division of binomials?

Question

i dont know how to solve this question:

(x^3+3x^2+3x+1)/(x+1)

and the answer is x^2+6x+9

its confusing the crap out of me because i got to
_x^2________________________
x+1)x^3+3x^2+3x+1
x^3+1x^2 |
------------------
+2x^2+3x


and i cant figure out what times (x+1)= x^2+3x


can someone help me and sorry if its so confusing

Answer 1

First of all, that answer is wrong. The answer is x²+2x+1 (it's fairly obvious, x³+3x²+3x+1 is the binomial expansion of (x+1)³, so dividing by x+1 you should get (x+1)², which is x²+2x+1)

Second, in the long division process, you only need to match up the leading terms, not the whole remainder. So figure out what times x is 2x², then multiply x+1 by that. You should get 2x²+2x, so subtracting that from the current remainder of 2x²+3x, you obtain x. Bring down the 1, and then your new remainder is x+1. Matching up the leading coefficients, you find that 1*x is x, so you multiply (x+1) by 1 and subtract it from x+1 to leave your final remainder of 0.

Answer 2

Here's how this works:

You only need to look at the leading terms. The leading term of x+1 is x, while the leading term of x^3 +3x^2+3x +1 is x^3. Thus, you need to figure out what to multiply x by to get x^3. Ofcourse this is x^2. So multiply (x+1) by x^2 to get x^3 + x^2 and subtract this from x^3 + 3x^2 +3x +1 to get 2x^2 +3x +1. Now, the leading term here is 2x^2, so you multiply x+1 by 2x to get 2x^2 +2x. Subtracting this from 2x^2 +3x +1 you get x + 1. Finally, you just need to multiply x+1 by 1 and then subtract to get 0. Therefore, the quotient is x^2 + 2x +1.

Answer 3

if you know how to do synthetic division you could reduce the equation to x^2 + 2x + 1 and factor that to (x+1)^2