Ex.14.3 Q3 Factorization - NCERT Maths Class 8

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Question

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

Text 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

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

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

(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

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

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

(v) \(\,96abc(3a - 12)(5b - 30) \div 144(a - 4)(b - 6)\)

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

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

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