Solve the following


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


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

$-\frac{3}{2} x^{2} y^{3} \times(2 x-y)$

$=\left(-\frac{3}{2} x^{2} y^{3} \times 2 x\right)-\left(-\frac{3}{2} x^{2} y^{3} \times y\right)$

$=\left(-3 x^{2+1} y^{3}\right)-\left(-\frac{3}{2} x^{2} y^{3+1}\right)$

$=-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{4}$

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

$-3 x^{3} y^{3}+\frac{3}{2} x^{2} y^{4}$


$=-3 \times 1 \times 8+\frac{3}{2} \times 1 \times 16$



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 .

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now