Exercise 14.4

Question 1
Find and correct the errors in the statement: 4(x − 5) = 4x − 5
Sol :
L.H.S. = 4(x − 5) = 4 × x − 4 × 5 = 4x − 20 ≠ R.H.S.
The correct statement is 4(x − 5) = 4x − 20

Q 2 - Ex 14.4 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 2
Find and correct the errors in the statement: x(3x + 2) = 3x² + 2
Sol :
L.H.S. = x(3x + 2) = x × 3x + x × 2 = 3x² + 2x ≠ R.H.S.
The correct statement is x(3x + 2) = 3x² + 2x

Question 3
Find and correct the errors in the statement: 2x + 3y = 5xy
Sol :
L.H.S = 2x + 3y ≠ R.H.S.
The correct statement is 2x + 3y = 2x + 3y

Question 4
Find and correct the errors in the statement: x + 2x + 3x = 5x
Sol :
L.H.S = x + 2x + 3x =1x + 2x + 3x = x(1 + 2 + 3) = 6x ≠ R.H.S.
The correct statement is x + 2x + 3x = 6x

Question 5
Find and correct the errors in the statement: 5y + 2y + y − 7y = 0
Sol :
L.H.S. = 5y + 2y + y − 7y = 8y − 7y = y ≠ R.H.S
The correct statement is 5y + 2y + y − 7y = y

Q 6 & Q 7 - Ex 14.4 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 6
Find and correct the errors in the statement: 3x + 2x = 5x²
Sol :
L.H.S. = 3x + 2x = 5x ≠ R.H.S
The correct statement is 3x + 2x = 5x

Question 7
Find and correct the errors in the statement: (2x)² + 4(2x) + 7 = 2x² + 8x + 7
Sol :
L.H.S = (2x)² + 4(2x) + 7 = 4x² + 8x + 7 ≠ R.H.S
The correct statement is (2x)² + 4(2x) + 7 = 4x² + 8x + 7

Q 8 - Ex 14.4 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 8
Find and correct the errors in the statement: (2x)² + 5x = 4x + 5x = 9x
Sol :
L.H.S = (2x)² + 5x = 4x² + 5x ≠ R.H.S.
The correct statement is (2x)² + 5x = 4x² + 5x

Question 9
Find and correct the errors in the statement: (3x + 2)² = 3x² + 6x + 4
Sol :
L.H.S. = (3x + 2)² = (3x)² + 2(3x)(2) + (2)² [(a + b)² = a² + 2ab + b²]
= 9x² + 12x + 4 ≠ R.H.S
The correct statement is (3x + 2)² = 9x² + 12x + 4

Q 10 - Ex 14.4 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 10
Find and correct the errors in the following mathematical statement. 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
Sol :
(a) For x = −3,
x² + 5x + 4 = (−3)² + 5 (−3) + 4 = 9 − 15 + 4 = 13 − 15 = −2
(b) For x = −3,
x² − 5x + 4 = (−3)² − 5 (−3) + 4 = 9 + 15 + 4 = 28
(c) For x = −3,
x² + 5x = (−3)² + 5(−3) = 9 − 15 = −6

Q 11 - Ex 14.4 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 11
Find and correct the errors in the statement: (y − 3)² = y² − 9
Sol :
L.H.S = (y − 3)² = (y)² − 2(y)(3) + (3)² [(a − b)² = a² − 2ab + b²]
= y² − 6y + 9 ≠ R.H.S
The correct statement is (y − 3)² = y² − 6y + 9

Question 12
Find and correct the errors in the statement: (z + 5)² = z² + 25
Sol :
L.H.S = (z + 5)² = (z)² + 2(z)(5) + (5)² [(a + b)² = a² + 2ab + b²]
= z² + 10z + 25 ≠ R.H.S
The correct statement is (z + 5)² = z² + 10z + 25

Q 13 - Ex 14.4 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 13
Find and correct the errors in the statement: (2a + 3b) (a − b) = 2a² − 3b²
Sol :
L.H.S. = (2a + 3b) (a − b) = 2a × a + 3b × a − 2a × b − 3b × b
= 2a² + 3ab − 2ab − 3b² = 2a² + ab − 3b² ≠ R.H.S.
The correct statement is (2a + 3b) (a − b) = 2a² + ab − 3b²

Q 14 - Ex 14.4 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 14
Find and correct the errors in the statement: (a + 4) (a + 2) = a² + 8
Sol :
L.H.S. = (a + 4) (a + 2) = (a)² + (4 + 2) (a) + 4 × 2
= a² + 6a + 8 ≠ R.H.S
The correct statement is (a + 4) (a + 2) = a² + 6a + 8

Q 15 & Q 16 - Ex 14.4 - Factorization - NCERT Maths Class 8th - Chapter 14

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Question 15
Find and correct the errors in the statement: (a − 4) (a − 2) = a² − 8
Sol :
L.H.S. = (a − 4) (a − 2) = (a)² + [(− 4) + (− 2)] (a) + (− 4) (− 2)
= a² − 6a + 8 ≠ R.H.S.
The correct statement is (a − 4) (a − 2) = a² − 6a + 8

Question 16
Find and correct the errors in the statement: