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Question
Multiply \[- \frac{3}{2} x^2 y^3 by (2x - y)\] and verify the answer for x = 1 and y = 2.
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Solution
To find the product, we will use distributive law as follows:
\[- \frac{3}{2} x^2 y^3 \times \left( 2x - y \right)\]
\[ = \left( - \frac{3}{2} x^2 y^3 \times 2x \right) - \left( - \frac{3}{2} x^2 y^3 \times y \right)\]
\[ = \left( - 3 x^{2 + 1} y^3 \right) - \left( - \frac{3}{2} x^2 y^{3 + 1} \right)\]
\[ = - 3 x^3 y^3 + \frac{3}{2} x^2 y^4\]
Substituting x = 1 and y = 2 in the result, we get:
\[- 3 x^3 y^3 + \frac{3}{2} x^2 y^4 \]
\[ = - 3 \left( 1 \right)^3 \left( 2 \right)^3 + \frac{3}{2} \left( 1 \right)^2 \left( 2 \right)^4 \]
\[ = - 3 \times 1 \times 8 + \frac{3}{2} \times 1 \times 16\]
\[ = - 24 + 24\]
\[ = 0\]
Thus, the product is \[- 3 x^3 y^3 + \frac{3}{2} x^2 y^4\],and its value for x = 1 and y = 2 is 0.
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