Question

Help please I’m not good at rational functions.

Answer 1

Answer:

See below.

Step-by-step explanation:

$\frac{6z^2-12z}{4z^2-16z+16} +\frac{3z}{z^2-z-2}$

First, factor the numerators and the denominators:

$=\frac{6z(z-2)}{4(z^2-4x+4)}+\frac{3z}{z^2-z-2}\\=\frac{6z(z-2)}{4(z-2)^2}+\frac{3z}{(z-2)(z+1)}\\=\frac{3z}{2(z-2)}+\frac{3z}{(z-2)(z+1)}$

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:

$(\frac{z+1}{z+1} )(\frac{3z}{2(z-2)})+(\frac{2}{2})(\frac{3z}{(z-2)(z+1)})\\=\frac{3z(z+1)}{2(z-2)(z+1)}+\frac{6z}{2(z-2)(z+1)}$

Now, combine them since they have a common denominator:

$=\frac{3z^2+3z+6z}{2(z-2)(z+1)}\\ =\frac{3z^2+9z}{2(z-2)(z+1)} \\=\frac{3z(z+3)}{2(z-2)(z+1)}$

This cannot be simplified further.