x^3-3x^2+3x-9 / x^4-81
i cant seem to get this answer right
assuming your function is ??
$\displaystyle \frac{x^3-3x^2+3x-9}{ x^4-81}$
You need to factor both the numerator and denominator to determine any cancelling expressions.
Then the denominator will factor as follows $\displaystyle x^4-81 = (x^2)^2-9^2 = (x^2+9)(x^2-9) = (x^2+9)(x-3)(x+3)$
This can be further factored using complex numbers. Is this a requirement?
Do you know the factor theorem? You will need it to factor the numerator. Try to find $\displaystyle f(a) = 0 $ where a is an integer factor of 9.
$\displaystyle \frac{x^3-3x^2+3x-9}{x^4-81} = \frac{x^3-3x^2+3x-9}{(x^2+9)(x^2-9)} = \frac{x^3-3x^2+3x-9}{(x^2+9)(x+3)(x-3)}$
This is just using the difference of squares method on the denominator, which is that $\displaystyle {a^2-b^2} = (a+b)(a-b)$
$\displaystyle \frac{x^2(x-3)+3(x-3)}{(x^2+9)(x+3)(x-3)} = \frac{(x^2+3)(x-3)}{(x^2+9)(x+3)(x-3)} = \frac{(x^2+3)}{(x^2+9)(x+3)}$
This is just using that $\displaystyle {ax+ay}={a(x+y)}$. You see how both $\displaystyle {x^3-3x^2}$ can be divided by $\displaystyle {x^2}$ and likewise $\displaystyle {3x-9}$ can be divided by 3? Apply that rule again on $\displaystyle {x^2(x-3)+3(x-3)}$, where $\displaystyle (x-3)$ is the a and $\displaystyle x^2$ and 3 are the x and y respectively. Then, I just divided both sides of the fraction by $\displaystyle (x-3)$.
The numerato you can factor by grouping. If you group x^3 and -3x^2 together and 3x and -9 together. You take the common factor from each and get
x^2(x-3) + 3(x-3)
then by the laws of grouping you can put them together and get (x^2 + 3)(x-3) and in the denominator you have (x-3)(x+3)(x^2 + 9) and then you can cross out the (x-3) on the top and bottom and you're left with:
(x^2 + 3)
__________
(x+3)(x^2 + 9)