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) = 3x^{2} + 2

Solution:

x(3x + 2) = 3x^{2} + 2

L.H.S. = x(3x + 2) = 3x^{2} + 2x

3x^{2} + 2x ≠ 3x^{2} + 2

∴ Correct statement x(3x + 2) = 3x^{2} + 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 = 5x^{2}

Solution:

3x + 2x = 5x^{2}

L.H.S. = 3x + 2x = 5x

5x ≠ 5x^{2}

∴ Correct statement 3x + 2x = 5x

Question 7.

(2x)^{2} + 4(2x) + 7 = 2x^{2} + 8x + 7

Solution:

(2x)^{2} + 4(2x) + 7 = 2x^{2} + 8x + 7

L.H.S. = (2x)^{2} + 4(2x) + 7

= 4x^{2} + 8x + 7

4x^{2} + 8x + 7 ≠ 2x^{2} + 8x + 7

Correct statement (2x)^{2} + 4(2x) + 7 = 4x^{2} + 8x + 7

Question 8.

(2x)^{2} + 5x = 4x + 5x = 9x

Solution:

(2x)^{2} + 5x = 4x + 5x = 9x

L.H.S. = (2x)^{2} + 5x = 4x^{2} + 5x

4x^{2} + 5x ≠ 4x + 5x = 9x

∴ Correct statement (2x)^{2} + 5x = 4x^{2} + 5x

Question 9.

(3x + 2)^{2} = 3x^{2} + 6x + 4

Solution:

(3x + 2)^{2} = 3x^{2} + 6x + 4

L.H.S. = (3x + 2)^{2}

= (3x)^{2} + 2(3x) (2) + (2)^{2}

= 9x^{2} + 12x + 4

∴ Correct statement (3x + 2)^{2} = 9x^{2} + 12x + 4

Question 10.

Substituting x = -3 in

(a) x^{2} + 5x + 4 gives (-3)^{2} + 5(-3) + 4 = 9 + 2 + 4 = 15

(b) x^{2} – 5x + 4 gives (-3)^{2} – 5(-3) + 4 = 9 – 15 + 4 = -2

(c) x^{2} + 5x gives (-3)^{2} + 5 (-3) = -9 – 15 = -24

Solution:

(a) x^{2} + 5x + 4

When x = -3

x^{2} + 5x + 4

= (-3)^{2} + 5(-3) + 4

= 9 – 15 + 4

= 13 – 15

= -2

It gives -2

(b) x^{2} – 5x + 4

When x = -3

x^{2} – 5x + 4

= (-3)^{2} – 5(-3) + 4

= 9 + 15 + 4

= 28

It gives 28

(c) x^{2} + 5x

When x = -3

x^{2} + 5x

= (-3)^{2} + 5(-3)

= 9 – 15

= -6

It gives -6

Question 11.

(y – 3)^{2} = y^{2} – 9

Solution:

(y – 3)^{2
}= (y)^{2} – 2(y) (3) + (3)^{2}

= y^{2} – 6y + 9

Question 12.

[z + 5]^{2} = z^{2} + 25

Solution:

(z + 5)^{2}

= z^{2} + 2(z) (5) + (5)^{2}

= z^{2} + 10z + 25

Question 13.

(2a + 3b) (a – b) = 2a^{2} – 3b^{2}

Solution:

(2a + 3) (a – b)

= 2a × a – 2a × b + 3a – 3b

= 2a^{2} – 2ab + 3a – 3b

Question 14.

(a + 4) (a + 2) = a^{2} + 8

Solution:

(a + 4) (a + 2)

= a × a + 2a + 4a + 4 × 2

= a^{2} + 6a + 8

Question 15.

(a – 4) (a – 2) = a^{2} – 8

Solution:

(a – 4) (a – 2)

= a(a – 2) – 4(a – 2)

= a^{2} – 2a – 4a + 8

= a^{2} – 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:

\(\frac{7 x+5}{5}\)

= \(\frac{7 x}{5}+\frac{5}{5}\)

= \(\frac{7 x}{5}+1\)

∴ \(\frac{7 x+5}{5}=\frac{7 x}{5}+1\)