Class 8 | NCERT Solution Maths Chapter 14 | Factorisation | Exercise 14.4

Exercise 14.4 Page No: 228

1. 4(x–5) = 4x–5

Solution:

4(x- 5)= 4x – 20 ≠ 4x – 5 = RHS

The correct statement is 4(x-5) = 4x–20

2. x(3x+2) = 3x2+2

Solution:

LHS = x(3x+2) = 3x2+2x ≠ 3x2+2 = RHS

The correct solution is x(3x+2) = 3x2+2x

3. 2x+3y = 5xy

Solution:

LHS= 2x+3y ≠ R. H. S

The correct statement is 2x+3y = 2x+3 y

4. x+2x+3x = 5x

Solution:

LHS = x+2x+3x = 6x ≠ RHS

The correct statement is x+2x+3x = 6x

5. 5y+2y+y–7y = 0

Solution:

LHS = 5y+2y+y–7y = y ≠ RHS

The correct statement is 5y+2y+y–7y = y

6. 3x+2x = 5x2

Solution:

LHS = 3x+2x = 5x ≠ RHS

The correct statement is 3x+2x = 5x

7. (2x) 2+4(2x)+7 = 2x2+8x+7

Solution:

LHS = (2x) 2+4(2x)+7 = 4x2+8x+7 ≠ RHS

The correct statement is (2x) 2+4(2x)+7 = 4x2+8x+7

8. (2x) 2+5x = 4x+5x = 9x

Solution:

LHS = (2x) 2+5x = 4x2+5x ≠ 9x = RHS

The correct statement is(2x) 2+5x = 4x2+5x

9. (3x + 2) 2 = 3x2+6x+4

Solution:

LHS = (3x+2) 2 = (3x)2+22+2x2x3x = 9x2+4+12x ≠ RHS

The correct statement is (3x + 2) 2 = 9x2+4+12x

10. Substituting x = – 3 in
(a) x2 + 5x + 4 gives (– 3) 2+5(– 3)+4 = 9+2+4 = 15
(b) x2 – 5x + 4 gives (– 3) 2– 5( – 3)+4 = 9–15+4 = – 2
(c) x2 + 5x gives (– 3) 2+5(–3) = – 9–15 = – 24

Solution:

(a) Substituting x = – 3 in x2+5x+4, we have

x2+5x+4 = (– 3) 2+5(– 3)+4 = 9–15+4 = – 2. This is the correct answer.

(b) Substituting x = – 3 in x2–5x+4

x2–5x+4 = (–3) 2–5(– 3)+4 = 9+15+4 = 28. This is the correct answer

(c)Substituting x = – 3 in x2+5x

x2+5x = (– 3) 2+5(–3) = 9–15 = -6. This is the correct answer

11.(y–3)2 = y2–9

Solution:

LHS = (y–3)2 , which is similar to (a–b)2 identity, where (a–b) 2 = a2+b2-2ab.

(y – 3)2 = y2+(3) 2–2y×3 = y2+9 –6y ≠ y2 – 9 = RHS

The correct statement is (y–3)2 = y2 + 9 – 6y

12. (z+5) 2 = z2+25

Solution:

LHS = (z+5)2 , which is similar to (a +b)2 identity, where (a+b) 2 = a2+b2+2ab.

(z+5) 2 = z2+52+2×5×z = z2+25+10z ≠ z2+25 = RHS

The correct statement is (z+5) 2 = z2+25+10z

13. (2a+3b)(a–b) = 2a2–3b2

Solution:

LHS = (2a+3b)(a–b) = 2a(a–b)+3b(a–b)

= 2a2–2ab+3ab–3b2

= 2a2+ab–3b2

≠ 2a2–3b2 = RHS

The correct statement is (2a +3b)(a –b) = 2a2+ab–3b2

14. (a+4)(a+2) = a2+8

Solution:

LHS = (a+4)(a+2) = a(a+2)+4(a+2)

= a2+2a+4a+8

= a2+6a+8

≠ a2+8 = RHS

The correct statement is (a+4)(a+2) = a2+6a+8

15. (a–4)(a–2) = a2–8

Solution:

LHS = (a–4)(a–2) = a(a–2)–4(a–2)

= a2–2a–4a+8

= a2–6a+8

≠ a2-8 = RHS

The correct statement is (a–4)(a–2) = a2–6a+8

16. 3x2/3x2 = 0

Solution:

LHS = 3x2/3x2 = 1 ≠ 0 = RHS

The correct statement is 3x2/3x2 = 1

17. (3x2+1)/3x2 = 1 + 1 = 2

Solution:

LHS = (3x2+1)/3x2 = (3x2/3x2)+(1/3x2) = 1+(1/3x2) ≠ 2 = RHS

The correct statement is (3x2+1)/3x2 = 1+(1/3x2)

18. 3x/(3x+2) = ½

Solution:

LHS = 3x/(3x+2) ≠ 1/2 = RHS

The correct statement is 3x/(3x+2) = 3x/(3x+2)

19. 3/(4x+3) = 1/4x

Solution:

LHS = 3/(4x+3) ≠ 1/4x

The correct statement is 3/(4x+3) = 3/(4x+3)

20. (4x+5)/4x = 5

Solution:

LHS = (4x+5)/4x = 4x/4x + 5/4x = 1 + 5/4x ≠ 5 = RHS

The correct statement is (4x+5)/4x = 1 + (5/4x)

21.(7x+5)/5 = 7x

Solution:

LHS = (7x+5)/5 = (7x/5)+ 5/5 = (7x/5)+1 ≠ 7x = RHS

The correct statement is (7x+5)/5 = (7x/5) +1