |
| Subject | Maths |
| Chapter | Chapter 10 |
| Chapter Name | Factorization |
| Exercise | Exercise 10.2 |
| Number of Questions | 3 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Ex 10.2
Question 1.
Factorize the following expressions.
(i) a² – 4
(ii) a² – 49b²
(iii) p³ – 121p
(iv) (a – b)² – c²
(v) a4 – b4
(vi) 5x3 – 125x
(vii) 63a² – 112b²
(viii) 9x²y² – 16
(ix) (l + m)² – (l – m)²
Solution:
(i) a² – 4
a² – 4
= (a)² – (2)²
= (a – 2) (a + 2)
(ii) a² – 49b²
a² – 49b²
= (a)² – (7b)²
= (a – 7b) (a + 7b)
(iii) p3 – 121p
p3 – 121p
= p(p² – 121)
= p{(p)² – (11)²}
= p(p – 11)(p + 11)
(iv) (a – b)² – c²
(a – b)² – c²
= (a – b – c) (a – b + c)
(v) a4 – b4
a4 – b4
= (a²)² – (b²)²
= (a² – b²) (a² + b²)
= (a – b)(RBSESolutions.com) (a + b) (a² + b²)
(vi) 5x3 – 125x
5x3 – 125x
= 5x (x² – 25)
= 5x(x² – 5²)
= 5x (x – 5) (x + 5)
(vii) 63a² – 112b²
63a² – 112b²
= 7 (9a² – 16b²)
= 7 {(3a)² – (4b)²}
= 1 (3a- 4b) (3a + 4b)
(viii) 9x²y² – 16
9x²y² – 16
= (3xy)² – (4)²
= (3xy – 4) (3xy + 4)
(ix) (l + m)² – (l – m)²
(l + m)² – (l – m)²
= {(l + m) – (l – m)} {(l + m) + (l – m)}
= (l + m – l + m) (l + m + l – m)
= (2m) (2l)
= 4lm
Question 2.
Factorize the(RBSESolutions.com)following expressions.
(i) lx² + mx
(ii) 2x3 + 2xy² + 2xz²
(iii) a(a + b) + 4 (a + b)
(iv) (xy + y) + x + 1
(v) 5a² – 15a – 6c + 2ac
(vi) am² + bm² + bn² + an²
Solution:
(i) lx² + mx
lx² + mx
= x (lx + m)
(ii) 2x3 + 2xy² + 2xz²
2x3 + 2xy² + 2xz²
= 2x (x² + y² + z²)
(iii) a(a + b) + 4 (a + b)
a(a + b) + 4 (a + b)
= (a + b) (a + 4)
(iv) (xy + y) + x + 1
(xy + y) + x + 1
= y (x + 1) + 1 (x + 1)
= (x + 1) (y + 1)
(v) 5a² – 15a -6c + 2ac
5a² – 15a – 6c + 2ac
= 5a² – 15a + 2ac – 6c
= 5a (a – 3) + 2c (a – 3)
= (a – 3)(RBSESolutions.com) (5a + 2c)
(vi) am² + bm² + bn² + an²
am² + bm² + bn² + an²
= am² + bm² + an² + bn²
= m² (a + b) + n² (a + n)
= (a + b) (m² + n²)
Question 3.
Factorize the following expressions.
(i) x² + 5x + 6
(ii) q² + 11q + 24
(iii) m² – 10m + 21
(iv) x² + 6x – 16
(v) x² – 7x – 18
(vi) k² – 11k – 102
(vii) y² + 2y – 48
(viii) d² – 4d – 45
(ix) m² + 16m + 63
(x) n² – 19n – 92
(xi) p² – 10p + 16
(xii) x² + 4x – 45
Solution:
(i) x² + 5x + 6
x² + 5x + 6
= x² + 2x + 3x + 6
= x (x + 2)(RBSESolutions.com) + 3(x +2)
= (x + 2) (x + 3)
(ii) q² + 11q + 24
q² + 11q + 24
= q² + 3q + 8q + 24
= q (q + 3) + 8 (q + 3)
= (q + 3) (q + 8)
(iii) m² – 10m + 21
m² – 10m + 21
= m² – 3 m – 7m + 21
= m (m – n) – 7 (m – 3)
= (m – 3) (m – 7)
(iv) x² + 6x – 16
x² + 6x – 16
= x² + 8x – 2x – 16
= x (x + 8)(RBSESolutions.com) – 2 (x + 8)
= (x + 8) (x – 2)
(v) x² – 7x – 18
x² – 7x – 18
= x² – 9x + 2x – 18
= x (x – 9) + 2 (x – 9)
= (x – 9) (x + 2)
(vi) k² – 11 k – 102
k² – 11k – 102
= k² – 17k + 6k – 102
= k (k – 17) + 6 (k – 17)
= (k – 17) (k + 6)
(vii) y² + 2y – 48
y² + 2y – 48
= y² + 8y(RBSESolutions.com)- 6y – 48
= y(y + 8) – 6(y + 8)
= (y + 8) (y – 6)
(viii) d² – 4d – 45
d² – 4d – 45
= d² – 9d + 5 d – 45
= d (d – 9) + 5 (d – 9)
= (d – 9) (d + 5)
(ix) m² + 16m + 63
m² + 16m + 63
= m² + 9m + 7m + 63
= m (m + 9) + 7 (m + 9)
= (m + 9) (m + 7)
(x) n² – 19n – 9²
n² – 19n – 92
= n² – 23n(RBSESolutions.com) + 4n – 92
= n (n – 23) + 4 (n – 23)
= (n – 23) (n + 4)
(xi) p² – 10p + 16
p² – 10p + 16
= p² – 8p – 2p + 16
= p (p – 8) – 2(p – 8)
= (p – 8) (p – 2)
(xii) x² + 4x – 45
x² + 4x – 45
= x² + 9x – 5x – 45
= x (x + 9) – 5(x + 9)
= (x + 9) (x – 5)
We hope the given RBSE Solutions for Class 8 Maths Chapter 10 Factorization Ex 10.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 10 Factorization Exercise 10.2, drop a comment below and we will get back to you at the earliest.