Odisha State Board BSE Odisha 8th Class Maths Solutions Algebra Chapter 4 ଉତ୍ପାଦକୀକରଣ Ex 4(a) Textbook Exercise Questions and Answers.

## BSE Odisha Class 8 Maths Solutions Algebra Chapter 4 ଉତ୍ପାଦକୀକରଣ Ex 4(a)

Question 1.

12x + 36

ଡ –

12x + 36 = 12(x + 3)

Question 2.

8a + 4b

ଡ –

8a + 4b = 4(2a + b)

Question 3.

22y – 33z

ଡ –

33z = 11(2y – 3z)

Question 4.

14pq + 35pqr

ଡ –

14pq + 35pqr = 7pq (2 + 5r)

Question 5.

10a²b + 5a

ଡ –

10a²b + 5a = 5a(2ab + 1)

Question 6.

15a²bc – 10ab²c

ଡ –

15a²bc – 10ab²c = 5 abc(3a – 2b)

Question 7.

8a³ + 4a² + 2a

ଡ –

8a³ + 4a² + 2a = 2a(4a² + 2a + 1)

Question 8.

30a³b³c³ + 25a^{5}b³c^{6} – 15a^{6}b^{6}c^{6}

ଡ –

30a³b³c³ + 25a^{5}b³c^{6} – 15a^{6}b^{6}c^{6}

= 5a³b³c³ (6 + 5a²c³ – 3 a³b³c³)

Question 9.

7(2x+5)+3(2x+5)

ଡ –

7(2x + 5) + 3(2x + 5) = (2x + 5) (7 + 3) (2x + 5)10 = 10(2x + 5)

Question 10.

5a(2x + 3y) – 2b(2x + 3y)

ଡ –

5a(2x + 3y) – 2b(2x + 3y) = (2x + 3y) (5a – 2b)

Question 11.

8(5x + 9y)² + 12(5x + 9y)

ଡ –

8(5x 9y)² + 12(5x + 9y)

= 4(5x + 9y) {2(5x + 9y) + 3} = 4(5x + 9y)(10x + 18y + 3)

Question 12.

9a (6a – 5b) – 12a²(6a – 5b)

ଡ –

9a (6a – 5b) – 12a²(6a – 5b) = 3a (6a – 5b) (3 – 4a)

Question 13.

5(x – 2y)² + 3(x – 2y)

ଡ –

5(X2y)²+3(x-2y)5(x-2y)(x-2y)+3(x_2y)5(x_2y)(x_2y+3)

Question 14.

6(a+2b)-4(a+2b)²

ଡ –

6(a + 2b) – 4(a + 2b)² = 6(a + 2b) – 4(a + 2b) (a + 2b)

= 2(a + 2b){3 – 2(a + 2b)} = 2(a + 2b){3 – 2a – 4b) = 2(a + 2b)(3 – 2a – 4b)

Question 15.

a(a – 1) + b(a – 1)

ଡ –

a(a – 1) + b(a – 1) = (a – 1) (a + b)

Question 16.

(x – y)² + (x – y)

ଡ –

(x – y)² + (x – y) = (x – y)(x – y) + (x – y)(x – y)(x – y + 1)

Question 17.

a(x – y) + 2b(y – x) + c(x – y)

ଡ –

a(x – y) + 2b(y – x) + c(x – y) = a(x – y) + 2b{-(x – y)} + c(x – y)

=a(x – y) – 2b(x – y) + c(x – y) = (x – y)(a – 2b + c)

Question 18.

a(b – c) + b(b – c) + c(b – c)

ଡ –

a(b – c) + b(b – c) + c(b – c) = (b – c)(a + b + c)

Question 19.

x³(a – 2b) + x²(a – 2b)

ଡ –

x³(a – 2b) + x²(a – 2b) = x² (a – 2b) (x + 1)

Question 20.

4(x + y) (3a – b) + 6(x + y) (2b – 3a)

ଡ –

4(x + y) (3a – b) + 6(x + y) (2b – 3a) = 2(x + y) {2(3a – b) + 3(2b – 3a)}

= 2(x + y)(6a – 2b + 6b – 9a) = 2(x + y)(4b³a)