Question

Multiply $-\frac{3}{2}{x}^2{y}^3\mathrm{by}(2x-y)$ and verify the answer for x = 1 and y = 2.

Answer 1

To find the product, we will use distributive law as follows:

$$\begin{gathered} -\frac{3}{2}{x}^2{y}^3\times \left(2x-y\right) \\ =\left(-\frac{3}{2}{x}^2{y}^3\times 2x\right)-\left(-\frac{3}{2}{x}^2{y}^3\times y\right) \\ =\left(-3x^{2+1}{y}^3\right)-\left(-\frac{3}{2}{x}^2{y}^{3+1}\right) \\ =-3x^3{y}^3+\frac{3}{2}{x}^2{y}^4 \end{gathered}$$

Substituting x = 1 and y = 2 in the result, we get:

$$\begin{gathered} -3x^3{y}^3+\frac{3}{2}{x}^2{y}^4 \\ =-3{\left(1\right)}^3{\left(2\right)}^3+\frac{3}{2}{\left(1\right)}^2{\left(2\right)}^4 \\ =-3\times 1\times 8+\frac{3}{2}\times 1\times 16 \\ =-24+24 \\ =0 \end{gathered}$$

Thus, the product is $-3x^3{y}^3+\frac{3}{2}{x}^2{y}^4$, and its value for ​x = 1 and y = 2 is 0.