Answer:
x cannot be 5 or -3.
[tex]\frac{x+5}{x+3}[/tex]
Step-by-step explanation:
The restrictions for a fraction is that the bottom cannot be 0.
So if we find when the bottom is 0 we have found the values that x cannot be.
Let's solve x^2-2x-15=0.
Since the coefficient of x^2 is 1 all we have to do is find two numbers whose product is -15 and whose sum is -2.
Those numbers are -5 and 3 since (-5)(3)=-15 and (-5)+(3)=-2.
So the factored form of the equation is:
(x-5)(x+3)=0
This means either x-5=0 or x+3=0.
We do have to solve both.
x-5=0 can be solved by adding 5 on both sides.
x-5+5=0+5
x+0=0+5
x=5
x+3=0 can be solved by subtracting 3 on both sides.
x+3-3=0-3
x+0=0-3
x=-3
So x can be any number except x=5 or x=-3.
We already factored the bottom as (x-5)(x+3).
The top is a difference of squares, x^2-a^2,
which can be factored as (x-a)(x+a).
So the top factors as (x-5)(x+5).
The fraction can then be written as:
[tex]\frac{x^2-25}{x^2-2x-15}=\frac{(x-5)(x+5)}{(x-5)(x+3)}[/tex]
This can further simplified assuming x is not 5 we can write it as [tex]\frac{x+5}{x+3}[/tex]. I canceled the common factor of (x-5).
