Question

Help please I’m not good at rational functions.

Answer 1

See below.

Explanation:

`(6z^2-12z)/(4z^2-16z+16) +(3z)/(z^2-z-2) \\\\`

First, factor the numerators and the denominators:

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

Now, make the two denominators equivalent. To do this, we can multiply the first term by (z+1) and multiply the second term by 2. This will give us:

`((z+1)/(z+1) )((3z)/(2(z-2)))+((2)/(2))((3z)/((z-2)(z+1)))\\=(3z(z+1))/(2(z-2)(z+1))+(6z)/(2(z-2)(z+1)) \\`

Now, combine them since they have a common denominator:

`=(3z^2+3z+6z)/(2(z-2)(z+1))\\ =(3z^2+9z)/(2(z-2)(z+1)) \\=(3z(z+3))/(2(z-2)(z+1))`

This cannot be simplified further.