Question

perform the following operation and express in simplest form.
\\(\dfrac{2x + 14}{x^2 + 15x + 56} \cdot \dfrac{x^2 + 3x - 40}{6x}\\)

Explanation:

Step1: Factor each expression

Factor \(2x + 14\) as \(2(x + 7)\).
Factor \(x^2 + 15x + 56\) by finding two numbers that multiply to \(56\) and add to \(15\), which are \(7\) and \(8\), so it factors to \((x + 7)(x + 8)\).
Factor \(x^2 + 3x - 40\) by finding two numbers that multiply to \(-40\) and add to \(3\), which are \(8\) and \(-5\), so it factors to \((x + 8)(x - 5)\).

The expression becomes:
\(\frac{2(x + 7)}{(x + 7)(x + 8)} \cdot \frac{(x + 8)(x - 5)}{6x}\)

Step2: Cancel common factors

Cancel \((x + 7)\) from the numerator and denominator of the first fraction.
Cancel \((x + 8)\) from the numerator of the second fraction and the denominator of the first fraction.

After canceling, we have:
\(\frac{2}{1} \cdot \frac{(x - 5)}{6x}\)

Step3: Multiply the remaining factors and simplify

Multiply the numerators: \(2 \cdot (x - 5) = 2(x - 5)\)
Multiply the denominators: \(1 \cdot 6x = 6x\)

Simplify \(\frac{2(x - 5)}{6x}\) by dividing numerator and denominator by \(2\):
\(\frac{x - 5}{3x}\)

Answer:

\(\frac{x - 5}{3x}\)