Question

Multiplying and dividing radical expressions and leaving them in factored form. I am trying to find the best and easy way to get the factored form correctly. This is my problem: 3x+8 over 36-2x / 27x^2+72x over 3x^2-27.

Answer 1

Answer:

$\frac{3x + 8}{36 - 2x} / \frac{27x^2 + 72x}{3x^2 - 27} = \frac{(x - 3)(x+3)}{6x(18 - x)}$

Step-by-step explanation:

Given

$\frac{3x + 8}{36 - 2x} / \frac{27x^2 + 72x}{3x^2 - 27}$

Required

Solve

Change / to *

$\frac{3x + 8}{36 - 2x} * \frac{3x^2 - 27}{27x^2 + 72x}$

Factor out 3

$\frac{3x + 8}{36 - 2x} * \frac{3(x^2 - 9)}{3(9x^2 + 24x)}$

$\frac{3x + 8}{36 - 2x} * \frac{(x^2 - 9)}{(9x^2 + 24x)}$

Factorize:

$\frac{3x + 8}{36 - 2x} * \frac{(x^2 - 9)}{3x(3x + 8)}$

Cancel out 3x + 8

$\frac{1}{36 - 2x} * \frac{(x^2 - 9)}{3x}$

Factorize:

$\frac{1}{2(18 - x)} * \frac{(x^2 - 9)}{3x}$

Combine

$\frac{x^2 - 9}{2*3x(18 - x)}$

$\frac{x^2 - 9}{6x(18 - x)}$

Express the numerator as a difference of two squares

$\frac{(x - 3)(x+3)}{6x(18 - x)}$

Hence:

$\frac{3x + 8}{36 - 2x} / \frac{27x^2 + 72x}{3x^2 - 27} = \frac{(x - 3)(x+3)}{6x(18 - x)}$