NCERT Solutions for Class 8 Maths Chapter 14 - Exercise

Exercise 14.4

Refer the page 15 to 20 for the exercise 14.4 in the PDF

1. Find and correct the errors in the statement\[{\mathbf{4}}\left( {{\mathbf{x}} - {\mathbf{5}}} \right) = {\mathbf{4x}} - {\mathbf{5}}\].

Ans:  From the above question we get,

$4(x - 5) = 4x - 20$

Thus, $4x - 20 \ne 4x - 5$

So, the correct statement is $4x - 20$.

2. Find and correct the errors in the statement\[{\mathbf{x}}\left( {{\mathbf{3x}} + {\mathbf{2}}} \right) = {\mathbf{3}}{{\mathbf{x}}^{\mathbf{2}}} + {\mathbf{2}}\]

Ans: From the above question we get,

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

Again, $3{x^2} + 2x \ne 3{x^2} + 2$

So, the correct answer is $3{x^2} + 2x$

3. Find and correct the errors in the statement: \[{\mathbf{2x}} + {\mathbf{3y}} = {\mathbf{5xy}}\]

Ans: \[L.H.S = 2x + 3y \ne R.H.S\]

The correct statement is \[2x + 3y = 2x + 3y\]

4. Find and correct the errors in the statement: x + 2x + 3x = 5x

Ans: \[L.H.S = x + 2x + 3x = 1x + 2x + 3x = x\left( {1 + 2 + 3} \right) = 6x \ne {\text{R}}.H.S.\]

The correct statement is x + 2x + 3x = 6x

5. Find and correct the errors in the statement: \[{\mathbf{5y}} + {\mathbf{2y}} + {\mathbf{y}} - {\mathbf{7y}} = {\mathbf{0}}\]

Ans: \[L.H.S. = 5y + 2y + y - {\text{7}}y = 8y - 7y = y \ne R.H.S\]

The correct statement is 5y + 2y + y − 7y = y

6. Find and correct the errors in the statement: 3x + 2x = 5x 2

Ans: \[L.H.S. = 3x + 2x = 5x \ne {\text{R}}.H.S\]

The correct statement is \[3x + 2x = 5x\]

7. Find and correct the errors in the statement: \[{\left( {{\mathbf{2x}}} \right)^{\mathbf{2}}} + {\mathbf{4}}\left( {{\mathbf{2x}}} \right) + {\mathbf{7}} = {\mathbf{2}}{{\mathbf{x}}^{\mathbf{2}}} + {\mathbf{8x}} + {\mathbf{7}}\]

Answer:\[\;L.H.S = {\left( {2x} \right)^2} + 4\left( {2x} \right) + 7 = 4{x^2} + 8x + 7 \ne R.H.S\]

The correct statement is \[{\left( {2x} \right)^2} + 4\left( {2x} \right) + 7 = 4{x^2} + 8x + 7\]

8. Find and correct the errors in the statement:

Ans:\[L.H.S = {\left( {2x} \right)^2} + 5x = 4{x^2} + 5x \ne R.H.S.\]The correct statement is \[{\left( {2x} \right)^2} + 5x = 4{x^2} + 5x\]

9. Find and correct the errors in the statement: (3x + 2)2 = 3x 2 + 6x + 4

Ans: \[L.H.S. = {\left( {3x + 2} \right)^2} = {\left( {3x} \right)^2} + 2\left( {3x} \right)\left( 2 \right) + {\left( 2 \right)^2}\]

\[\left[ {{{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2}} \right] = 9{x^2} + 12x + 4 \ne R.H.S\]

The correct statement is \[{\left( {3x + 2} \right)^2} = 9{x^2} + {\text{1}}2x + 4\]

10. Find and correct the errors in the following mathematical statement. Substituting \[{\mathbf{x}} =  - {\mathbf{3}}\]in

a) \[{{\mathbf{x}}^{\mathbf{2}}} + {\mathbf{5x}} + {\mathbf{4}}\] gives \[{\left( { - {\mathbf{3}}} \right)^{\mathbf{2}}} + {\mathbf{5}}\left( { - {\mathbf{3}}} \right) + {\mathbf{4}} = {\mathbf{9}} + {\mathbf{2}} + {\mathbf{4}} = {\mathbf{15}}\]

Ans: For\[x =  - 3\], 

\[{x^2} + 5x + 4\]

\[ = {\left( { - 3} \right)^2} + 5\left( { - 3} \right) + 4\]

\[ = 9 - 15 + 4\]

\[ = 13 - 15\]

\[ =  - 2\]

b) \[{{\mathbf{x}}^{\mathbf{2}}} - {\mathbf{5x}} + {\mathbf{4}}\] gives \[{\left( { - {\mathbf{3}}} \right)^{\mathbf{2}}} - {\mathbf{5}}\left( { - {\mathbf{3}}} \right) + {\mathbf{4}} = {\mathbf{9}} - {\mathbf{15}} + {\mathbf{4}} =  - {\mathbf{2}}\]

Ans: For \[x =  - 3\],

\[\;{x^2} - 5x + 4\]

\[ = {\left( { - 3} \right)^2} - 5\left( { - 3} \right) + 4\]

\[ = 9 + 15 + 4\]

\[ = 28\]

c) \[{{\mathbf{x}}^{\mathbf{2}}} + {\mathbf{5x}}\] gives \[{\left( { - {\mathbf{3}}} \right)^{\mathbf{2}}} + {\mathbf{5}}\left( { - {\mathbf{3}}} \right) =  - {\mathbf{9}} - {\mathbf{15}} =  - {\mathbf{24}}\]

Ans:For \[x =  - 3\],

\[{x^2} + 5x\]

\[ = {\left( { - 3} \right)^2} + 5\left( { - 3} \right)\]

\[ = 9 - 15\]

\[ =  - 6\]

11. Find and correct the errors in the statement: \[{\left( {{\mathbf{y}} - {\mathbf{3}}} \right)^{\mathbf{2}}} = {\mathbf{y}} - {\mathbf{9}}\]

Ans: \[L.H.S = \left( {y - 3} \right)2 = {\left( y \right)^2} - 2\left( y \right)\left( 3 \right) + {\left( 3 \right)^2}\left[ {{{\left( {a - b} \right)}^2} = {a^2} - 2ab + {b^2}} \right] = {y^2} - 6y + 9 \ne R.H.S\,\]The correct statement is \[{\left( {y - 3} \right)^2} = {y^2} - 6y + 9\]

12. Find and correct the errors in the statement: \[{\left( {{\mathbf{z}} + {\mathbf{5}}} \right)^{\mathbf{2}}} = {{\mathbf{z}}^{\mathbf{2}}} + {\mathbf{25}}\]

Ans: \[L.H.S = {\left( {z + 5} \right)^2} = {\left( z \right)^2} + 2\left( z \right)\left( 5 \right) + {\left( 5 \right)^2}\left[ {{{\left( {a{\text{ }} + {\text{ }}b} \right)}^2} = {a^2} + 2ab + {b^2}} \right] = {z^2} + 10z + 25 \ne R.H.S\]The correct statement is \[{\left( {z + 5} \right)^2} = {z^2} + {\text{1}}0z + 25\]

13. Find and correct the errors in the statement: \[\left( {{\mathbf{2a}} + {\mathbf{3b}}} \right)\left( {{\mathbf{a}} - {\mathbf{b}}} \right) = {\mathbf{2}}{{\mathbf{a}}^{\mathbf{2}}} - {\mathbf{3}}{{\mathbf{b}}^{\mathbf{2}}}\]

Ans: \[L.H.S. = \left( {2a + 3b} \right)\left( {a - b} \right) = 2a \times a + 3b \times a - 2a \times b - 3b \times b = 2{a^2} + 3ab - 2ab - 3{b^2} = 2{a^2} + {\text{a}}b - 3{b^2} \ne R.H.S.\]The correct statement is \[\left( {2a + 3b} \right)\left( {a - b} \right) = 2a2 + {\text{a}}b - 3{b^2}\]

14. Find and correct the errors in the statement: \[\left( {{\mathbf{a}} + {\mathbf{4}}} \right)\left( {{\mathbf{a}} + {\mathbf{2}}} \right) = {{\mathbf{a}}^{\mathbf{2}}} + {\mathbf{8}}\]

Ans:\[L.H.S. = \left( {a + 4} \right)\left( {a + 2} \right) = {\left( a \right)^2} + \left( {4 + 2} \right)\left( a \right) + 4 \times 2 = {a^2} + 6a + 8 \ne R.H.S\]correct statement is \[\left( {a + 4} \right)\left( {a + 2} \right) = {a^2} + 6a + 8\]

15. Find and correct the errors in the statement: \[\left( {{\mathbf{a}} - {\mathbf{4}}} \right)\left( {{\mathbf{a}} - {\mathbf{2}}} \right) = {{\mathbf{a}}^{\mathbf{2}}} - {\mathbf{8}}\]

Ans: \[L.H.S. = \left( {a - 4} \right)\left( {a - 2} \right) = {\left( a \right)^2} + \left[ {\left( { - 4} \right) + \left( { - 2} \right)} \right]\left( a \right) + \left( { - 4} \right)\left( { - 2} \right) = {a^2} - 6a + 8 \ne R.H.S.\]

The correct statement is \[\left( {a - 4} \right)\left( {a - 2} \right) = {a^2} - {\text{6}}a + 8\]

16. Find and correct the errors in the statement $\dfrac{{3{x^2}}}{{3{x^2}}} = 0$

Ans: L.H.S = The correct statement is

LHS

$\dfrac{{3{x^2}}}{{3{x^2}}} = \dfrac{{3 \times x \times x}}{{3 \times x \times x}} = 1 \ne R.H.S.$

The correct statement is$\dfrac{{3{x^2}}}{{3{x^2}}} = 1$

17. Find and correct the errors in the statement: $\dfrac{{3{x^2} + 1}}{{3{x^2}}} = 1 + 1 = 2$

Ans: The correct statement is

$\dfrac{{3{x^2} + 1}}{{3{x^2}}}$

$ = \dfrac{{3{x^2}}}{{3{x^2}}} + \dfrac{1}{{3{x^2}}}$

$ = 1 + \dfrac{1}{{3{x^2}}}$

$ \ne R.H.S$

The correct statement is $\dfrac{{3{x^2} + 1}}{{3{x^2}}} = 1 + \dfrac{1}{{3{x^2}}}$

18. Find and correct the errors in the statement: $\dfrac{{3x}}{{3x + 2}} = \dfrac{1}{2}$

Ans: L.H.S =$\dfrac{{3x}}{{3x + 2}} \ne R.H.S.$

The correct statement is $\dfrac{{3x}}{{3x + 2}} = \dfrac{{3x}}{{3x + 2}}$

19. Find and correct the errors in the statement: $\dfrac{3}{{4x + 3}} = \dfrac{1}{{4x}}$

Ans: L.H.S. =$\dfrac{3}{{4x + 3}}$ ≠ R.H.S. 

The correct statementis $\dfrac{3}{{4x + 3}} = \dfrac{3}{{4x + 3}}$

20. Find and correct the errors in the statement: $\dfrac{{4x + 5}}{{4x}} = 5$

Ans: L.H.S. $\dfrac{{4x + 5}}{{4x}} = \dfrac{{4x}}{{4x}} + \dfrac{5}{{4x}} = 1 + \dfrac{5}{{4x}} \ne R.H.S$

The correct statement is$\dfrac{{4x + 5}}{{4x}} = 1 + \dfrac{5}{{4x}}$

21. Find and correct the errors in the statement: $\dfrac{{7x + 5}}{5} = 7x$

Ans: L.H.S. =$\dfrac{{7x + 5}}{5} = \dfrac{{7x}}{5} + \dfrac{5}{5} = \dfrac{{7x}}{5} + 1 \ne R.H.S$

The correct statement is$\dfrac{{7x + 5}}{5} = \dfrac{{7x}}{5} + 1$

NCERT Solutions for Class 8 Maths Chapter 14 Factorisation (Ex 14.4) Exercise 14.4

Opting for the NCERT solutions for Ex 14.4 Class 8 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 14.4 Class 8 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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