Question

1. find the product. if possible, write your answer in factored form.

\\(\frac{4xy^3}{x^2y} \cdot \frac{y}{8x} = \square , y \
eq \square\\)

Explanation:

Step1: Multiply numerators and denominators

Multiply the numerators \(4xy^3\) and \(y\), and the denominators \(x^2y\) and \(8x\). So we get \(\frac{4xy^3 \cdot y}{x^2y \cdot 8x}\).

Step2: Simplify exponents of like bases

For \(x\): \(x \cdot 1\) in numerator and \(x^2 \cdot x=x^{3}\) in denominator, so \(x/x^{3}=1/x^{2}\). For \(y\): \(y^3 \cdot y = y^{4}\) in numerator and \(y\) in denominator, so \(y^{4}/y = y^{3}\). For constants: \(4/8 = 1/2\). Combining these, we have \(\frac{4xy^{4}}{8x^{3}y}=\frac{y^{3}}{2x^{2}}\). Also, \(y
eq0\) (since \(y\) is in denominator of original fractions, \(y = 0\) would make denominators zero) and \(x
eq0\) but the question asks for \(y
eq\), so \(y
eq0\).

Answer:

\(\frac{y^{3}}{2x^{2}}\), \(y
eq 0\)