# Factorisation : Exercise 14.4 (Mathematics NCERT Class 8th)

Find and correct the errors in the following mathematical statements.
Q.1. $4(x - 5) = 4x - 5$
Sol. Given, $4(x - 5) = 4x - 5$
LHS = $4(x - 5)$
= $4x - 20$
≠RHS
Correct statement: $4(x - 5) = 4x - 20$

Q.2. $x(3x + 2) = 3{x^2} + 2$
Sol. Given, $x(3x + 2) = 3{x^2} + 2$
LHS = $x(3x + 2)$
= $3{x^2} + 2x$
≠RHS
Correct statement: $x(3x + 2) = 3{x^2} + 2x$

Q.3. $2x + 3y = 5xy$
Sol. Given, $2x + 3y = 5xy$
LHS = $2x + 3y$
≠RHS
Correct statement: $2x + 3y = 2x + 3y$

Q.4. $x + 2x + 3x = 5x$
Sol. Given, $x + 2x + 3x = 5x$
LHS = $x + 2x + 3x$
= $6x$
≠RHS
Correct statement: $x + 2x + 3x = 6x$

Q.5 $5y + 2y + y - 7y = 0$
Sol. Given, $5y + 2y + y - 7y = 0$
LHS = $5y + 2y + y - 7y$
= $8y - 7y$
= $y$
≠RHS
Correct statement: $5y + 2y + y - 7y = y$

Q.6 $3x + 2x = 5{x^2}$
Sol. Given, $3x + 2x = 5{x^2}$
LHS = $3x + 2x$
= $5x$
≠RHS
Correct statement: $3x + 2x = 5x$

Q.7 ${(2x)^2} + 4(2x) + 7 = 2{x^2} + 8x + 7$
Sol. Given, ${(2x)^2} + 4(2x) + 7 = 2{x^2} + 8x + 7$
LHS = ${(2x)^2} + 4(2x) + 7$
= $4{x^2} + 8x + 7$
≠RHS
Correct statement: ${(2x)^2} + 4(2x) + 7 = 4{x^2} + 8x + 7$

Q.8 ${(2x)^2} + 5x = 4x + 5x = 9x$
Sol. Given, ${(2x)^2} + 5x = 4x + 5x = 9x$
LHS = ${(2x)^2} + 5x$
= $4{x^2} + 5x$
≠RHS
Correct statement: ${(2x)^2} + 5x = 4{x^2} + 5x$

Q.9 ${(3x + 2)^2} = 3{x^2} + 6x + 4$
Sol. Given, ${(3x + 2)^2} = 3{x^2} + 6x + 4$
LHS = ${(3x + 2)^2}$
= $9{x^2} + 12x + 4$
≠RHS
Correct statement: ${(3x + 2)^2} = 9{x^2} + 12x + 4$

Q.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 ${( - 3)^2} + 5( - 3) = - 9 - 15 = - 24$
Sol. (a) ${x^2} + 5x + 4$gives ${( - 3)^2} + 5( - 3) + 4 = 9 + 2 + 4 = 15$
LHS = ${x^2} + 5x + 4$
Substituting , we get,
= ${( - 3)^2} + 5( - 3) + 4$
= $9 - 15 + 4$
= $- 2$
≠RHS
Correct statement: ${x^2} + 5x + 4$gives ${( - 3)^2} + 5( - 3) + 4 = 9 - 15 + 4 = - 2$

(b) ${x^2} - 5x + 4$gives ${( - 3)^2} - 5( - 3) + 4 = 9 - 15 + 4 = - 2$
LHS = ${x^2} - 5x + 4$
Substituting , we get,
= ${( - 3)^2} - 5( - 3) + 4$
= $9 + 15 + 4$
= $28$
≠RHS
Correct statement: ${x^2} - 5x + 4$gives ${( - 3)^2} - 5( - 3) + 4 = 9 + 15 + 4 = 28$

(c) ${x^2} + 5x$gives ${( - 3)^2} + 5( - 3) = - 9 - 15 = - 24$
LHS = ${x^2} + 5x$
Substituting , we get,
= ${( - 3)^2} + 5( - 3)$
= $9 - 15$
= $- 6$
≠RHS
Correct statement: ${x^2} + 5x$gives ${( - 3)^2} + 5( - 3) = 9 - 15 = - 6$

Q.11 ${(y - 3)^2} = {y^2} - 9$
Sol. LHS = ${(y - 3)^2}$
= ${y^2} - 6y + 9$
≠RHS
Correct statement: ${(y - 3)^2} = {y^2} - 6y + 9$

Q.12 ${(z + 5)^2} = {z^2} + 25$
Sol. LHS = ${(z + 5)^2}$
= ${z^2} + 10z + 25$
≠RHS
Correct statement: ${(y - 3)^2} = {y^2} - 6y + 9$

Q.13 $(2a + 3b)(a - b) = 2{a^2} - 3{b^2}$
Sol. LHS = $(2a + 3b)(a - b)$
= $2{a^2} - 2ab + 3ab - 3{b^2}$
= $2{a^2} + ab - 3{b^2}$
≠RHS
Correct statement: $(2a + 3b)(a - b) = 2{a^2} + ab - 3{b^2}$

Q.14 $(a + 4)(a + 2) = {a^2} + 8$
Sol. LHS = $(a + 4)(a + 2)$
= ${a^2} + 2a + 4a + 8$
= ${a^2} + 6a + 8$
≠RHS
Correct statement: $(a + 4)(a + 2) = {a^2} + 6a + 8$

Q.15 $(a - 4)(a - 2) = {a^2} - 8$
Sol. LHS = $(a - 4)(a - 2)$
= ${a^2} - 2a - 4a + 8$
= ${a^2} - 6a + 8$
≠RHS
Correct statement: $(a - 4)(a - 2) = {a^2} - 6a + 8$

Q.16 ${{3{x^2}} \over {3{x^2}}} = 0$
Sol. LHS = ${{3{x^2}} \over {3{x^2}}}$
= $1$
≠RHS
Correct statement: ${{3{x^2}} \over {3{x^2}}} = 1$

Q.17 ${{3{x^2} + 1} \over {3{x^2}}} = 1 + 1 = 2$
Sol. LHS = ${{3{x^2} + 1} \over {3{x^2}}}$
= ${{3{x^2}} \over {3{x^2}}} + {1 \over {3{x^2}}}$
= $1 + {1 \over {3{x^2}}}$
≠RHS
Correct statement: ${{3{x^2} + 1} \over {3{x^2}}} = 1 + {1 \over {3{x^2}}}$

Q.18 ${{3x} \over {3x + 2}} = {1 \over 2}$
Sol. LHS = ${{3x} \over {3x + 2}}$
≠RHS
Correct statement: ${{3x} \over {3x + 2}} = {{3x} \over {3x + 2}}$

Q.19 ${3 \over {4x + 3}} = {1 \over {4x}}$
Sol. LHS = ${3 \over {4x + 3}}$
≠RHS
Correct statement: ${3 \over {4x + 3}} = {3 \over {4x + 3}}$

Q.20 ${{4x + 5} \over {4x}} = 5$
Sol. LHS = ${{4x + 5} \over {4x}}$
= ${{4x} \over {4x}} + {5 \over {4x}}$
= $1 + {5 \over {4x}}$
≠RHS
Correct statement: ${{4x + 5} \over {4x}} = 1 + {5 \over {4x}}$

Q.21 ${{7x + 5} \over 5} = 7x$
Sol. LHS = ${{7x + 5} \over 5}$
= ${{7x} \over 5} + {5 \over 5}$
= ${{7x} \over 5} + 1$
≠RHS
Correct statement: ${{7x + 5} \over 5} = {{7x} \over 5} + 1$

• Praveen

Thank you so much

• Nandini

Thanks you so much

• Gokulavarshini

Thanks for giving step by step answers

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