Answer:
[tex](\frac{f}{g})(x) = \frac{x- 3}{2x + 1}[/tex]
Step-by-step explanation:
Given
[tex]f(x) =x^2 -x - 6[/tex]
[tex]g(x) = 2x^2 + 5x + 2[/tex]
Required
[tex](\frac{f}{g})(x)[/tex]
This is calculated as:
[tex](\frac{f}{g})(x) = \frac{f(x)}{g(x)}[/tex]
So, we have:
[tex](\frac{f}{g})(x) = \frac{x^2 - x - 6}{2x^2 + 5x + 2}[/tex]
Expand
[tex](\frac{f}{g})(x) = \frac{x^2 +2x - 3x - 6}{2x^2 + 4x+x + 2}[/tex]
Factorize
[tex](\frac{f}{g})(x) = \frac{x(x +2) - 3(x + 2)}{2x(x + 2)+1(x + 2)}[/tex]
Factor out x + 2
[tex](\frac{f}{g})(x) = \frac{(x- 3)(x + 2)}{(2x + 1)(x + 2)}[/tex]
Cancel out x + 2
[tex](\frac{f}{g})(x) = \frac{x- 3}{2x + 1}[/tex]
