Hi Lizzie;
You are obviously working on the premise that only x^{2} can be factored. This is incorrect. A factor can be...
(x^{2} )(x )
or x^{3}, x^{4}, etc.
Henceforth
27x^{3}1
is factorable.
To factor, we must do the following...
We know that only
1x1=1
Henceforth...
(x^{2} 1)(x 1)
We also know that one of the 1s must be negative to yield 1. Therefore, one of the parenthetical equations must engage in subtraction, while the other addition.
To yield 27...
9 x 3
is the only possibility, and both numbers must either be negative or positive.
We also know that the FOIL equation must yield 0x. The above will produce 1x^{2} and 1x.
So we must start again as follows...
(x 1)(x 1)
is the only equation which will work. However, we cannot forget about the third x. Therefore...
x(x + 1)(x  1)
We also cannot forget about the 27. Furthermore, both parenthetical xs must be multiplied by the same number to yield 0x.
3x3x3 is the only mathematical formula which will yield 27.
3x(3x+1)(3x1)
FOIL
FIRST...3x(9x^{3})
OUTER...3x(3x)
INNER...3x(3x)
LAST...1
27x^{3}9x^{2}+9x^{2}1=27x^{3}1
Please let me know if you need anything else.
I LOVE MATH!!!!
10/3/2013

Vivian L.