Ex.14.3 Q5 Factorization - NCERT Maths Class 8

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Question

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

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

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

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

Text 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:

\(\begin{align}\left( {{m^2} - 14m - 32} \right)\text{can be written as}\quad&{m^2} + 2m - 16m - 32\\&= m(m + 2) - 16(m + 2)\\&= (m + 2)(m - 16)\end{align}\)

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:

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

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:

\(\begin{align}4yz\left( {{z^2} + 6z - 16} \right)\quad\text{can be written as}\quad &4yz\left( {{z^2} - 2z + 8z - 16} \right)\\&= 4yz\left[ {z(z - 2) + 8(z - 2)} \right]\\&= 4yz(z - 2)(z + 8)\end{align}\)

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:

\(\begin{align}12xy\left( {9{x^2} - 16{y^2}} \right)\,\,\text{can be written as}\,\,12xy[ {{(3x)}^2} - {(4y)^2}]\\ &= 12xy(3x - 4y)(3x + 4y)\\& = 2 \times 2 \times 3 \times x \times y \times (3x - 4y) \times (3x + 4y)\end{align}\)

Then,

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

(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:

\(\begin{align}39{y^3}\left( {50{y^2} - 98} \right)\, \text{can be written as}\quad&3 \times 13 \times y \times y \times y \times 2\left[ {\left( {25{y^2} - 49} \right)} \right]\\&= 3 \times 13 \times 2 \times y \times y \times y \times \left[ {{{(5y)}^2} - {{(7)}^2}} \right]\\ &= 3 \times 13 \times 2 \times y \times y \times y(5y - 7)(5y + 7)\end{align}\)

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

Then,

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