Exercise 14.3 Factorization- NCERT Solutions Class 8

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Chapter 14 Ex.14.3 Question 1

Carry out the following divisions.

(i)\(\begin{align}\quad 28{x^4} \div 56x\end{align}\)

(ii)\(\begin{align}\quad - 36{y^3} \div 9{y^2}\end{align}\)

(iii)\(\begin{align}\quad 66p{q^2}{r^3} \div 11q{r^2}\end{align}\)

(iv)\(\begin{align}\quad 34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3}\end{align}\)

(v)\(\begin{align}\quad 12{a^8}{b^8} \div ( { - 6{a^6}{b^4}} )\end{align}\)

Solution

Video Solution

(i)\(\,28{x^4} \div 56x\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \(28{x}^4\) and \(56{x}\) then cancel out common factor of \(28{x}^4\) and \(56{x}\).

Steps:

\(28{x^4}\) can be written as

\(28{x^4} = 2 \times 2 \times 7 \times x \times x \times x \times x\) 

and \(56x\) can be written as

\(56x = 2 \times 2 \times 2 \times 7 \times x\)

Then,

\[\begin{align} & 28{x^4} \div 56x \\ \\ &= \frac{2 \times 2 \times 7 \times x \times x \times x \times x  }{{2 \times 2 \times 2 \times 7 \times x}}\\ &= \frac{{{x^3}}}{2}\\ &= \frac{1}{2}{x^3}\end{align}\]

(ii) \(\,- 36{y^3} \div 9{y^2}\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \(-36{y^3}\) and \(9{y^2}\) then cancel out common factor of \(-36{y^3}\) and \(9{y^2}\)

\(- 36{y^3}\) can be written as

\(- 2 \times 2 \times 3 \times 3 \times y \times y \times y\)

and \(9{y^2}\) can be written as

\(3 \times 3 \times y \times y\)

Then,

\[\begin{align} & - 36{y^3} \div 9{y^2}\\ \\ &= \frac{{ - 2 \times 2 \times 3 \times 3 \times y \times y \times y}}{{3 \times 3 \times y \times y}}\\ &= - 4y\end{align}\]

(iii) \(\;66p{q^2}{r^3} \div 11q{r^2}\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \(66p{q^2}\) and \(11p{r^2}\) then cancel out common factor of \(66p{q^2}\) and \(11q{r^2}\)

\(66p{q^2}{r^3}\) can be written as

\(2 \times 3 \times 11 \times p \times q \times q \times r \times r \times r\)

and \(11q{r^2}\) can be written as

\(11 \times q \times r \times r\)

Then,

\[\begin{align} & 66p{q^2}{r^3} \div 11q{r^2} \\ \\ &= \frac{{\begin{pmatrix} 2 \times 3 \times 11 \times p \times \\ q \times q \times r \times r \times r \end{pmatrix} }}{{11 \times q \times r \times r}}\\&= 6pqr\end{align}\]

(iv) \(\,34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3}\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \(34{x^3}{y^3}{z^3}\) and \(51x{y^2}{z^3}\) then cancel out common factor of \(34{x^3}{y^3}{z^3}\) and \(51x{y^2}{z^3}\)

\(34{x^3}{y^3}{z^3}\) can be written as

\(\begin{pmatrix} 2 \times 17 \times x \times x \times x \times  \\ y \times y \times y \times z \times z \times z \end{pmatrix} \)

and \(51x{y^2}{z^3}\) can be written as

\(3 \times 17 \times x \times y \times y \times z \times z \times z\)

Then,

\[\begin{align}& 34{x^3}{y^3}{z^3} \div 51x{y^2}{z^3} \\ \\&= \frac{{ \begin{pmatrix}2 \times 17 \times x \times x \times x \times \\ y \times y \times y \times z \times z \times z \end{pmatrix} }}{{\begin{pmatrix}3 \times 17 \times x \times \\ y \times y \times z \times z \times z \end{pmatrix} }}\\&= \frac{2}{3}{x^2}y\end{align}\]

(v) \(\;12{a^8}{b^8} \div \left( { - 6{a^6}{b^4}} \right)\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \(12{a^8}{b^8}\) and \(- 6{a^6}{b^4}\) then cancel out common factor of \(- 6{a^6}{b^4}\) and \(51x{y^2}{z^3}\)

\(12{a^8}{b^8}\) can be written as

\(2 \times 2 \times 3 \times {a^8} \times {b^8}\)

and \(- 6{a^6}{b^4}\) can be written as

\(- 2 \times 3 \times {a^6} \times {b^4}\)

Then,

\[\begin{align} &12{a^8}{b^8} \div \left( { - 6{a^6}{b^4}} \right) \\ &= \frac{{2 \times 2 \times 3 \times {a^8} \times {b^8}}}{{ - 2 \times 3 \times {a^6} \times {b^4}}}\\&= - 2{a^2}{b^4}\end{align}\]

Chapter 14 Ex.14.3 Question 2

Divide the given polynomial by the given monomial.

(i) \(\begin{align} \quad \left( {5{x^2} - 6x} \right) \div 3x\end{align}\)

(ii) \(\begin{align} \quad \left( {3{y^8} - 4{y^6} + 5{y^4}} \right) \div {y^4}\end{align}\)

(iii) \(\begin{align} \begin{Bmatrix} 8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right) \\ \div 4{x^2}{y^2}{z^2} \end{Bmatrix} \end{align}\)

(iv) \(\begin{align} \quad \left( {{x^3} + 2{x^2} + 3x} \right) \div 2x\end{align}\)

(v) \(\begin{align} \quad \left( {{p^3}{q^6} - {p^6}{q^3}} \right) \div {p^3}{q^3}\end{align}\)

Solution

Video Solution

\({\rm{(i) }} \;\left( {5{x^2} - 6x} \right) \div 3x\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor \(({5{x^2} - 6x})\) and \(3{x}\) then cancel out common factors of \(({5{x^2} - 6x})\) and \(3{x}\).

Steps:

\(5{x^2} - 6x\) can be written as \(x(5x - 6)\)

Then,

\[\begin{align} \left( {5{x^2} - 6x} \right) \div 3x &= \frac{{x(5x - 6)}}{{3x}}\\&= \frac{1}{3}(5x - 6)\end{align}\]

\({\rm{(ii) }}\;( {3{y^8} - 4{y^6} + 5{y^4}}) \div {y^4}\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor \(( {3{y^8} - 4{y^6} + 5{y^4}})\) and \({y^4}\) then cancel out common factors of \(( {3{y^8} - 4{y^6} + 5{y^4}})\) and \({y^4}\).

Steps:

\(3{y^8} - 4{y^6} + 5{y^4}\) can be written as \({y^4}\left( {3{y^4} - 4{y^2} + 5} \right)\)

Then,

\[\begin{align} & \left( {3{y^8} - 4{y^6} + 5{y^4}} \right) \div {y^4} \\ & = \frac{{{y^4}\left( {3{y^4} - 4{y^2} + 5} \right)}}{{{y^4}}}\\&= 3{y^4} - 4{y^2} + 5\end{align}\]

(iii) 

\(\ \begin{align} \begin{Bmatrix} 8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right) \\ \div 4{x^2}{y^2}{z^2} \end{Bmatrix} \end{align}\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor \(8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right)\) and \(4{x^2}{y^2}{z^2}\) then cancel out common factors of \(8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right)\) and \(4{x^2}{y^2}{z^2}\).

Steps:

\(8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right)\) can be written as \(8{x^2}{y^2}{z^2}(x + y + z)\)

Then,

\[\begin{align}& \begin{Bmatrix} 8\left( {{x^3}{y^2}{z^2} + {x^2}{y^3}{z^2} + {x^2}{y^2}{z^3}} \right) \\ \div 4{x^2}{y^2}{z^2}\end{Bmatrix} \\ \\ &= \frac{{8{x^2}{y^2}{z^2}(x + y + z)}}{{4{x^2}{y^2}{z^2}}}\\&= 2(x + y + z)\end{align}\]

\({\rm{(iv)}} \,({{x^3} + 2{x^2} + 3x}) \div 2x\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor \(({{x^3} + 2{x^2} + 3x})\) and \(2{x}\) then cancel out common factors of \(({{x^3} + 2{x^2} + 3x})\) and \(2{x}\).

Steps:

\({x^3} + 2{x^2} + 3x\) can be written as \(x\left( {{x^2} + 2x + 3} \right)\)

Then,

\[\begin{align} & \left( {{x^3} + 2{x^2} + 3x} \right) \div 2x\\ \\ &= \frac{{x\left( {{x^2} + 2x + 3} \right)}}{{2x}}\\&= \frac{1}{2}\left( {{x^2} + 2x + 3} \right)\end{align}\]

\(({\rm{v}})\;( {{p^3}{q^6} - {p^6}{q^3}}) \div {p^3}{q^3}\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor \(({{p^3}{q^6} - {p^6}{q^3}})\) and \({p^3}{q^3}\) then cancel out common factors of \(( {{p^3}{q^6} - {p^6}{q^3}})\) and \({p^3}{q^3}\).

Steps:

\({p^3}{q^6} - {p^6}{q^3}\) can be written as \({p^3}{q^3}\left( {{q^3} - {p^3}} \right)\)

Then,

\[\begin{align} & \left( {{p^3}{q^6} - {p^6}{q^3}} \right) \div {p^3}{q^3} \\ \\ &= \frac{{{p^3}{q^3}\left( {{q^3} - {p^3}} \right)}}{{{p^3}{q^3}}}\\&= {q^3} - {p^3}\end{align}\]

Chapter 14 Ex.14.3 Question 3

 Work out the following divisions.

(i) \(\begin{align} \quad (10x - 25) \div 5\end{align}\)

(ii) \(\begin{align}\quad (10x - 25) \div (2x - 5)\end{align}\)

(iii) \(\begin{align}\quad 10y(6y + 21) \div 5(2y + 7)\end{align}\)

(iv) \(\begin{align}\quad 9{x^2}{y^2}(3z - 24) \div 27xy(z - 8)\end{align}\)

(v) \(\begin{align}\quad \begin{Bmatrix} 96abc(3a - 12)(5b - 30) \\ \div 144(a - 4)(b - 6) \end{Bmatrix} \end{align}\)

Solution

Video Solution

(i) \(\;(10x - 25) \div 5\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \((10x-25)\) then cancel out common factors of \((10x-25)\) and \(5\)

Steps:

Factorising \((10x-25)\), We get

\[\begin{align}(10x - 25) &= 5 \times 2 \times x - 5 \times 5\\&= 5\left( {2x - 5} \right)\end{align}\]

\[\begin{align}(10x - 25) \div 5 &= \frac{{5(2x - 5)}}{5}\\& = 2x - 5\end{align}\]

(ii)  \((10x - 25) \div (2x - 5)\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \((10x-25)\) then cancel out common factors of \((10x-25)\) and \((2x-5)\).

Steps:

Factorising \((10x-25)\), We get

\[\begin{align}(10x - 25) &= 5 \times 2 \times x - 5 \times 5\\&= 5\left( {2x - 5} \right)\end{align}\]

\[\begin{align} & (10x - 25) \div (2x -5) \\ &=\frac{{5(2x - 5)}}{{2x - 5}}\\&= 5\end{align}\]

(iii) \(\;10y(6y + 21) \div 5(2y + 7)\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \(10y(6y + 21)\) then cancel out common factors of \(10y(6y + 21)\) and \(5(2y + 7)\).

Steps:

Factorising \(10y(6y + 21)\), We get

\[\begin{align} & 10y(6y + 21) \\ &= \begin{Bmatrix} 5 \times 2 \times y \times \\ \left( {2 \times 3 \times y + 3 \times 7} \right) \end{Bmatrix} \\&= 5 \times 2 \times y \times 3\left( {2 \times y + 7} \right)\\&= 30y\left( {2y + 7} \right)\end{align}\]

\[\begin{align} & 10y(6y + 21) \div 5(2y + 7)\\ \\ &= \frac{{30y(2y + 7)}}{{5(2y+7)}}\\&=6y\end{align}\]

(iv) \(\;9{x^2}{y^2}(3z - 24) \div 27xy(z - 8)\)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \(9{x^2}{y^2}(3z - 24)\) then cancel out common factors of \(9{x^2}{y^2}(3z - 24)\) and \(27xy(z - 8)\).

Steps:

Factorising \(9{x^2}{y^2}(3z - 24)\), We get

\[\begin{align}&9{x^2}{y^2}(3z - 24) \\ &= \begin{Bmatrix} 3 \times 3 \times x \times x \times y \times y \times \\ \left( {3 \times z - 2 \times 2 \times 2 \times 3} \right) \end{Bmatrix} \\&= \begin{Bmatrix} 3 \times 3 \times x \times x \times y \times y \times 3 \\ \left( {z - 2 \times 2 \times 2} \right) \end{Bmatrix} \\&= 27{x^2}{y^2}\left( {z - 8} \right)\end{align}\]

\[\begin{align} & 9{x^2}{y^2}(3z - 24) \div 27xy(z - 8) \\ \\ &= \frac{{27{x^2}{y^2} \times (z - 8)}}{{27xy(z - 8)}}\\&= xy\end{align}\]

(v) \(\, \begin{Bmatrix} 96abc(3a - 12)(5b - 30) \\ \div 144(a - 4)(b - 6) \end{Bmatrix} \)

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Find out factor of \(96abc(3a - 12)(5b - 30)\) then cancel out common factors of \(96abc(3a - 12)(5b - 30)\) and \(144(a - 4)(b - 6)\).

Steps:

Factorising \(96abc(3a - 12)(5b - 30)\), We get

\[\begin{align} & 96abc(3a - 12)(5b - 30) \\ &= \begin{Bmatrix} 96abc \times \left( {3 \times a - 2 \times 2 \times 3} \right) \\\times \left( {5 \times b - 5 \times 2 \times 3} \right) \end{Bmatrix} \\&= \begin{Bmatrix} 96abc \times 3\left( {a - 2 \times 2} \right) \\ \times 5\left( {b - 2 \times 3} \right) \end{Bmatrix} \\&= 1440abc\left( {a - 4} \right)\left( {b - 6} \right)\end{align}\]

\[\begin{align}  &\begin{Bmatrix} 96abc(3a - 12)(5b - 30) \\ \div 144(a - 4)(b - 6) \end{Bmatrix} \\ \\&={\frac{{1440abc\left( {a - 4} \right)\left( {b - 6} \right)}}{{144(a - 4)(b - 6)}}}\\&= {10abc}\end{align}\]

Chapter 14 Ex.14.3 Question 4

Divide as directed.

(i)\(\begin{align}\quad 5(2x + 1)(3x + 5) \div (2x + 1)\end{align}\)

(ii)\(\begin{align}\quad \begin{bmatrix} 26xy(x + 5)(y - 4) \\ \div 13x(y - 4) \end{bmatrix} \end{align}\)

(iii)\(\begin{align}\quad \begin{bmatrix} 52 pqr\; (p + q)(q + r)(r + p) \\ \div 104pq(q + r)(r + p) \end{bmatrix} \end{align}\)

(iv)\(\begin{align} \quad \begin{bmatrix} 20(y + 4)\left( {{y^2} + 5y + 3} \right) \\ \div 5(y + 4) \end{bmatrix} \end{align}\)

(v)\(\begin{align} \quad \begin{bmatrix} x(x + 1)(x + 2)(x + 3) \\ \div x(x + 1) \end{bmatrix} \end{align}\)

Solution

Video Solution

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Cancel out common factor of the followinng.

Steps:

\(\begin{align}\left( {\rm{i}} \right) \quad & 5(2x + 1)(3x + 5) \div (2x + 1) \\ &= \frac{{5(2x + 1)(3x + 1)}}{{(2x + 1)}}\\ & = 5(3x + 1)\end{align}\)

\(\begin{align}  \left( {{\rm{ii}}} \right) \quad & \begin{bmatrix} 26xy(x + 5)(y - 4) \\ \div 13x(y - 4) \end{bmatrix} \\ \\&= \frac{{\begin{bmatrix} 2 \times 13 \times \\ xy(x + 5)(y - 4) \end{bmatrix} }}{{13x(y - 4)}}\\ & = 2y(x + 5)\end{align}\)

\(\begin{align}{\rm{(iii)}} \quad & \begin{bmatrix} 52pqr(p + q)(q + r)(r + p) \\ \div 104pq(q + r)(r + p) \end{bmatrix}\\ \\ &= \frac{{\begin{bmatrix} 2 \times 2 \times 13 \times p \times q \times r  \times\\ (p + q) \times (q + r) \times (r + p) \end{bmatrix} }}{{ \begin{bmatrix} 2 \times 2 \times 2 \times 13 \times p \times  \\q\times (q + r) \times (r + p) \end{bmatrix} }}\\&= \frac{1}{2}r(p + q)\end{align}\)

\(\begin{align}(\rm{iv})\quad &\begin{bmatrix} 20(y + 4)\left( {{y^2} + 5y + 3} \right) \\ \div 5(y + 4) \end{bmatrix} \\ \\&= \frac{{ \begin{bmatrix}2 \times 2 \times 5 \times (y + 4) \\ \times \left( {{y^2} + 5y + 3} \right) \end{bmatrix} }}{{5 \times (y + 4)}}\\&= 4\left( {{y^2} + 5y + 3} \right)\end{align}\)

\(\begin{align} \left( {\rm{v}} \right) \quad  & \begin{bmatrix} x(x + 1)(x + 2)(x + 3) \\ \div x(x + 1) \end{bmatrix}\\  \\ &= \frac{{x(x + 1)(x + 2)(x + 3)}}{{x(x + 1)}}\\&= (x + 2)(x + 3)\end{align}\)

Chapter 14 Ex.14.3 Question 5

Factorize the expressions and divide them as directed.

(i)\(\begin{align} \quad \left( {{y^2} + 7y + 10} \right) \div (y + 5)\end{align}\)

(ii)\(\begin{align} \quad \left( {{m^2} - 14m - 32} \right) \div (m + 2)\end{align}\)

(iii)\(\begin{align}\quad   \left( {5{p^2} - 25p + 20} \right) \div (p - 1)\end{align}\)

(iv)\(\begin{align} \quad  \begin{Bmatrix} 4yz\left( {{z^2} + 6z - 16} \right) \\ \div 2y(z + 8) \end{Bmatrix} \end{align}\)

(v)\(\begin{align} \quad  5pq\left( {{p^2} - {q^2}} \right) \div 2p(p + q)\end{align}\)

(vi)\(\begin{align} \quad \begin{Bmatrix} 12xy\left( {9{x^2} - 16{y^2}} \right) \\ \div 4xy(3x + 4y) \end{Bmatrix} \end{align}\)

(vii)\(\begin{align} \quad  \begin{Bmatrix} 39{y^3}\left( {50{y^2} - 98} \right) \\ \div 26{y^2}(5y + 7) \end{Bmatrix} \end{align}\)

Solution

Video Solution

(i) \(\,\left( {{y^2} + 7y + 10} \right) \div (y + 5)\)

What is known?

Algebraic expression.

What is known?

Division of the algebraic expression.

Reasoning:

Factorise \(({{y^2} + 7y + 10})\) then cancel out common factor of \(( {{y^2} + 7y + 10})\) and \(({y+5})\).

Steps:

\(\begin{align} &({{y^2} + 7y + 10})\, \text{can be written as }\\  & \quad \; {y^2} + 2y + 5y + 10\\&= y(y + 2) + 5(y + 2)\\&= (y + 2)(y + 5) \end{align}\)

Then,

\[\begin{align} & \left( {{y^2} + 7y + 10} \right) \div (y + 5) \\ \\ &= \frac{{(y + 2)(y + 5)}}{{(y + 5)}}\\& = y + 2\end{align}\]

(ii)  \(\left( {{m^2} - 14m - 32} \right) \div (m + 2)\)

What is known?

Algebraic expression.

What is known?

Division of the algebraic expression.

Reasoning:

Factorise \(( {{m^2} - 14m - 32})\) then cancel out common factor of \(( {{m^2} - 14m - 32})\) and \((m + 2)\).

Steps:

\(\left( {{m^2} - 14m - 32} \right) \)  can be written as

 \(\begin{align} & {m^2} + 2m - 16m - 32\\&= m(m + 2) - 16(m + 2)\\&= (m + 2)(m - 16)\end{align}\)

Then,

\[\begin{align} & \left( {{m^2} - 14m - 32} \right) \div (m + 2) \\ \\&= \frac{{(m + 2)(m - 16)}}{{(m + 2)}}\\
&= m - 16\end{align}\]

(iii)  \(\left( {5{p^2} - 25p + 20} \right) \div (p - 1)\)

What is known?

Algebraic expression.

What is known?

Division of the algebraic expression.

Reasoning:

Factorise \({5{p^2} - 25p + 20}\) then cancel out common factor of \({5{p^2} - 25p + 20}\) and \((p - 1)\).

Steps:

\(\begin{align} & 5{p^2} - 25p + 20 \; \text{ can be written as } \\ &\quad 5\left( {{p^2} - 5p + 4} \right) \\&= {5\left( {{p^2} - p - 4p + 4} \right)}\\&= {5\left[ {p(p - 1) - 4(p - 1)} \right]}\\&= 5(p - 1)(p - 4)\\\end{align}\)

Then,

\[\begin{align} & \left( {5{p^2} - 25p + 20} \right) \div (p - 1) \\ \\ &= \frac{{5(p - 1)(p - 4)}}{{(p - 1)}}\\&= 5(p - 4)\end{align}\]

(iv)\(\,4yz\left( {{z^2} + 6z - 16} \right) \div 2y(z + 8)\)

What is known?

Algebraic expression.

What is known?

Division of the algebraic expression.

Reasoning:

Factorise \(4yz\left( {{z^2} + 6z - 16} \right)\) then cancel out common factor of \(4yz\left( {{z^2} + 6z - 16} \right)\) and \(2y(z + 8)\).

Steps:

\(4yz\left( {{z^2} + 6z - 16} \right)\)  can be written as 

\(\begin{align} &  4yz\left( {{z^2} - 2z + 8z - 16} \right)\\&= 4yz\left[ {z(z - 2) + 8(z - 2)} \right]\\&= 4yz(z - 2)(z + 8)\end{align}\)

Then,

\[\begin{align} & 4yz\left( {{z^2} + 6z - 16} \right) \div 2y(z + 8) \\ &= \frac{{4yz(z - 2)(z + 8)}}{{2y(z + 8)}}\\&= 2z(z - 2)\end{align}\]

(v) \(\,5pq\left( {{p^2} - {q^2}} \right) \div 2p(p + q)\)

What is known?

Algebraic expression.

What is known?

Division of the algebraic expression.

Reasoning:

Factorise \(5pq\left( {{p^2} - {q^2}} \right)\) by using identity \({a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\) then cancel out common factor of \(5pq\left( {{p^2} - {q^2}} \right)\) and \(2p(p + q)\).

Steps:

 \(5pq\left( {{p^2} - {q^2}} \right)\) can be written as \(5pq(p - q)(p + q)\)

Then, 

\[\begin{align} & 5pq\left( {{p^2} - {q^2}} \right) \div 2p(p + q) \\ \\ &= \frac{{5pq(p - q)(p + q)}}{{2p(p + q)}}\\&= \frac{5}{2}q(p - q)\end{align}\]

(vi)\(12xy\left( {9{x^2} - 16{y^2}} \right) \div 4xy(3x + 4y)\)

What is known?

Algebraic expression.

What is known?

Division of the algebraic expression.

Reasoning:

Factorise \(12xy\left( {9{x^2} - 16{y^2}} \right)\) by using identity \({a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\) then cancel out common factor of \(12xy\left( {9{x^2} - 16{y^2}} \right)\) and \(4xy(3x + 4y)\).

Steps:

\(12xy\left( {9{x^2} - 16{y^2}} \right)\) can be written as

\(\begin{align} & 12xy[ {{(3x)}^2} - {(4y)^2}]\\ &= 12xy(3x - 4y)(3x + 4y)\\& = \begin{Bmatrix} 2 \times 2 \times 3 \times x \times  y \times \\  (3x - 4y) \times (3x + 4y) \end{Bmatrix} \end{align}\)

Then,

\[\begin{align}& 12xy\left( {9{x^2} - 16{y^2}} \right)  \div 4xy(3x + 4y)  \\ \\& = \frac{{ \begin{Bmatrix} 2 \times 2 \times 3 \times x \times y \times \\  (3x - 4y) \times (3x + 4y) \end{Bmatrix} }}{{ 2 \times 2 \times x \times y  \times (3x + 4y)  }}\\ &= 3(3x - 4y)\end{align}\]

(vii)\(\;39{y^3}\left( {50{y^2} - 98} \right) \div 26{y^2}(5y + 7)\)

What is known?

Algebraic expression.

What is known?

Division of the algebraic expression.

Reasoning:

Factorise \(39{y^3}\left( {50{y^2} - 98} \right)\) by using identity \({a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\) then cancel out common factor of \(39{y^3}\left( {50{y^2} - 98} \right)\) and \(26{y^2}(5y + 7)\).

Steps:

\(39{y^3}\left( {50{y^2} - 98} \right)\) can be written as

\(\begin{align} &  3 \times 13 \times y \times y \times  y \times 2\left[ {\left( {25{y^2} - 49} \right)} \right]  \\&= \begin{Bmatrix} 3 \times 13 \times 2 \times y \times y \\ \times y \times \left[ {{{(5y)}^2} - {{(7)}^2}} \right] \end{Bmatrix} \\ &= \begin{Bmatrix} 3 \times 13 \times 2 \times y \times y \times\\  y(5y - 7)(5y + 7) \end{Bmatrix} \end{align}\)

\(26{y^2}(5y + 7)\) can be written as \(2 \times 13 \times y \times y \times (5y + 7)\)

Then,

\[\begin{align} & 39{y^3}\left( {50{y^2} - 98} \right) \div 26{y^2}(5y + 7) \\ \\&= \frac{{\begin{Bmatrix}3 \times 13 \times 2 \times y \times y \times\\ y(5y - 7)(5y + 7) \end{Bmatrix} }}{{ 2 \times 13 \times y \times y \times (5y + 7) }}\\& = 3y(5y - 7)\end{align}\]

Chapter 14 Ex.14.3 Question 6

Find and correct the errors in the statement: \(2x + 3y = 5xy\)

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(\rm L.H.S.\)

Steps:

\[\begin{align} {\rm{L.H.S.}} &= 2x + 3y = 2x + 3y\\ {\rm{L.H.S.}} &\ne {\rm{R.H.S.}}.\end{align}\]

Chapter 14 Ex.14.3 Question 7

Find and correct the errors in the statement: \(x+2x+3x=5x\)

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(\rm L.H.S.\)

Steps:

\[\begin{align} {\rm {L.H.S.}}  &= x + 2x + 3x = 1x + 2x + 3x\\ & = x(1 + 2 + 3) = 6x\\ {\rm {L.H.S.}} &\ne {\rm {R.H.S.}}\end{align}\]

The correct statement is \(x+2x+3x=6x\)

Chapter 14 Ex.14.3 Question 8

Find and correct the errors in the statement:

 \(5y + 2y + y - 7y = 0\)

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S. &= 5y + 2y + y - 7y \\&= 8y - 7y \\&= y \ne R.H.S. \end{align}\]

The correct statement is \(5y + 2y + y{\rm{ }} - 7y = y\)

Chapter 14 Ex.14.3 Question 9

Find and correct the errors in the statement:

\[3x + 2x = 5{x^2}\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S. &= 3x + 2x = 5x\\L.H.S. &\ne R.H.S.\end{align}\]

The correct statement is \(3x + 2x = 5x\)

Chapter 14 Ex.14.3 Question 10

Find and correct the errors in the statement:

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

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S. &= {(2x)^2} + 4(2x) + 7 \\&= 4{x^2} + 8x + 7\\L.H.S.& \ne R.H.S\end{align}\]

The correct statement is

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

Chapter 14 Ex.14.3 Question 11

Find and correct the errors in the statement:

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

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S.& = {(2x)^2} + 5x \\&= 4{x^2} + 5x\\L.H.S. &\ne R.H.S.\end{align}\]

The correct statement is

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

Chapter 14 Ex.14.3 Question 12

Find and correct the errors in the statement:

\[{\left( {3x + 2} \right)^2} = 3{x^2} + 6x + 4\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S. &= {(3x + 2)}^2 \\&= {(3x)}^2 + 2(3x)(2) + {(2)}^2 \\ {\text{Using identity}}\\{(a + b)}^2 &= a^2 + 2ab + b^2 \\&= 9{x^2} + 12x + 4\\\\L.H.S. &\ne R.H.S\end{align}\]

The correct statement is

\[{\left( {3x{\rm{ }} + {\rm{ }}2} \right)^2} = {\rm{ }}9{x^2}{\rm{ }} + {\rm{ }}12x{\rm{ }} + {\rm{ }}4\]

Chapter 14 Ex.14.3 Question 13

Find and correct the errors in the following mathematical statement.

Substituting \(x = - 3\) in

(a)\(\begin{align}\quad{x^2} + 5x + 4\end{align}\) gives

\(\begin{align}&={{( - 3)}^2} + 5( - 3) + 4 \\&= 9 + 2 + 4 \\&= 15\end{align}\)

(b)\(\begin{align}\quad{x^2} - 5x + 4\end{align}\) gives

\(\begin{align}&={( - 3)}^2 - 5( - 3) + 4 \\&= 9 - 15 + 4 \\&= - 2\end{align}\)

(c)\(\begin{align}\quad{x^2} + 5x\end{align}\) gives

\(\begin{align}&={{( - 3)}^2} + 5( - 3)\\&= - 9 - 15 \\&= - 24\end{align}\)

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Put value of \(x\) in \(L.H.S\) and find correct solution.

Steps:

(a) For \(x = -\, 3\)

\[\begin{align}L.H.S &= {x^2} + 5x + 4 \\&= {( - 3)^2} + 5( - 3) + 4\\&= 9 - 15 + 4\\& = 13 - 15\\& =  - 2\\L.H.S &\ne R.H.S\\\end{align}\]

The correct statement is \({x^2} + 5x + 4=- 2\)

(b) For \(x = -\, 3\)

\[\begin{align}{x^2} - 5x + 4 &= {( - 3)^2} - 5( - 3) + 4\\& = 9 + 15 + 4\\&= 28\\\end{align}\]

The correct statement is\({x^2} - 5x + 4 = 28\)

(c) For \(x = - 3\)

\[\begin{align}{x^2} + 5x & = {( - 3)^2} + 5( - 3)\\& = 9 - 15\\&  =  - 6\end{align}\]

The correct statement is \({x^2} + 5x =  - 6\)

Chapter 14 Ex.14.3 Question 14

Find and correct the errors in the statement:

\[{(y - 3)^2} = {y^2} - 9\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Use identity \({(a - b)^2} = {a^2} - 2ab + {b^2}\)

Steps:

\[\begin{align}L.H.S. &= {(y - 3)}^2 \\&= {{(y)}^2} - 2(y)(3) + {{(3)}^2} \quad \\\text {Using identity}\\(a - b)^2 &= {a^2} - 2ab + {b^2}\\&= {y^2} - 6y + 9\\L.H.S. &\ne R.H.S.\end{align}\]

The correct statement is

\[{(y - 3)^2} = {y^2} - 6y + 9\]

Chapter 14 Ex.14.3 Question 15

Find and correct the errors in the statement:

\[{\left( {z + 5} \right)^2} = {z^2} + 25\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Use identity \({(a + b)^2} = {a^2} + 2ab + {b^2}\)

Steps:

\[\begin{align}L.H.S.&= {{(z + 5)}^2} \\&= {{(z)}^2} + 2(z)(5) + {(5)}^2 \quad\\ \text{Using identity}\\{(a + b)}^2 &= {a^2} + 2ab + {b^2}\\&= {z^2} + 10z + 25\\L.H.S. &\ne R.H.S.\end{align}\]

The correct statement is \({\left( {z + 5} \right)^2} = {z^2} + 10z + 25\)

Chapter 14 Ex.14.3 Question 16

Find and correct the errors in the statement:

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

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}{L.H.S.}&{ = (2a + 3b)(a - b)}\\{}&={\left( \begin{array}{l}  2a \times a - 2a \times b + \\3b \times a - 3b \times b\end{array} \right)}\\{}&{ = 2{a^2} - 2ab + 3ab - 3{b^2}}\\{}&{ = 2{a^2} + ab - 3{b^2}}\\{L.H.S.}&{ \ne {\rm{R}}.{\rm{H}}.{\rm{S}}.}\end{align}\]

The correct statement is

\[\left( {2a + 3b} \right)\left( {a - b} \right) = 2{a^2} + ab - 3{b^2}\]

Chapter 14 Ex.14.3 Question 17

Find and correct the errors in the statement:

\[\left( {a + 4} \right)\left( {a + 2} \right) = {\rm{ }}{a^2} + 8\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S.&= (a + 4)(a + 2)\\ &=\!a\!\times\!a\!+\!2\!\times\!a\!+\! 4\!\times\!a\!+\!4\!\times\!2\\&{ = {{(a)}^2} + a(4 + 2) + (4 \times 2)}\\&{ = {a^2} + 6a + 8}\\{L.H.S.}&{ \ne R.H.S.}\end{align}\]

The correct statement is

\[\left( {a + 4} \right)\left( {a + 2} \right) = {a^2} + 6a + 8\]

Chapter 14 Ex.14.3 Question 18

Find and correct the errors in the statement:

\[\left( {a - 4} \right)\left( {a - 2} \right) = {a^2} - 8\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}{L.H.S.}&{ = (a - 4)(a - 2)}\\{}&{ = \left\{ \begin{array}{l}\!\!a\!\times\!a\!+\!\left( { - 2} \right) \!\times\!a +\!\left( { - 4} \right)\!\times\!\!\!\\a + \left[ {( - 4) \times ( - 2)} \right]\end{array} \right\}}\\&{ = {a^2} - 2a - 4a + 8}\\&{ = {a^2} - 6a + 8}\\{L.H.S.}&{ \ne R.H.S.}\end{align}\]

The correct statement is

\[\left( {a - 4} \right)\left( {a - 2} \right) = {a^2} - 6a + 8\]

Chapter 14 Ex.14.3 Question 19

Find and correct the errors in the statement:

\[\begin{align}\frac{{3{x^2}}}{{3{x^2}}} = 0\end{align}\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S. &= \frac{{3{x^2}}}{{3{x^2}}} = \frac{{3 \times x \times x}}{{3 \times x \times x}} = 1\\
L.H.S. &\ne R.H.S.\end{align}\]

The correct statement is \(\begin{align}\frac{{3{x^2}}}{{3{x^2}}} = 1\end{align}\)

Chapter 14 Ex.14.3 Question 20

Find and correct the errors in the statement:

\[\begin{align}\frac{{3{x^2} + 1}}{{3{x^2}}} = 1 + 1 = 2\end{align}\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S. &= \frac{{3{x^2} + 1}}{{3{x^2}}} = \frac{{3{x^2}}}{{3{x^2}}} + \frac{1}{{3{x^2}}}\\
&= 1 + \frac{1}{{3{x^2}}}\\L.H.S. &\ne R.H.S.\end{align}\]

The correct statement is

\[\begin{align}\frac{{3{x^2} + 1}}{{3{x^2}}} = 1 + \frac{1}{{3{x^2}}}\end{align}\]

Chapter 14 Ex.14.3 Question 21

Find and correct the errors in the statement:

\[\frac{{3x}}{{3x + 2}} = \frac{1}{2}\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[L.H.S. = \frac{{3x}}{{3x + 2}} \ne R.H.S.\]

The correct statement is

\[\begin{align}\frac{{3x}}{{3x + 2}} = \frac{{3x}}{{3x + 2}} \end{align}\]

Chapter 14 Ex.14.3 Question 22

Find and correct the errors in the statement:

\[\begin{align}\frac{3}{{4x + 3}} = \frac{1}{{4x}}\end{align}\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[L.H.S = \frac{3}{{4x + 3}} \ne R.H.S\]

The correct statement is

\[\begin{align} \frac{3}{{4x + 3}} = \frac{3}{{4x + 3}} \end{align}\]

Chapter 14 Ex.14.3 Question 23

Find and correct the errors in the statement:

\[\frac{{4x + 5}}{{4x}} = 5\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S. &= \frac{{4x + 5}}{{4x}} = \frac{{4x}}{{4x}} + \frac{5}{{4x}}\\&= 1 + \frac{5}{{4x}}\\L.H.S. &\ne R.H.S.\end{align}\]

The correct statement is

\[\begin{align}\frac{{4x + 5}}{{4x}} = 1 + \frac{5}{{4x}} \end{align}\]

Chapter 14 Ex.14.3 Question 24

Find and correct the errors in the statement:

\[\frac{{7x + 5}}{5} = 7x\]

Solution

What is known?

Incorrect mathematical statement.

What is unknown?

Correct mathematical statement.

Reasoning:

Solve \(L.H.S.\)

Steps:

\[\begin{align}L.H.S. &= \frac{{7x + 5}}{5} = \frac{{7x}}{5} + \frac{5}{5}\\&= \frac{{7x}}{5} + 1\\L.H.S. &\ne R.H.S.\end{align}\]

The correct statement is

\[\begin{align}\frac{{7x + 5}}{5} = \frac{{7x}}{5} + 1 \end{align}\]

  
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