Consider the following relation involving polynomials:

\[\left( {{x^3} + x + 1} \right) = \left( {x + 1} \right)\left( {{x^2} - x + 2} \right) + \left( { - 1} \right)\]

Let us denote the different polynomials involved in this relation as follows:

\[\begin{align}&a\left( x \right)\quad: \quad {x^3} + x + 1\\&b\left( x \right)\quad:\quad x + 1\\&q\left( x \right)\quad:\quad{x^2} - x\\&r\qquad\;\;: \quad- 1\end{align}\]

In terms of these, we can write the polynomial relation as follows:

\[a\left( x \right) = b\left( x \right)q\left( x \right) + r\]

We can look at this in the following manner: if we divide the polynomial \(a\left( x \right)\) by the linear polynomial \(b\left( x \right)\), the **quotient polynomial** is \(q\left( x \right)\), while the **remainder** (which is a constant) is *r*. The polynomial \(a\left( x \right)\) is the **dividend**, while the polynomial \(b\left( x \right)\) is the **divisor**. Let’s look at another example:

\[\left( {4{x^4} - {x^2} + 1} \right) = \left( {2x + 1} \right)\left( {2{x^3} - {x^2}} \right) + \left( 1 \right)\]

The dividend in this case is

\[a\left( x \right):4{x^4} - {x^2} + 1\]

while we are taking the divisor to be

\[b\left( x \right):2x + 1\]

Thus, the quotient is

\[q\left( x \right):2{x^3} - {x^2}\]

and the remainder is

\[r:1\]

We make the observation that whenever a polynomial is divided by a linear divisor, the remainder *will always be a constant*. Can you see why?

Now suppose that we have to divide the polynomial

\[a\left( x \right):{x^4} - {x^2} + x + 1\]

by the quadratic polynomial

\[b\left( x \right):{x^2} + 1\]

Once again, there will be a quotient polynomial and a remainder – and in this case the remainder *will not be a constant*; it will itself be a polynomial. The answer to this division problem is:

\[a\left( x \right) = b\left( x \right)\left( {{x^2} - 2} \right) + \left( {x + 3} \right)\]

Verify that this is indeed correct. We will soon see how to actually carry out this division. Thus, the quotient is the quadratic polynomial

\[q\left( x \right):{x^2} - 2,\]

while the remainder turns out to be a linear polynomial:

\[r\left( x \right):x + 3\]

Let us now consider the general division problem of two arbitrary polynomials. Suppose that \(a\left( x \right)\) is of degree *m*, and \(b\left( x \right)\) is of degree *n*, where *m* is greater than *n*. When we divide \(a\left( x \right)\) by \(b\left( x \right)\), the quotient polynomial \(q\left( x \right)\) will be of degree \(m - n\). Why? Because, the degree of the divisor \(b\left( x \right)\) is *n*, while the degree of the dividend \(a\left( x \right)\) is *m*, so the difference between the two degrees must be *bridged* by the degree of the quotient:

\[\begin{align}&{\rm{Deg}}\left( {b\left( x \right)} \right) + {\rm{Deg}}\left( {q\left( x \right)} \right) = {\rm{Deg}}\left( {a\left( x \right)} \right)\\&n\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,m - n\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,\,\,\,\,\,\,\,\,\,m\end{align}\]

The degree of the remainder \(r\left( x \right)\) *will always be less than* *n*. Let us understand why.