# 3.2.1 Algebraic Expressions II, PT3 Focus Practice

3.2.1 Algebraic Expressions II, PT3 Focus Practice

Question 1:
Calculate the product of each of the following pairs of algebraic terms.
(a) 2rs × 4r2s3t
$\begin{array}{l}\text{(b)}1\frac{2}{5}{a}^{2}{b}^{2}×\frac{5}{14}a{b}^{3}{c}^{2}\\ \text{(c)}1\frac{1}{2}xy×\frac{4}{9}{x}^{2}z\end{array}$

Solution:
(a)
2rs × 4r2s3t = 2 × r × s× 4 × r × r × s × s × s × t = 8 r3s4t

$\begin{array}{l}\text{(b)}1\frac{2}{5}{a}^{2}{b}^{2}×\frac{5}{14}a{b}^{3}{c}^{2}\\ =\frac{{\overline{)7}}^{1}}{{\overline{)5}}^{1}}aabb×\frac{{\overline{)5}}^{1}}{{\overline{)14}}^{2}}abbbcc\\ =\frac{1}{2}{a}^{3}{b}^{5}{c}^{2}\end{array}$

$\begin{array}{l}\text{(c)}1\frac{1}{2}xy×\frac{4}{9}{x}^{2}z\\ =\frac{{\overline{)3}}^{1}}{{\overline{)2}}^{1}}xy×\frac{{\overline{)4}}^{2}}{{\overline{)9}}^{3}}xxz\\ =\frac{2}{3}{x}^{3}yz\end{array}$

Question 2:
Find the quotients of each of the following pairs of algebraic terms.
$\text{(a)}\frac{-48{a}^{2}{b}^{3}{c}^{4}}{-16a{b}^{2}{c}^{2}}$
$\begin{array}{l}\text{(b)}-15e{f}^{3}{g}^{2}÷\left(-40e{f}^{2}g\right)\\ \text{(c)}5{a}^{3}{c}^{2}÷\frac{1}{2}ac\end{array}$

Solution:
$\begin{array}{l}\text{(a)}\frac{-48{a}^{2}{b}^{3}{c}^{4}}{-16a{b}^{2}{c}^{2}}\\ =\frac{-{\overline{)48}}^{3}\overline{)a}×a×\overline{)b}×\overline{)b}×b×\overline{)c}×\overline{)c}×c×c}{-{\overline{)16}}^{1}\overline{)a}×\overline{)b}×\overline{)b}×\overline{)c}×\overline{)c}}\\ =3a{c}^{2}\end{array}$
$\begin{array}{l}\text{(b)}-15e{f}^{3}{g}^{2}÷\left(-40e{f}^{2}g\right)\\ =\frac{-{\overline{)15}}^{3}\overline{)e}×\overline{)f}×\overline{)f}×f×\overline{)g}×g}{-{\overline{)40}}^{8}\overline{)e}×\overline{)f}×\overline{)\overline{)f}}×\overline{)g}}\\ =\frac{3}{8}fg\end{array}$

$\begin{array}{l}\text{(c)}5{a}^{3}{c}^{2}÷\frac{1}{2}ac=5a{\overline{){}^{3}}}^{2}c\overline{){}^{2}}×\frac{2}{\overline{)a}\overline{)c}}\\ \text{}=10{a}^{2}c\end{array}$

Question 3:
(– 4a2b) ÷ 3b2c2× 9ac =

Solution:
$\begin{array}{l}\left(-4{a}^{2}b\right)÷3{b}^{2}{c}^{2}×9ac\\ =\frac{-4{a}^{2}\overline{)b}}{\overline{)3}{b}^{\overline{)2}1}{c}^{\overline{)2}1}}×{\overline{)9}}^{3}a\overline{)c}\\ =\frac{-12{a}^{3}}{bc}\end{array}$

Question 4:
2a2b × 14b3c ÷ 56ab2c2 =

Solution:
$\begin{array}{l}2{a}^{2}b×14{b}^{3}c÷56a{b}^{2}{c}^{2}\\ =\overline{)2}{a}^{\overline{)2}1}b×\frac{\overline{)14}{b}^{\overline{)3}1}\overline{)c}}{{\overline{)56}}^{2}\overline{)a}{\overline{)b}}^{\overline{)2}}{c}^{\overline{)2}1}}\\ =\frac{a{b}^{2}}{2c}\end{array}$

Question 5:
$-\frac{2}{3}\left(3a-6b+\frac{3}{4}c\right)=$

Solution:
$\begin{array}{l}-\frac{2}{3}\left(3a-6b+\frac{3}{4}c\right)\\ =-\frac{2}{\overline{)3}}×\overline{)3}a-\frac{2}{\overline{)3}}\left(-{\overline{)6}}^{2}b\right)-\frac{\overline{)2}}{\overline{)3}}\left(\frac{\overline{)3}}{{\overline{)4}}^{2}}c\right)\\ =-2a+4b-\frac{1}{2}c\end{array}$