Read the introductory example

Let $A=x^{3}+2x^{2}-1$ and $B=x^{2}-x+1$. Then we can write


where $Q=x+3$ and $R=2x-4$ are called the quotient and the remainder, respectively. You may verify the above equation by expanding and simplifying the right-hand side.


In general, if $A$ and $B$ are two polynomials such that $\text{degree}(B)\leq \text{degree}(A)$, then there are unique polynomials $Q$ and $R$ such that


where $\text{degree}(R)<\text{degree}(B)$.

  • The process of finding $Q$ and $R$ is called the process of dividing $A$ by $B$.
  • In this process, $A$ is called the dividend, $B$ the divisor, $Q$ the quotient, and $R$ the remainder. If $R=0$, we say $A$ is divisible by $B$.

How the quotient is obtained is best explained in the following example.

 Divide $2x^{3}-32x-15$ by $x-3$ and find the quotient and the remainder.

First we make sure that the dividend and the divisor are written in descending powers of $x$. Next we divide the first term of the dividend by the first term of the divisor
\[\frac{2x^{3}}{x}=2x^{2}\] then multiply $2x^{2}$ by the divisor and subtract the result from the dividend

(2x^{3}-32x-15)-2x^{2}(x-3) & =\cancel{2x^{3}}-32x-15\cancel{-2x^{3}}+6x^{2}\\
& =6x^{2}-32x-15

or using the long division we have

To simplify calculations, we can reverse the signs of the product of the multiplication and then add it to the dividend; namely

Now we divide $6x^{2}-32x-15$ by $x-3$ and follow the same steps; that is, we write it in descending power of $x$ and divide its first term, $6x^{2}$, by the first term of the divisor, $x$: $6x^{2}/x=6x$

and again repeat until the degree of the remainder becomes less than the degree of the divisor



Here the quotient is $Q=2x^{2}+6x-14$ and the remainder is $R=-57.$

Now try to solve the following example. 


Divide $x^{5}-4x^{3}+x^{2}+1$ by $x^{2}+2x-1$ and find the quotient and the remainder


The quotient and the remainder are $Q=x^{3}-2x^{2}+x-3$ and $R=7x-2$, respectively.