Question

the expression \\(\frac{x + 8}{x^2} \div \frac{2x + 16}{2x^3}\\) is equivalent to \\(\frac{x + 8}{x^2} \cdot \frac{2x^3}{2x + 16}\\). the simplified quotient of \\(\frac{x + 8}{x^2} \div \frac{2x + 16}{2x^3}\\) is \\(\square\\).

Explanation:

Step1: Factor the numerator of the second fraction

We have \(2x + 16\), which can be factored as \(2(x + 8)\). So the expression becomes \(\frac{x + 8}{x^{2}}\cdot\frac{2x^{2}}{2(x + 8)}\).

Step2: Cancel out common factors

We can cancel out the common factors \((x + 8)\) from the numerator and the denominator, and also cancel out the common factor \(2\) from the numerator and the denominator. After canceling, we are left with \(\frac{x^{2}}{x^{2}}\) (since the remaining terms are \(\frac{1}{x^{2}}\cdot\frac{2x^{2}}{1}\) after canceling \(2\) and \((x + 8)\), and then \(x^{2}\) in the numerator and denominator cancels out).

Step3: Simplify the remaining expression

\(\frac{x^{2}}{x^{2}}=1\) (for \(x
eq - 8\) and \(x
eq0\) to avoid division by zero).

Answer:

\(1\)