Multiply −3 2 x^2y^3 by (2x − y) and verify the answer for x = 1 and y = 2.

To find the product, we will use distributive law as follows:−3 2 x^2y^3 (2x − y) = (−3 2 x^2y^3×2x )− (−3 2 x^2y^3×y ) = (−3x^ 2+1 y^3 )− (−3 2 x^2y^ 3+1 ) =−3x^3y^3+3 2 x^2y^4 Substituting x = 1 and y = 2 in the result, we get:=−3x^3y^3+3 2 x^2y^4 =−3(1)^3(2)^3+3 2 (1)^2(2)^4 =−3×1×8+3 2 ×1×16 =−24+24 =0 Thus, the product is −3x^3y^3+3 2 x^2y^4 and its value for x = 1 and y = 2 is 0.

Multiply −3 2 x^2y^3 by (2x − y) and verify the answer for x = 1 …

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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.

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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.