# Dividing Two Polynomials

Dividing Two Polynomials

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

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

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

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.

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Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus