Question

(c) problem 19.
multiply these fractions and select the simplified answer.
\\(\frac{2x^2 - 10x}{3x^2} \cdot \frac{x^5}{4x^2 - 20x}\\)
after you pick your answer press go.
a. \\(\frac{x^5 + 2x^2 - 10x}{7x^2 - 20x}\\) b. \\(\frac{x^3}{6}\\) c. \\(\frac{1 + x^3}{5}\\)
d. \\(\frac{x^5}{3x + 2}\\) e. \\(\frac{2}{3x^7}\\)

Explanation:

Step1: Factor numerators and denominators

Factor \(2x^2 - 10x\) as \(2x(x - 5)\), \(4x^2 - 20x\) as \(4x(x - 5)\). So the expression becomes \(\frac{2x(x - 5)}{3x^2} \cdot \frac{x^5}{4x(x - 5)}\).

Step2: Cancel common factors

Cancel out \(2x\) with \(4x\) (leaving \(\frac{1}{2}\) and \(\frac{2}{4}\) simplification), \(x - 5\) from numerator and denominator, and \(x\) from \(x\) (in \(2x\)) and \(x^2\) (leaving \(x\)) and \(x^5\) (leaving \(x^4\) initially, but after more cancellations: \(x\) in numerator (from \(2x\)) and \(x\) in denominator (from \(4x\)) cancel, \(x - 5\) cancels, then \(x\) from \(x\) (in \(2x\)) and \(x^2\) (leaves \(x\)) and \(x^5\) (leaves \(x^4\))? Wait, let's do it step by step.

First, rewrite: \(\frac{2x(x - 5)}{3x^2} \times \frac{x^5}{4x(x - 5)}\). Now, cancel \(x - 5\) (both numerator and denominator), cancel \(x\) (from \(2x\) and \(4x\), so \(2x/4x = 1/2\)), cancel \(x\) (from \(x\) in \(2x\) and \(x^2\) in denominator: \(x^2\) becomes \(x\)), and \(x^5\) in numerator with \(x\) in denominator (from \(4x\))? Wait, better to multiply numerators and denominators first:

Numerator: \(2x(x - 5) \times x^5 = 2x^6(x - 5)\)

Denominator: \(3x^2 \times 4x(x - 5) = 12x^3(x - 5)\)

Now cancel \(2\) and \(12\) (divide by 2: \(2/12 = 1/6\)), cancel \(x^6\) and \(x^3\) (leaves \(x^3\)), cancel \(x - 5\). So we get \(\frac{x^3}{6}\).

Answer:

B. \(\frac{x^3}{6}\)