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 learn how to actually carry out this division later. 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:

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

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