Question

Work out the following division
(i)$\left( {10x - 25} \right) \div 5$
(ii)$\left( {10x - 25} \right) \div \left( {2x - 5} \right)$
(iii)$10y\left( {6y + 21} \right) \div 5\left( {2y + 7} \right)$
(iv)$9{x^2}{y^2}\left( {3z - 24} \right) \div 27xy\left( {z - 8} \right)$
(v)$96abc\left( {3a - 12} \right)\left( {5b - 30} \right) \div 144\left( {a - 4} \right)\left( {b - 6} \right)$

Answer 1

Hint:Here, we will use the factorization process. All the terms in the question can be expanded into their factors which will result in simplified expression.

Complete step-by-step solution
(i)Expand the terms in the expression $\left( {10x - 25} \right) \div 5$ into their factors.
$\begin{array}{c}
\left( {10x - 25} \right) \div 5 = \dfrac{{5 \times 2 \times x - 5 \times 5}}{5}\\
 = \dfrac{{5\left( {2x - 5} \right)}}{5}\\
 = 2x - 5
\end{array}$
Therefore, $2x - 5$ is the factor.
(ii)Expand the terms in the expression $\left( {10x - 25} \right) \div \left( {2x - 5} \right)$ into their factors.
$\begin{array}{c}
\left( {10x - 25} \right) \div \left( {2x - 5} \right) = \dfrac{{2 \times 5 \times x - 5 \times 5}}{{2x - 5}}\\
 = \dfrac{{5\left( {2x - 5} \right)}}{{2x - 5}}\\
 = 5
\end{array}$
Therefore, $5$ is the factor.
(iii) Expand the terms in the expression $10y\left( {6y + 21} \right) \div 5\left( {2y + 7} \right)$ into their factors.
$\begin{array}{c}
10y\left( {6y + 21} \right) \div 5\left( {2y + 7} \right) = \dfrac{{2 \times 5 \times y\left( {2 \times 3 \times y + \left( {3 \times 7} \right)} \right)}}{{5\left( {2y + 7} \right)}}\\
 = \dfrac{{2 \times 5 \times y \times 3\left( {2y + 7} \right)}}{{5\left( {2y + 7} \right)}}\\
 = 6y
\end{array}$
Therefore, $6y$ is the factor.
(iv)Expand the terms in the expression $9{x^2}{y^2}\left( {3z - 24} \right) \div 27xy\left( {z - 8} \right)$ into their factors.
\[\begin{array}{c}
9{x^2}{y^2}\left( {3z - 24} \right) \div 27xy\left( {z - 8} \right) = \dfrac{{9{x^2}{y^2}\left[ {\left( {3 \times z} \right) - \left( {2 \times 2 \times 2 \times 3} \right)} \right]}}{{27xy\left( {z - 8} \right)}}\\
 = \dfrac{{{x^2}{y^2} \times 3\left[ {z - 8} \right]}}{{3xy\left( {z - 8} \right)}}\\
 = xy
\end{array}\]
Therefore, $xy$ is the factor.
(v) Expand the terms in the expression $96abc\left( {3a - 12} \right)\left( {5b - 30} \right) \div 144\left( {a - 4} \right)\left( {b - 6} \right)$ into their factors.
\[\begin{array}{c}
96abc\left( {3a - 12} \right)\left( {5b - 30} \right) \div 144\left( {a - 4} \right)\left( {b - 6} \right) = \dfrac{{96abc\left( {3 \times a - 3 \times 4} \right)\left( {5 \times b - 2 \times 3 \times 5} \right)}}{{144\left( {a - 4} \right)\left( {b - 6} \right)}}\\
 = \dfrac{{2abc \times 3 \times 5\left( {a - 4} \right)\left( {b - 6} \right)}}{{3\left( {a - 4} \right)\left( {b - 6} \right)}}\\
 = 10abc
\end{array}\]
Therefore, $10abc$ is the factor.

Note: In such types of problems, we will factorize the dividend. The common factors will be cancelled considering that the value of the divisor is not zero. Make sure to check the common terms in the expression so that these terms can be resolved to get the answer.