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

`(3x + 8)/(36 - 2x) / (27x^2 + 72x)/(3x^2 - 27) = ((x - 3)(x+3))/(6x(18 - x))`

Explanation:

Given

`(3x + 8)/(36 - 2x) / (27x^2 + 72x)/(3x^2 - 27)`

Required

Solve

Change / to *

`(3x + 8)/(36 - 2x) * (3x^2 - 27)/(27x^2 + 72x)`

Factor out 3

`(3x + 8)/(36 - 2x) * (3(x^2 - 9))/(3(9x^2 + 24x))`

`(3x + 8)/(36 - 2x) * ((x^2 - 9))/((9x^2 + 24x))`

Factorize:

`(3x + 8)/(36 - 2x) * ((x^2 - 9))/(3x(3x + 8))`

Cancel out 3x + 8

`(1)/(36 - 2x) * ((x^2 - 9))/(3x)`

Factorize:

`(1)/(2(18 - x)) * ((x^2 - 9))/(3x)`

Combine

`(x^2 - 9)/(2*3x(18 - x))`

`(x^2 - 9)/(6x(18 - x))`

Express the numerator as a difference of two squares

`((x - 3)(x+3))/(6x(18 - x))`

Hence:

`(3x + 8)/(36 - 2x) / (27x^2 + 72x)/(3x^2 - 27) = ((x - 3)(x+3))/(6x(18 - x))`