Question

Can you combine `\frac { 4x + 2} { 3x ^ { 2} - 6x - 24} + \frac { 9} { 3x ^ { 2} - 15x + 12}` into a single term?

Answer 1

`(4x^2+7x+16)/(3x^3-9x^2-18x+24)`

Explanation:

First things first, we need to create a common denominator. Let's factor everything first and see if anything divides out

`(2(2x-1))/(3(x^2-2x-8)) + (3 xx 3)/(3(x^2-5x+4))`

`(2(2x-1))/(color(green)(3)color(green)((x-4))(x+2)) + (3 xx 3)/(color(green)(3)(x-1)color(green)((x-4)))`

Nothing divides out, but it looks like the denominators are mostly the same, except that the right side doesn't have `(x+2)`, and the left side is missing `(x-1)`. Let's multiply both sides by the term they're missing (remember, multiply both the numerator and denominator)

`(x-1)/(x-1) xx (4x+2)/(3(x-4)(x+2)) + (9)/(3(x-1)(x-4)(x+2)) xx (x+2)/(x+2)`

`(4x^2-4x+2x-2)/(3(x-4)(x+2)(x-1)) + (9x+18)/(3(x-1)(x-4)(x+2))`

Now that the denominators match, let's combine the numerators:

`(4x^2-2x-2+9x+18)/(3(x-4)(x+2)(x-1))`

That denominator could use some work...

`3(x-4)(x+2)(x-1)`

`(3x-12)(x+2)(x-1)`

`(3x^2+6x-12x-24)(x-1)`

`(3x^2-6x-24)(x-1)`

`3x^3-3x^2-6x^2+6x-24x+24`

`color(green)(3x^3-9x^2-18x+24)`

Now we have

`(4x^2+7x+16)/(color(green)(3x^3-9x^2-18x+24))`

I hope that helps!