f(x)/g(x)
(x2-1)/(x-1)
x cannot equal 1 because this would render the denominator, also known as the divisor, as zero.
x can be greater than 1.
If you would like, we can factor the numerator...
x2-1
(x+1)(x-1)
FIRST...(x)(x)=x2
OUTER...(x)(-1)=-1x=-x
INNER...(1)(x)=1x=x
LAST...(1)(-1)=-1
x2-x+x-1
x2-1
So...
f(x)/g(x)=[(x+1)(x-1)]/(x-1)
(x-1) is in the numerator and denominator. It cancels...
f(x)/g(x)=(x+1)
