KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4

You can Download KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4 Questions and Answers helps you to revise the complete syllabus.

KSEEB Solutions for Class 8 Maths Chapter 14 Factorization Ex 14.4

Find and correct the errors in the following mathematical statements.

Question 1.
4(x – 5) = 4x – 5
Solution:
4(x – 5) = 4x – 5
L.H.S. = 4(x – 5) = 4x – 20
4x – 20 ≠ 4x – 5
∴ 4(x – 5) = 4x – 20

Question 2.
x(3x + 2) = 3x² + 2
Solution:
x(3x + 2) = 3x² + 2
L.H.S. = x(3x + 2) = 3x² + 2x
3x² + 2x ≠ 3x² + 2
∴ Correct statement x(3x + 2) = 3x² + 2x

Question 3.
2x + 3y = 5xy
Solution:
2x + 3y = 5xy
L.H.S. = 2x + 3y = 2x + 3y
Incorrect statement
2x + 3y ≠ 5xy
∴ Correct statement 2x + 3y = 2x + 3y

Question 4.
x + 2x + 3x = 5x
Solution:
x + 2x + 3x = 5x
L.H.S. = x + 2x + 3x = 6x
6x ≠ 5x
Correct statement x + 2x + 3x = 6x

Question 5.
5y + 2y + y – 7y = 0
Solution:
5y + 2y + y – 7y = 0
L.H.S. = 5y + 2y + y – 7y
= 8y – 7y
= y
y ≠ 0
∴ Correct statement 5y + 2y + y – 7y = y

Question 6.
3x + 2x = 5x²
Solution:
3x + 2x = 5x²
L.H.S. = 3x + 2x = 5x
5x ≠ 5x²
∴ Correct statement 3x + 2x = 5x

Question 7.
(2x)² + 4(2x) + 7 = 2x² + 8x + 7
Solution:
(2x)² + 4(2x) + 7 = 2x² + 8x + 7
L.H.S. = (2x)² + 4(2x) + 7
= 4x² + 8x + 7
4x² + 8x + 7 ≠ 2x² + 8x + 7
Correct statement (2x)² + 4(2x) + 7 = 4x² + 8x + 7

Question 8.
(2x)² + 5x = 4x + 5x = 9x
Solution:
(2x)² + 5x = 4x + 5x = 9x
L.H.S. = (2x)² + 5x = 4x² + 5x
4x² + 5x ≠ 4x + 5x = 9x
∴ Correct statement (2x)² + 5x = 4x² + 5x

Question 9.
(3x + 2)² = 3x² + 6x + 4
Solution:
(3x + 2)² = 3x² + 6x + 4
L.H.S. = (3x + 2)²
= (3x)² + 2(3x) (2) + (2)²
= 9x² + 12x + 4
∴ Correct statement (3x + 2)² = 9x² + 12x + 4

Question 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
When x = -3
x² + 5x + 4
= (-3)² + 5(-3) + 4
= 9 – 15 + 4
= 13 – 15
= -2
It gives -2

(b) x² – 5x + 4
When x = -3
x² – 5x + 4
= (-3)² – 5(-3) + 4
= 9 + 15 + 4
= 28
It gives 28

(c) x² + 5x
When x = -3
x² + 5x
= (-3)² + 5(-3)
= 9 – 15
= -6
It gives -6

Question 11.
(y – 3)² = y² – 9
Solution:
(y – 3)²
= (y)² – 2(y) (3) + (3)²
= y² – 6y + 9

Question 12.
[z + 5]² = z² + 25
Solution:
(z + 5)²
= z² + 2(z) (5) + (5)²
= z² + 10z + 25

Question 13.
(2a + 3b) (a – b) = 2a² – 3b²
Solution:
(2a + 3) (a – b)
= 2a × a – 2a × b + 3a – 3b
= 2a² – 2ab + 3a – 3b

Question 14.
(a + 4) (a + 2) = a² + 8
Solution:
(a + 4) (a + 2)
= a × a + 2a + 4a + 4 × 2
= a² + 6a + 8

Question 15.
(a – 4) (a – 2) = a² – 8
Solution:
(a – 4) (a – 2)
= a(a – 2) – 4(a – 2)
= a² – 2a – 4a + 8
= a² – 6a + 8

Question 16.
\(\frac{3 x^{2}}{3 x^{2}}=0\)
Solution:
\(\frac{3 x^{2}}{3 x^{2}}=1\)

Question 17.
\(\frac{3 x^{2}+1}{3 x^{2}}\) = 1 + 1 = 2
Solution:
\(\frac{3 x^{2}+1}{3 x^{2}}\)
= \(\frac{3 x^{2}}{3 x^{2}}+\frac{1}{3 x^{2}}\)
= \(1+\frac{1}{3 x^{2}}\)
∴ \(\frac{3 x^{2}+1}{3 x^{2}}=1+\frac{1}{3 x^{2}}\)

Question 18.
\(\frac{3 x}{3 x+2}=\frac{1}{2}\)
Solution:
\(\frac{3 x}{3 x+2}=\frac{3 x}{3 x+2}\)

Question 19.
\(\frac{3}{4 x+3}=\frac{1}{4 x}\)
Solution:
\(\frac{3}{4 x+3}=\frac{3}{4 x+3}\)

Question 20.
\(\frac{4 x+5}{4 x}=5\)
Solution:
\(\frac{4 x+5}{4 x}\)
= \(\frac{4 x}{4 x}+\frac{5}{4 x}\)
= \(1+\frac{5}{4 x}\)
∴ \(\frac{4 x+5}{4 x}=1+\frac{5}{4 x}\)

Question 21.
\(\frac{7 x+5}{5}=7 x\)
Solution:
\(\frac{7 x+5}{5}\)
= \(\frac{7 x}{5}+\frac{5}{5}\)
= \(\frac{7 x}{5}+1\)
∴ \(\frac{7 x+5}{5}=\frac{7 x}{5}+1\)