The simplified expression for h(x) = f(x)/g(x) is [tex]h(x) = \frac{1}{3}\cdot \frac{x-4}{x-\frac{2}{3} }[/tex] and the domain of the function is all real numbers except x = 2/3.
How to determine the domain of a rational function
In this problem we have a rational function, the domain of rational functions is the set of all real numbers except the values such that the denominator becomes zero. First, we need to simplify the division of two rational functions:
[tex]h(x) = \frac{\frac{3\cdot x^{2} + 17\cdot x + 20}{3\cdot x^{2} + 10\cdot x - 8} }{\frac{3 \cdot x^{2} + 8\cdot x + 5}{x^{2}-3\cdot x - 4} }[/tex]
[tex]h(x) = \frac{(3\cdot x^{2}+17\cdot x + 20)\cdot (x^{2}-3\cdot x - 4)}{(3\cdot x^{2}+10\cdot x - 8)\cdot (3\cdot x^{2}+8\cdot x + 5)}[/tex]
[tex]h(x) = \frac{1}{3}\cdot \frac{\left(x^{2}+\frac{17}{3}\cdot x + \frac{20}{3} \right)\cdot (x^{2}-3\cdot x - 4)}{\left(x^{2}+\frac{10}{3}\cdot x -\frac{8}{3}\right)\cdot \left(x^{2}+\frac{8}{3}\cdot x +\frac{5}{3} \right)}[/tex]
[tex]h(x) = \frac{1}{3}\cdot \frac{\left(x + \frac{5}{3} \right)\cdot (x+4)\cdot (x - 4)\cdot (x+ 1)}{\left(x-\frac{2}{3} \right)\cdot (x + 4)\cdot (x + 1)\cdot \left(x + \frac{5}{3} \right)}[/tex]
[tex]h(x) = \frac{1}{3}\cdot \frac{x-4}{x-\frac{2}{3} }[/tex]
The simplified expression for h(x) = f(x)/g(x) is [tex]h(x) = \frac{1}{3}\cdot \frac{x-4}{x-\frac{2}{3} }[/tex] and the domain of the function is all real numbers except x = 2/3.
To learn more on rational functions: https://brainly.com/question/27914791
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