Add. Enter your answer as a single rational expression where the numerator and denominator have no common factors.

[tex]\[ \frac{z}{z^2+9z+8} + \frac{8}{z^2+12z+32} \][/tex]



Answer :

To add the given rational expressions, we first need to find a common denominator. The given fractions are:

[tex]\[ \frac{z}{z^2 + 9z + 8} + \frac{8}{z^2 + 12z + 32} \][/tex]

First, let's factor the denominators.

1. For [tex]\(z^2 + 9z + 8\)[/tex]:
[tex]\[ z^2 + 9z + 8 = (z + 1)(z + 8) \][/tex]

2. For [tex]\(z^2 + 12z + 32\)[/tex]:
[tex]\[ z^2 + 12z + 32 = (z + 4)(z + 8) \][/tex]

So, the expressions become:

[tex]\[ \frac{z}{(z + 1)(z + 8)} + \frac{8}{(z + 4)(z + 8)} \][/tex]

The common denominator for these fractions is the product of all different factors appearing in the denominators:

[tex]\[ (z + 1)(z + 4)(z + 8) \][/tex]

Next, we need to rewrite each fraction with this common denominator.

1. For the first fraction:
[tex]\[ \frac{z}{(z + 1)(z + 8)} \cdot \frac{z + 4}{z + 4} = \frac{z(z + 4)}{(z + 1)(z + 4)(z + 8)} = \frac{z^2 + 4z}{(z + 1)(z + 4)(z + 8)} \][/tex]

2. For the second fraction:
[tex]\[ \frac{8}{(z + 4)(z + 8)} \cdot \frac{z + 1}{z + 1} = \frac{8(z + 1)}{(z + 1)(z + 4)(z + 8)} = \frac{8z + 8}{(z + 1)(z + 4)(z + 8)} \][/tex]

Now we can add these fractions together:

[tex]\[ \frac{z^2 + 4z}{(z + 1)(z + 4)(z + 8)} + \frac{8z + 8}{(z + 1)(z + 4)(z + 8)} = \frac{z^2 + 4z + 8z + 8}{(z + 1)(z + 4)(z + 8)} \][/tex]

Combine like terms in the numerator:

[tex]\[ \frac{z^2 + 12z + 8}{(z + 1)(z + 4)(z + 8)} \][/tex]

We know (z+1)(z+4)(z+8) expands to [tex]\(z^3 + 13z^2 + 44z + 32\)[/tex]. Therefore, our final expression is:

[tex]\[ \frac{z^2 + 12z + 8}{z^3 + 13z^2 + 44z + 32} \][/tex]

Thus, the sum of the given rational expressions, simplified, is:

[tex]\[ \boxed{\frac{z^2 + 12z + 8}{z^3 + 13z^2 + 44z + 32}} \][/tex]