Question
Multiply $−\frac{3}{2}\text{x}^2\text{y}^3 \ \text{by} \ (2\text{x} − \text{y})$ and verify the answer for $x = 1$ and $y = 2$.
✓
Answer
To find the product, we will use distributive law as follows:$−\frac{3}{2}\text{x}^2\text{y}^3 \ \times \ (2\text{x} − \text{y})$
$=\big(−\frac{3}{2}\text{x}^2\text{y}^3×2\text{x}\big)−\big(−\frac{3}{2}\text{x}^2\text{y}^3×\text{y}\big)$
$=\big(−3\text{x}^{2+1}\text{y}^3\big)−\big(−\frac{3}{2}\text{x}^2\text{y}^{3+1}\big)$
$=−3\text{x}^3\text{y}^3+\frac{3}{2}\text{x}^2\text{y}^4$
Substituting $x = 1$ and $y = 2$ in the result,
we get:$=−3\text{x}^3\text{y}^3+\frac{3}{2}\text{x}^2\text{y}^4$
$=−3(1)^3(2)^3+\frac{3}{2}(1)^2(2)^4$
$=−3×1×8+\frac{3}{2}×1×16$
$=−24+24$
$=0$
Thus, the product is $−3\text{x}^3\text{y}^3+\frac{3}{2}\text{x}^2\text{y}^4$ and its value for $x = 1$ and $y = 2$ is $0$.
$=\big(−\frac{3}{2}\text{x}^2\text{y}^3×2\text{x}\big)−\big(−\frac{3}{2}\text{x}^2\text{y}^3×\text{y}\big)$
$=\big(−3\text{x}^{2+1}\text{y}^3\big)−\big(−\frac{3}{2}\text{x}^2\text{y}^{3+1}\big)$
$=−3\text{x}^3\text{y}^3+\frac{3}{2}\text{x}^2\text{y}^4$
Substituting $x = 1$ and $y = 2$ in the result,
we get:$=−3\text{x}^3\text{y}^3+\frac{3}{2}\text{x}^2\text{y}^4$
$=−3(1)^3(2)^3+\frac{3}{2}(1)^2(2)^4$
$=−3×1×8+\frac{3}{2}×1×16$
$=−24+24$
$=0$
Thus, the product is $−3\text{x}^3\text{y}^3+\frac{3}{2}\text{x}^2\text{y}^4$ and its value for $x = 1$ and $y = 2$ is $0$.
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