Answer:
[tex]\frac{x-5}{x+1}[/tex] , x ≠ −1, x ≠ −9
Step-by-step explanation:
To simplify this, we must first factor both the numerator and the denominator:
[tex]\frac{x^2+4x-45}{x^2+10x+9}[/tex]
They can be factored by using grouping.
The numerator expands to:
x^2 + 9x - 5x - 45
(x^2+9x) + (-5x-45)
x(x+9) - 5(x+9)
(x-5)(x+9)
The denominator expands to:
x^2 + 9x + x + 9
(x^2 + 9x) + (x + 9)
x(x+9) + 1(x+9)
(x+1)(x+9)
So now it looks like this:
[tex]\frac{(x-5)(x+9)}{(x+1)(x+9)}[/tex]
Now that it is fully factored (albeit not simplified), we can determine which values of x would be restricted from this function.
Since the factors of the denominator are x + 1 and x + 9, we know that values of x that would make them 0 are going to be restrictions.
(x + 1)(x + 9) = 0
x + 1 = 0; x = -1
x + 9 = 0; x = -9
Because these values of x make the denominator undefined, they are restrictions. An undefined denominator makes the entire answer undefined, which proves to be a problem, right?
And lastly, to simplify the answer, we can divide (x + 9) by (x + 9) and they will cancel out because they are the same. That leaves the answer to be:
x - 5 / x + 1
~~~~~~~~~~~~~
Have a nice day, you all. Remember, you're not alone - always someone there for you.
