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NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4 Factorisation

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4 Factorisation, has been designed by the NCERT to test the knowledge of the student on the topic Finding the Error. Class 8 maths chapter 14 exercise 14.4 contains 21 questions and each question has explained stepwise. If you are class 8th student and currently preparing ex 14.4 then you must be looking for the class 8th exercise 14.4 solution for your exams preparation. Here we are providing complete solutions for class 8 maths exercise 14.4.

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4 – Factorisation

Find and correct the errors in the following mathematical statements.

1. 4(x – 5) = 4x – 5

Sol.: –

L.H.S. = 4(x – 5)
= 4×x – 4×5
= 4x – 20
≠ R.H.S.
Hence, the correct statement is 4(x – 5) = 4x – 20

2. x(3x + 2) = 3x² + 2

Sol.: –

L.H.S. = x(3x + 2)
= x×3x + x×2
= 3x² +2x
≠ R.H.S.
Hence, the correct statement is x(3x + 2) = 3x² +2x

3. 2x + 3y = 5xy

Sol.: –

L.H.S. = 2x + 3y
= 2x + 3y
≠ R.H.S.
Hence, the correct statement is 2x + 3y = 2x + 3y

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

Sol.: –

L.H.S. = x + 2x + 3x
= 1x + 2x + 3x
= 6x
≠ R.H.S.
Hence, the correct statement is x + 2x + 3x = 6x

5. 5y + 2y + y – 7y = 0

Sol.: –

L.H.S. = 5y + 2y + y – 7y
= 5y + 2y + 1y – 7y
= y
≠ R.H.S.
Hence, the correct statement is 5y + 2y + y – 7y = y

6. 3x + 2x = 5x²

Sol.: –

L.H.S. = 3x + 2x
= 5x
≠ R.H.S.
Hence, the correct statement is 3x + 2x = 5x

7. (2x)² + 4(2x) + 7 = 2x² + 8x + 7

Sol.: –

L.H.S. = (2x)² + 4(2x) + 7
= 4x² + 8x +7
≠ R.H.S.
Hence, the correct statement is (2x)² + 4(2x) + 7 = 4x² + 8x +7

8. (2x)² + 5x = 4x + 5x = 9x

Sol.: –

L.H.S. = (2x)² + 5x
= 4x² + 5x
≠ R.H.S.
Hence, the correct statement is (2x)² + 5x = 4x² + 5x

9. (3x + 2)² = 3x² + 6x + 4

Sol.: –

L.H.S. = (3x + 2)²= (3x)² + 2×3x×2 + 2² .. { (a + b)² = a² + 2ab + b² }
= 9x² + 12x + 4
≠ R.H.S.
Hence, the correct statement is (3x + 2)² = 9x² + 12x + 4

10. Substituting x = – 3 in

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

Sol.: –

L.H.S. = x² + 5x + 4
For x = -3
= (-3)² + 5 × (-3) + 4
= 9 – 15 + 4
= -2
≠ R.H.S.
Hence, the correct statement is x² + 5x + 4 = -2 for x = -3.

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

Sol.: –

L.H.S. = x² – 5x + 4
For x = -3
= (-3)² -5 × (-3) + 4
= 9 + 15 + 4
= 28
≠ R.H.S.
Hence, the correct statement is x² – 5x + 4= 28 for x = -3.

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

Sol.: –

L.H.S. = x² + 5x
For x = -3
= (-3)² + 5 × (-3)
= 9 – 15
= -6
≠ R.H.S.
Hence, the correct statement is x² + 5x = -6 for x = -3.

11. (y – 3)² = y² – 9

Sol.: –

L.H.S. = (y – 3)²= y² – 2×y×3 + 3² .. { (a – b)² = a² – 2ab + b² }
= y² – 6y +9
≠ R.H.S.
Hence, the correct statement is (y – 3)² = y² – 6y +9

12. (z + 5)² = z² + 25

Sol.: –

L.H.S. = (z + 5)²= z² + 2×z×5 + 5² .. { (a + b)² = a² + 2ab + b² }
= z² + 10z + 25
≠ R.H.S.
Hence, the correct statement is (z + 5)² = z² + 10z + 25

13. (2a + 3b) (a – b) = 2a² – 3b²

Sol.: –

L.H.S. = (2a + 3b) (a – b)
= 2a × a + 2a × (-b) + 3b × a + 3b × (-b)
= 2a² – 2ab + 3ab – 3b²
= 2a² + ab – 3b²≠ R.H.S.
Hence, the correct statement is (2a + 3b) (a – b) = 2a² + ab – 3b²

14. (a + 4) (a + 2) = a² + 8

Sol.: –

L.H.S. = (a + 4) (a + 2)
= a × a + a × 2 + 4 × a + 4 × 2
= a² + 2a + 4a + 8
= a² + 6a + 8
≠ R.H.S.
Hence, the correct statement is (a + 4) (a + 2) = a² + 6a + 8

15. (a – 4) (a – 2) = a² – 8

Sol.: –

L.H.S. = (a – 4) (a – 2)
= a × a + a × (-2) + (-4) × a + (-4) × (-2)
= a² – 2a – 4a + 8 .. { (x + a)(x + b) = x² + (a + b)x + ab }
= a² – 6a + 8
≠ R.H.S.
Hence, the correct statement is (a – 4) (a – 2) = a² – 6a + 8

16. $\cfrac { { 3x }^{ 2 } }{ { 3x }^{ 2 } }$ = 0

Sol.: –

L.H.S. = $\cfrac { { 3x }^{ 2 } }{ { 3x }^{ 2 } }$
= 1
≠ R.H.S.
Hence, the correct statement is $\cfrac { { 3x }^{ 2 } }{ { 3x }^{ 2 } }$ = 1

17. $\cfrac { { 3x }^{ 2 }+1 }{ { 3x }^{ 2 } }$ = 1+1 = 2

Sol.: –

L.H.S. = $\cfrac { { 3x }^{ 2 }+1 }{ { 3x }^{ 2 } }$
= $\cfrac { { 3x }^{ 2 } }{ { 3x }^{ 2 } }$ + $\cfrac { 1 }{ { 3x }^{ 2 } }$
= 1 + $\cfrac { 1 }{ { 3x }^{ 2 } }$
≠ R.H.S.
Hence, the correct statement is $\cfrac { { 3x }^{ 2 }+1 }{ { 3x }^{ 2 } }$ = 1 + $\cfrac { 1 }{ { 3x }^{ 2 } }$

18. $\cfrac { { 3x } }{ { 3x+2 } }$ = $\cfrac { 1 }{ 2 }$

Sol.: –

L.H.S. = $\cfrac { { 3x } }{ { 3x+2 } }$
≠ R.H.S.
Hence, the correct statement is $\cfrac { { 3x } }{ { 3x+2 } }$ = $\cfrac { { 3x } }{ { 3x+2 } }$

19. $\cfrac { { 3 } }{ { 4x+3 } }$ = $\cfrac { 1 }{ 4x }$

Sol.: –

L.H.S. = $\cfrac { { 3 } }{ { 4x+3 } }$
≠ R.H.S.
Hence, the correct statement is $\cfrac { { 3 } }{ { 4x+3 } }$ = $\cfrac { { 3 } }{ { 4x+3 } }$

20. $\cfrac { 4x+5 }{ { 4x } }$ = 5

Sol.: –

L.H.S. = $\cfrac { 4x+5 }{ { 4x } }$
= $\cfrac { 4x }{ { 4x } }$ + $\cfrac { 5 }{ { 4x } }$
= 1 + $\cfrac { 5 }{ { 4x } }$
≠ R.H.S.
Hence, the correct statement is

21. $\cfrac { 7x+5 }{ { 5 } }$ = 7x

Sol.: –

L.H.S. = $\cfrac { 7x+5 }{ { 5 } }$
= $\cfrac { 7x }{ { 5 } }$ + $\cfrac { 5 }{ { 5 } }$
= $\cfrac { 7x }{ { 5 } }$ + 1
≠ R.H.S.
Hence, the correct statement is $\cfrac { 7x+5 }{ { 5 } }$ = $\cfrac { 7x }{ { 5 } }$ + 1.

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4 Factorisation

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4 – Factorisation

Find and correct the errors in the following mathematical statements.

NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4

Download NCERT Solutions for Class 8 Maths Chapter 14 Exercise 14.4 Factorisation

NCERT Solutions for Class 8

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