Question

How do you simplify `(3x)/(x^2-x-12) - (x-1)/(x^2+6x+9) + (x-6)/(2x+6)`?

Answer 1

`(3x)/(x^2-x-12)-(x-1)/(x^2+6x+9)+(x-6)/(2x+6)=(x^3-3x^2+22x+64)/(2(x+3)(x+3)(x-4))`

Explanation:

For simplifying this, we first need to factorize denominators (as numerators are already in simplest terms).

`x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x+3)(x-4)`

`x^2+6x9=x^2+3x+3x+9=x(x+3)+3(x+3)=(x+3)(x+3)=(x+3)^2`

`2x+6=2(x+3)`

And LCD of three denominators is `2(x+3)(x+3)(x-4)`

Hence `(3x)/(x^2-x-12)-(x-1)/(x^2+6x+9)+(x-6)/(2x+6)`

= `(3x)/((x+3)(x-4))-(x-1)/(x+3)^2+(x-6)/(2(x+3))`

= `(3x xx2xx(x+3)-(x-1)xx2(x-4)+(x-6)(x+3)(x-4))/(2(x+3)(x+3)(x-4))`

= `((6x^2+18x)-2(x^2-5x+4)+(x-6)(x^2-x-12))/(2(x+3)(x+3)(x-4))`

= `((6x^2+18x)-2(x^2-5x+4)+(x^3-x^2-12x-6x^2+6x+72))/(2(x+3)(x+3)(x-4))`

= `(x^3-3x^2+22x+64)/(2(x+3)(x+3)(x-4))`

Answer 2

`(x^3-3x^2+22x+64)/(2(x-4)(x+3)^2)`

Explanation:

Look for common factors. So we need to investigate factorisation of the denominators.

`x^2-x-12 = (x+3)(x-4)`

`x^2+6x+9=(x+3)(x+3)`

`2x+6=2(x+3)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Factor out `(x+3)` from the denominators giving:

`1/(x+3) [color(white)(.)(3x)/(x-4)-(x-1)/(x+3)+(x-6)/2color(white)(.)]color(red)("... Eqn (1)")`

Using a common denominator of `2(x-4)(x+3)`

'.................................................................................................................
Consider `(3x)/(x-4) -> (3x xx2xx(x+3))/(2(x-4)(x+3)) = (6x^2+18x)/(2(x-4)(x+3))`
,...................................................................................................................

Consider `-(x-1)/(x+3)-> ((x-1)xx2xx(x-4))/(2(x-4)(x+3)) = -(2x^2-10x+8)/(2(x-4)(x+3))` .
'.....................................................................................................................

Consider `(x-6)/2 -> ((x-6)(x-4)(x+3))/(2(x-4)(x+3))`

`=(x^3-7x^2-6x+72)/(2(x-4)(x+3)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Putting it all together")`

`1/(x+3)[(x^3-3x^2+22x+64)/(2(x-4)(x+3))]`

`(x^3-3x^2+22x+64)/(2(x+3)(x-4)(x+3))`

Factoring this further

`((x+2)(x^2-5x+32))/(2(x-4)(x+3)^2) larr" don't think this helps much"`