NCERT Solutions Class 8 Maths Chapter 14 Factorisation Ex 14.4

NCERT Solutions Class 8 Mathematics 
Chapter – 14 (Factorisation) 

The NCERT Solutions in English Language for Class 8 Mathematics Chapter – 14 Factorisation  Exercise 14.4 has been provided here to help the students in solving the questions from this exercise. 

Chapter 14: Factorisation

Exercise – 14.4

Find and correct the errors in the following mathematical statements.

1. 4(x–5) = 4x–5

Solution –
4(x–5) = 4x–5
⇒ 4x – 20 ≠ 4x – 5
Thus, LHS ≠ RHS
The correct statement is 4(x – 5) = 4x – 20

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

Solution –
x(3x+2)
⇒ 3x2+2x ≠ 3x2+2
Thus, L.H.S ≠ R.H.S.
The correct solution is x(3x+2) = 3x2+2x

3. 2x+3y = 5xy

Solution –
2x+3y ≠ 5xy
Thus, L.H.S ≠ R.H.S.
The correct statement is 2x + 3y = 2x + 3y

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

Solution –
x + 2x + 3x
⇒ 6x ≠ 5x
Thus, L.H.S ≠ R.H.S.
The correct statement is x + 2x + 3x = 6x

5. 5y+2y+y–7y = 0

Solution –
5y+2y+y–7y
= y ≠ 0
Thus, L.H.S ≠ R.H.S.
The correct statement is 5y+2y+y–7y = y

6. 3x+2x = 5x2

Solution –
3x + 2x
⇒ 5x ≠ 5x2
Thus, L.H.S ≠ R.H.S.
The correct statement is 3x+2x = 5x

7. (2x) 2+4(2x)+7 = 2x2+8x+7

Solution –
(2x) 2+4(2x) + 7
⇒ 4x2 + 8x + 7 ≠ 2x2+8x+7
Thus, L.H.S ≠ R.H.S.
The correct statement is (2x) 2+4(2x)+7 = 4x2+8x+7

8. (2x)2 + 5x = 4x+5x = 9x

Solution –
(2x) 2 + 5x
⇒ 4x2 + 5x ≠ 9x
Thus, L.H.S ≠ R.H.S.
The correct statement is(2x) 2+5x = 4x2+5x

9. (3x + 2)2 = 3x2 + 6x + 4

Solution –
(3x+2) 2
= (3x)2 + 22 + 2 × 2 × 3x
= 9x2 + 4 + 12x ≠ 3x2 + 6x + 4
Thus, L.H.S ≠ R.H.S.
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) x2 + 5x + 4 gives (– 3)2 + 5(–3) + 4 = 9+2+4 = 15
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) x2 – 5x + 4 gives (– 3) 2– 5( – 3)+4 = 9–15+4 = – 2
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) x2 + 5x gives (– 3) 2+5(–3) = – 9–15 = – 24
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
⇒ y2 + (3) 2 – 2y × 3
⇒ y2 + 9 – 6y ≠ y2 – 9
Thus, L.H.S ≠ R.H.S
The correct statement is (y–3)2 = y2 + 9 – 6y

12. (z+5) 2 = z2+25

Solution –
LHS = (z+5)2
(z + 5) 2
⇒ z2 + 52 + 2×5×z
⇒ z2 + 25 + 10z ≠ z2 + 25
Thus, L.H.S ≠ R.H.S
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 =
Thus, L.H.S ≠ R.H.S
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
Thus, L.H.S ≠ R.H.S
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
Thus, L.H.S ≠ R.H.S
The correct statement is (a–4)(a–2) = a2–6a+8

16. \mathbf{\frac{3x^2}{3x^2} = 0}

Solution –
LHS = \frac{3x^2}{3x^2}
⇒ 1 ≠ 0
Thus, L.H.S ≠ R.H.S
The correct statement is \frac{3x^2}{3x^2} = 1

17. \mathbf{\frac{(3x^2+1)}{3x^2}} = 1 + 1 = 2

Solution –
LHS = \frac{(3x^2+1)}{3x^2} 

⇒ \frac{3x^2}{3x^2} + \frac{1}{3x^2}

⇒ 1 + \frac{1}{3x^2} ≠ 2
Thus, L.H.S ≠ R.H.S
The correct statement is \frac{(3x^2+1)}{3x^2} = 1 + \frac{1}{3x^2}

18. \mathbf{\frac{3x}{(3x+2)} = \frac{1}{2}}

Solution –
LHS = \frac{3x}{(3x+2)}  ≠ \frac{1}{2}
Thus, L.H.S ≠ R.H.S
The correct statement is \frac{3x}{(3x+2)} = \frac{3x}{(3x+2)}

19. \mathbf{\frac{3}{(4x+3)} = \frac{1}{4x}}

Solution –
LHS = \frac{3}{(4x+3)} ≠ \frac{1}{4x}
Thus, L.H.S ≠ R.H.S
The correct statement is \frac{3}{(4x+3)} = \frac{3}{(4x+3)}

20. \mathbf{\frac{(4x+5)}{4x} = 5}

Solution –
LHS = \frac{(4x+5)}{4x}
⇒ \frac{4x}{4x} +\frac{5}{4x}   

⇒ 1 + \frac{5}{4x} ≠ 5

Thus, L.H.S ≠ R.H.S
The correct statement is \frac{(4x+5)}{4x} = 1 + \frac{5}{4x}

21. \mathbf{\frac{7x +5}{5} = 7x}

Solution –

LHS = \frac{7x +5}{5}

⇒  \frac{7x}{5} + \frac{5}{5}

⇒ \frac{7x}{5} + 1 ≠ 7x
Thus, L.H.S ≠ R.H.S
The correct statement is \frac{7x +5}{5} = \frac{7x}{5} + 1

 

 

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