Question

factor the numerator of the third fraction. keep the other fractions in mind when selecting the factors.
\\(\left(\frac{x + 4}{x^2} \cdot \frac{2x + 1}{x - 4}\
ight) \div \frac{(\square)^2}{x^2 - 4x}\\)

factor the denominator of the third fraction. keep the other fractions in mind when selecting the factors.
\\(\left(\frac{x + 4}{x^2} \cdot \frac{2x + 1}{x - 4}\
ight) \div \frac{(2x + 1)^2}{\square (x - 4)}\\)

Explanation:

Step1: Factor the numerator term

We look at the existing numerators in the product: $\frac{x+4}{x^2} \cdot \frac{2x+1}{x-4}$. The squared term in the third fraction's numerator should match a factor from these, so it is $2x+1$.

Step2: Factor the denominator term

Factor $x^2-4x$:
$x^2-4x = x(x-4)$
We already have $(x-4)$ shown, so the missing factor is $x$.

Answer:

First blank: $2x+1$
Second blank: $x$