# 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 \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}

## 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:

$$\left( {{m^2} - 14m - 32} \right)$$  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}

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