# Ex.14.3 Q4 Factorization - NCERT Maths Class 8

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## Question

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}

## Text 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}

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