# How does synthetic division work?

I've read that we can divide any polynomial by a linear polynomial by synthetic division considerably faster than that by long division method.

Now, I've learnt the steps to do so but I don't quite understand how it works.

Because here rather than dividing by the factor we are dividing by the zero ( i.e.* root*) and apparently these two cases are quite different.

But, in a way, the steps involved in this method seem to be equivalent to that in the long division but I haven't been to fully grasp how and why that is so.

Bonus Question:- Also, is there a similar relatively easy method for dividing by higher degrees polynomials too?

• If you know $f(x) = \sum_{n=0}^N a_n x^n$ how do you compute the $b_n$ such that $f(x+c) = \sum_{n=0}^N b_n x^n$ ? Sep 24 '17 at 17:54
• Have you done any research? Sep 24 '17 at 18:05
• @Chase, I've learnt how to apply the method. On a website (Purplemath) they compared it to long division method visually, without an explanation. From that I got that these two methods are equivalent in some way but I don't know exactly how. Sep 24 '17 at 18:08
• Can anyone attempt to answer the question? Sep 25 '17 at 10:58
• @reuns, I tried but I don't know how to compute $b_n$ there. Sep 25 '17 at 17:09

I have now moved on (with much of my confusion cleared). So here, for posterity's sake I'm providing two links that helped me to understand why 'synthetic division' works :- I think This and This will be helpful to anyone who has the same doubt in the future...

I'm aware the above is not much of answer but I decided to write it as one anyway because I think comments tend to get deleted, so the links I provided would disappear if I write them in comments and I didn't include this in the question details as I don't think most of the visitors on the forum actually read whole of the question.

Now, if anyone wants to give their own answer or make up a "fully-fledged" answer based on the information in the links I provided, I welcome them to do so and in that case I'll be happy to delete this answer.