Answer: 3) [tex]\frac{x+3}{x}[/tex]
Step-by-step explanation:
Start by factoring the numerator by grouping
[tex]\frac{x^3+2x^2-9x-18}{x^3-x^2-6x} \\\frac{(x^3+2x^2)+(-9x-18)}{x^3-x^2-6x} \\\frac{x^2(x+2)-9(x+2)}{x^3-x^2-6x} \\\frac{(x^2-9)(x+2)}{x^3-x^2-6x} \\\frac{(x-3)(x+3)(x+2)}{x^3-x^2-6x}[/tex]
Now factor the denominator
[tex]\\\frac{(x-3)(x+3)(x+2)}{x^3-x^2-6x}\\ \\\frac{(x-3)(x+3)(x+2)}{x(x^2-x-6)}\\[/tex]
Factor the trinomials by finding two numbers that add to get -1 and multiply to get -6. In this case the numbers are -3 and 2
[tex]\\\frac{(x-3)(x+3)(x+2)}{x(x^2-x-6)}\\\\\\\frac{(x-3)(x+3)(x+2)}{x(x-3)(x+2)}\\[/tex]
Cancel out the terms that are both in the numerator and denominator. In this case they are (x-3) and (x+2)
[tex]\\\frac{(x-3)(x+3)(x+2)}{x(x-3)(x+2)}\\\\\\\frac{(x+3)}{x}\\[/tex]
