Question
Find the product: $75\text{x}^2\text{y}\big(\frac{3}{5}\text{xy}^2+\frac{2}{5}\text{x}\big)$
✓
Answer
To find the product, we will use distributive law as follows:$75\text{x}^2\text{y}\big(\frac{3}{5}\text{xy}^2+\frac{2}{5}\text{x}\big)$
$=\frac{7}{5}\text{x}^2\text{y}×\frac{3}{5}\text{xy}^2+\frac{7}{5}\text{x}^2\text{y}×\frac{2}{5}\text{x}$
$=\frac{21}{25}\text{x}^{2+1}\text{y}^{1+2}+\frac{14}{25}\text{x}^{2+1}\text{y}$
$=\frac{21}{25}\text{x}^3\text{y}^3+\frac{14}{25}\text{x}^3\text{y}$
Thus, the answer is $\frac{21}{25}\text{x}^3\text{y}^3+\frac{14}{25}\text{x}^3\text{y}$
$=\frac{7}{5}\text{x}^2\text{y}×\frac{3}{5}\text{xy}^2+\frac{7}{5}\text{x}^2\text{y}×\frac{2}{5}\text{x}$
$=\frac{21}{25}\text{x}^{2+1}\text{y}^{1+2}+\frac{14}{25}\text{x}^{2+1}\text{y}$
$=\frac{21}{25}\text{x}^3\text{y}^3+\frac{14}{25}\text{x}^3\text{y}$
Thus, the answer is $\frac{21}{25}\text{x}^3\text{y}^3+\frac{14}{25}\text{x}^3\text{y}$
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