so we know 
has roots equal to the cube roots of 1, not including 
itself, which are
Any polynomial of the form 
is divisible by if both 
and 
(this is the polynomial remainder theorem).
This means
But since 
is a root to , it follows that
and since 
, we have 
so that
From here, notice that if 
for some integer 
, then
which is to say, 
is divisible by for all 
in the given range that are *not* multiples of 3, i.e. the integers 
and 
for 
.
Since 2005 = 668*3 + 1, it follows that there are 
integers such that 
.
Finally, 
.
I think it best to actually do the problem to determine the correct answer.
Clearing out the fractions, I found that
x(x-2) - (x -1) = 2x -5. This results in x=2 and x=3.
However, 2 is not a solution (but is an extraneous solution), because substituting 2 for x in the 2nd term results in div. by zero.
Thus, only the second answer is correct.
There are many equations that equal 7 but here are a few
3+4=7
15-8=7
3.5*2=7
49/7=7
-35/-5=7
Write the equations:
Substitute for h:
Solve for s:
There were 280 downloads of the standard version. There were 1120 downloads of the high-quality version.
If you <em>deposit</em> $35, you can imagine drawing a green line (representing the money gained) to the 35 on the number line, which can be seen as <em>moving </em>35 to the right<em />. Fo <em>w</em><em>ithdrawing, </em>imagine drawing a red line back 50 notches on the number line (representing money lost). The net effect that the deposit and withdrawal will have on your account balance is a
dollar <em>loss</em>, which can be represented on the number line as a line drawn to the point -15.
