# Division of Polynomials in General

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Now suppose that we have to divide the polynomial

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

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

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.

Polynomials
Polynomials
Grade 10 | Questions Set 1
Polynomials