How do you simplify $\frac{x^{2} y}{4} \cdot \frac{16 x}{y}$ $(x^2y)/4*(16x)/y$ ?

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Answer 1 · 2016-12-03T04:03:38

$4 x^{3}$ $4x^3$

First, multiply the numerator and denominator and group terms using the rules for exponents:

$\frac{16 x 2 x^{1} y}{4 y}$ $(16x2x^1y)/(4y)$

$\frac{16 x^{2 + 1} y}{4 y}$ $(16x^(2 + 1)y)/(4y)$

$\frac{16 x^{3} y}{4 y}$ $(16x^3y)/(4y)$

Next, separate terms and factor:

$(\frac{16}{4}) x^{3} (\frac{y}{y})$ $(16/4)x^3(y/y)$

$4 \cdot x^{3} \cdot 1$ $4*x^3*1$

$4 x^{3}$ $4x^3$

How do you simplify x2y4⋅16xy(x^2y)/4*(16x)/y?