# Why do we only consider the highest term of the divisor when dividing polynomials?

In every explanation for polynomial division they say that in every step we should only divide the highest term in the dividend with the highest term in the divisor, but I never see an explanation as to why this is the case. How can it be that we ignore all the other terms and still get an accurate result? We'd never do that with natural numbers..

You can see what I mean in step 1 in this Wikipedia page:
https://en.wikipedia.org/wiki/Polynomial_long_division#Example

• math.stackexchange.com/questions/496233/… I like this answer even though it hasn't been accepted. – randomgirl Jan 21 '17 at 0:01
• Thanks but I really don't understand the answer given in that post. Why do we want to "eliminate" the highest x term? And how subtraction is related to this? Here is the answer given in that post: So how to attain progress? All we need to do for that, is eliminate the highest power of x ("the variable") by subtracting a suitable multiple of the denominator, without introducing new, higher powers. As it turns out, by the nature of polynomial multiplication, we can use only a multiple of the highest power of the denominator to eliminate this highest power of the numerator. – Pineapple29 Jan 21 '17 at 0:06

## 1 Answer

It is for the same reason in division of natural numbers: if the dividend is less than the divisor, the division is trivial: the quotient is $0$ and the remainder is the dividend.

So if you have to divide the $n+1$-digit number $A=a_na_{n-1}\dots a_1a_0$ (in base $10$) by the $m+1$-digit number $D=d_md_{m-1}\dots d_1d_0$, i.e. divide $a_n10^n+a_{n-1}+\dots +10a_1+a_0$ by $d_m10^m+d_{m-1}10^{m-1}+10d_1+d_0$, and $A\ge D$, you begin with dividing the leftmost digit $a_n$ of the dividend by the leftmost digit $d_m$ of the dividend, to determine roughly how many times $10^{n-m}d$ are within $A$, and subtract this many times $D$ from $A$ to obtain a number $A_1<A$. Then same procedure is repeated with $A_1$, yielding a number $A_2<A_1$, and so on, until there remains a natural number $R<D$.

Replace the powers of $10$ with powers of $x$, and you obtain the Euclidean division of polynomials.

Remark:

There exists another division of polynomials: the division by increasing powers, in which we eliminate the lowest degree term of the polynomial by dividing it by the lowest degree term of the divisor. This division doesn't stop, or, more accurately, you decide a priori at which step the procedure stops.

• Thanks but I think the answer is too abstract.. can you please add an example that connects what you said with natural numbers? I don't see how I would ever ignore a number in the divisor when I use simple natural numbers in long division – Pineapple29 Jan 21 '17 at 0:56
• In long division in the naturals you consider enough of the highest places to know what the next digit of the quotient is and continue. If you are dividing $37$ into $12345$ you can see that $37$ doesn't divide into $12$, so you try $123$ and find it divides in $3$ times. Then you multiply $37 \cdot 3=111$ and subtract $111$. The polynomial division is the same way. You divide the largest terms so that when you do the subtraction the largest term gets canceled. You have reduced it to a simpler problem because the degree of the polynomial you are dividing has been reduced by $1$. – Ross Millikan Jan 21 '17 at 17:22