Originally posted by: sao123
I performed this polynomial division myself, which results in a remainder, which doesnt appear you dealt with.
Yes I did, the remainder is always zero when you divide by a root factor.
In general you have,
P(x) = Q(x) (x-a) + R (where Q is the quotient and R the remainder). It follows (by substituting
x=a in the above) that R = P(a). This is a well know result called the
Polynomial remainder theorem. It is true in general (whether or not "a" is a root), but naturally for the case when "
a" is a root we simply have
R = P(a)[/b] which is
zero by definition.
How did you ensure you actually have the integer root?
We were told that one of the three roots was integer, I simply choose to name that root as "
a". There is no loss of generality involved here. Eventually I was able to prove that
a was in fact 1200, no doubt an integer.
furthermore doing long poly division by
x - 1200 which we have verified as a root leaves an remainder N - M + 1200 ^ 2 which in turn must be zero for 1200 to be the integer root. Leaving the equation N = M - 1200 ^ 2 which we must satisfy also.
No, you made an error in your calcuations. If you re-do that calc you will suely find that the remainder is,
R = n + 1200m - 1200^3 .
You may do this the hard way if you wish, you'll get the same answer (eventually), but remember that by the above "remainder theorem" that this remainder is none other than
P(1200). That is,the original polynomial evaluated at x=1200.
Oh and btw, just take a look at
step_8 in my solution and you will see precisely that equation, and yes it was set to equal zero as required.
Facebook
Twitter