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

(iii)\(\begin{align}\quad 52{\mathop{\rm pqr}\nolimits} (p + q)(q + r)(r + p) \div 104pq(q + r)(r + p)\end{align}\)

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

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

Text Solution

What is known?

Algebraic expression.

What is unknown?

Division of the algebraic expression.

Reasoning:

Cancel out common factor of the followinng.

Steps:

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

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

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

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

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