Mathematics – Class 8 – Chapter 14 – Factorisation – Exercise 14.4 – NCERT Exercise Solution

Find and correct the errors in the following mathematical statements.

1. 4(x–5) = 4x–5

Solution:

Let us take LHS

= 4(x- 5)

= 4x – 20 ≠ 4x – 5 = RHS

Hence, the correct statement is 4(x-5) = 4x–20

2. x(3x+2) = 3x²+2

Solution:

Let us take LHS

= x(3x+2)

= 3x²+2x ≠ 3×2+2 = RHS

Hence, the correct solution is x(3x+2) = 3x²+2x

3. 2x+3y = 5xy

Solution:

Here,

LHS= 2x+3y

RHS = 5xy

Hence, LHS ≠ RHS

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

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

Solution:

We take LHS:

LHS = x+2x+3x = 6x

Hence, LHS≠ RHS

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

5. 5y+2y+y–7y = 0

Solution:

We take LHS

LHS = 5y+2y+y–7y = y

LHS ≠ RHS

Hence, the correct statement is 5y+2y+y–7y = y

6. 3x+2x = 5x²

Solution:

We take LHS

LHS = 3x+2x = 5x

LHS ≠ RHS

Hence, the correct statement is 3x+2x = 5x

7. (2x)²+4(2x)+7 = 2×2+8x+7

Solution:

We take LHS

LHS = (2x)²+4(2x)+7

       = 4x²+8x+7

LHS ≠ RHS

Hence, the correct statement is (2x)²+4(2x)+7 = 4x²+8x+7

8. (2x)²+5x = 4x+5x = 9x

Solution:

We take LHS

LHS = (2x)²+5x

       = 4x²+5x ≠ 9x

LHS ≠ RHS

Hence, the correct statement is(2x)²+5x = 4x²+5x

9. (3x + 2)² = 3x²+6x+4

Solution:

We take LHS

LHS = (3x+2)²

     = (3x)²+2²+2x2x3x

    = 9x²+4+12x

LHS ≠ RHS

Hence, 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

Solution:

= (– 3) ²+5(– 3) + 4

= 9–15+4

= – 2.

Hence, this is the correct answer.

(b) x² – 5x + 4 gives (– 3) 2– 5( – 3)+4 = 9–15+4 = – 2

Solution:

= (–3) 2–5(– 3) +4

= 9+15+4

= 28.

Hence, this is the correct answer

(c) x² + 5x gives (– 3) 2+5(–3) = – 9–15 = – 24

Solution:

= (– 3) 2+5(–3)

= 9–15

= -6.

Hence, this is the correct answer

11.(y–3)² = y²–9

Solution:

LHS = (y–3)²

Using identity: (a–b) ² = a²+b²-2ab

(y – 3)²

= y²+(3)²–2y×3

= y²+9 –6y ≠ y² – 9

LHS ≠ RHS

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

12. (z+5)² = z²+25

Solution:

LHS = (z+5)²  

 Using identity: (a+b)² = a²+b²+2ab.

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

LHS ≠ 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

LHS ≠ 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

LHS ≠ RHS

Hence, 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

LHS ≠ RHS

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

16. 3x²/3x² = 0

Solution:

LHS = 3x²/3x² = 1

LHS ≠ 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

LHS ≠ RHS

The correct statement is (3×2+1)/3×2 = 1+(1/3×2)

18. 3x/(3x+2) = ½

Solution:

LHS = 3x/(3x+2) ≠ 1/2

LHS ≠ RHS

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

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

Solution:

LHS = 3/(4x+3) ≠ 1/4x

LHS ≠ RHS

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

LHS ≠ 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

LHS ≠ RHS

Hence, the correct statement is (7x+5)/5 = (7x/5) +1