7
$\begingroup$

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?

$\endgroup$
6
  • $\begingroup$ If you know $f(x) = \sum_{n=0}^N a_n x^n$ how do you compute the $b_n$ such that $f(x+c) = \sum_{n=0}^N b_n x^n$ ? $\endgroup$
    – reuns
    Sep 24, 2017 at 17:54
  • $\begingroup$ Have you done any research? $\endgroup$ Sep 24, 2017 at 18:05
  • $\begingroup$ @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. $\endgroup$ Sep 24, 2017 at 18:08
  • $\begingroup$ Can anyone attempt to answer the question? $\endgroup$ Sep 25, 2017 at 10:58
  • $\begingroup$ @reuns, I tried but I don't know how to compute $b_n$ there. $\endgroup$ Sep 25, 2017 at 17:09

2 Answers 2

0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

I urge you to look at the answer I posted to a similar question here.

That question explains how to re-write the polynomial in nested form (Horner's form), which is a necessary first step to understanding how this algorithm works and also to appreciating its beauty and its efficiency.

The other key concept is the Remainder Theorem, which is true for polynomials of any degree, but is only being applied at each step in synthetic division to a linear factor of the form $(Ax+B)$. The theorem states that the remainder for $\frac{Ax+B}{x-a}$ is simply $Aa+B$. This can also be seen by considering a linear equation in point-slope form. Also very important to note is that the quotient for the prior expression is simply $A$, because the divisor $x-a$ is monic (has a coefficient of $1$ on its leading term).

Put more simply: $Ax+B = A(x-a) + (Aa+B)$

The algorithm starts with the two leading terms and then proceeds, incorporating one new term at at time, by using the remainder from the previous step as the "A" for the next step. Each remainder becomes, in turn, the leading coefficient for the next step, which then becomes the coefficient of the corresponding quotient term, after division. Remainders are found at each step by substituting $a$ for the next $x$ in the Horner's form polynomial, according to the Remainder Theorem.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .