We are starting with division and congruence in my algebra course... this is one of the first exercises for the division algorithm. I've done the first that were given with fixed values but now I have a few with different expressions and don't know how to get pass after a certain point.
Calculate the quotient and the rest for the division of a by b in the case: $$a = n^2 + 5$$ $$b = n + 2$$
I've done this: $$n+ 2 | n^2 + 5$$ $$n+ 2 | n^2 + 5 - n^2 - 2n$$ $$n+ 2 | 5 - 2n + 2n + 4$$ $$n+ 2 | 9$$
So I know that $|n+2|$ should be smaller or equal to 9 and that narrows down the possibilities, but is there a way to continue that doesn't involve calculating all of the possible remainders/quotients for each value up to 7? Thanks!
There is an easier way :
$$n^2+5=n^2-4+9=(n+2)(n-2)+9$$
So, if we divide $n^2+5$ by $n+2$, the remainder is $9$.
Note, that as soon we have a polynomial with degree smaller than the degree of the divisor (here $n+2$), we have found the remainder (here $9$).
I do not know if you had already polynomial division. This works as follows :
$(n^2+5):(n+2)=n-2$
$-(n^2+2n)$
$-2n+5$
$-(-2n-4)$
$ 9$
You divide the largest power of the dividend by the largest power of the divisor. Then you multiply the term you got with the divisor and subtract from the last dividend you used. You continue until the degree of the remainder is smaller than the degree of the divisor.