Question

Multiply $(3x-\frac{4}{5}{y}^{2}x)$ by $\frac{1}{2}xy$

• A. $\frac{x^2}{10}(3y-4y^3)$
• B. $\frac{2x^2}{10}(3y-y^3)$
• C. $\frac{x^2}{10}(15y-4y^3)$
• D. $\frac{x^2}{10}(3y-4x{y}^3)$

Answer 1

Answer: (C)

$\frac{x^2}{10}(15y-4y^3)$

Let us multiply the given binomials as shown below:

$(3x-\frac{4}{5}{y}^{2}x)\left(\frac{1}{2}xy\right)=(\frac{3\times 5}{1\times 5}x-\frac{4}{5}{y}^{2}x)\left(\frac{1}{2}xy\right)\left(TakingLCM\right)=\left(\frac{15x-4y^{2}x}{5}\right)\left(\frac{xy}{2}\right)=\frac{xy}{10}(15x-4y^{2}x)$

$=\frac{x^{2}y}{10}(15-4y^2)=\frac{x^2}{10}(15y-4y^3)$

Hence, $(3x-\frac{4}{5}{y}^{2}x)\left(\frac{1}{2}xy\right)=\frac{x^2}{10}(15y-4y^3)$.