Suppose that a polynomial of degree \(n \ge 1\) is divided by a linear polynomial. How do we find out the quotient polynomial and the (constant) remainder? Note that the quotient polynomial will have a degree equal to \(n - 1\). The process of finding these out is straightforward. We call it the **division algorithm**, where the word algorithm means *a step-by-step procedure for calculating or evaluating something*. For now, we are discussing this algorithm only for the case where the divisor is a linear polynomial.

Let us take a concrete example. Suppose that the dividend \(a\left( x \right)\) and the divisor \(b\left( x \right)\) are given by

\[\begin{array}{l}a\left(x\right):2x^3-x^2+x-1\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;b\left(x\right):x+7\end{array}\]

We write these out as follows:

\(x + 7\left){\vphantom{1{2{x^3} - {x^2} + x - 1}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^3} - {x^2} + x - 1}}}\)

Now, we think of that term with which we multiply the first term of \(b\left( x \right)\) (which is *x*) to generate the first term of \(a\left( x \right)\):

\[x\,\, \times \,\,\left( ? \right) = 2{x^3}\]

Obviously, the multiplier is \(2{x^2},\) and so we write this as follows:

\[x + 7\mathop{\left){\vphantom{1{2{x^3} - {x^2} + x - 1}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^3} - {x^2} + x - 1}}}}\limits^{\displaystyle\,\,\, {2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\]

Now, we multiply the divisor by this multiplier which we have figured out, and write the result below the dividend, so that terms of the same degree align with each other:

\[x + 7\mathop{\left){\vphantom{1\begin{array}{l}2{x^3} - {x^2} + x - 1\\2{x^3} + 14{x^2}\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}2{x^3} - {x^2} + x - 1\\2{x^3} + 14{x^2}\end{array}}}}\limits^{\displaystyle\,\,\, {2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\]

Next, we subtract this new polynomial from the original dividend, and obtain our dividend for the next step of the algorithm:

\[x + 7\mathop{\left){\vphantom{1\begin{array}{l} 2{x^3} - {x^2} + x - 1\\ \underline {2{x^3} + 14{x^2}\,\,\,\,\,\,\,\,\,} \\ \, & - 15{x^2} + x \end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l} 2{x^3} - {x^2} + x - 1\\ \underline {2{x^3} + 14{x^2}\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,- 15{x^2} + x \end{array}}}} \limits^{\displaystyle {2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\]

Once we have this, we figure out the next multiplier: that term with which we multiply the first term of \(b\left( x \right)\) (which is *x*) to generate the highest degree term of the *new dividend*:

\[x\,\, \times \,\,(?) = - 15{x^2}\]

That multiplier is \( - 15x,\)and so we now have

\[\begin{array}{l}x + 7\!\!\!\!\mathop{\left){\vphantom{1{2{x^3} - {x^2} + x - 1}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^3} - {x^2} + x - 1}}}} \limits^{\displaystyle\,\,\;\;{2{x^2} - 15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\\ {\qquad\underline {2{x^3} + 14{x^2}\,\,\,\,\,\,\,\,\,} \\\, \qquad \qquad- 15{x^2} + x}\\{\qquad\qquad\underline {- 15{x^2} -105 x}}\\ {\qquad\qquad\qquad {106x - 1}}\\\end{array}\]

Finally, the last multiplier will be 106, and in the last step of the algorithm, we have

\[\begin{array}{l}x + 7\kern-3em\mathop{\left){\vphantom{1{2{x^3} - {x^2} + x - 1}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{2{x^3} - {x^2} + x - 1}}}} \limits^{\displaystyle\quad\quad\quad {2{x^2} - 15x + 106\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\\ {\qquad\qquad\underline {2{x^3} + 14{x^2}\,\,\,\,\,\,\,\,\,} \\\, \qquad \qquad- 15{x^2} + x}\\{\qquad\qquad\underline {\;\;- 15{x^2} -105 x}}\\ {\qquad\qquad\qquad{106x - 1\,\,\,\,\,\,\,\,\,} \\\,\qquad\qquad\qquad\underline {106x + 742}}\\{\qquad\qquad\qquad\qquad - 743}\\ \\\end{array}\]

Thus, the quotient polynomial and the remainder are

\[\begin{array}{l}q\left(x\right):2x^2-15x+106\\\,\,\,\,\,\,\;\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;r:-743\end{array}\]

**Example 1:** Consider the following two polynomials:

\[\begin{array}{l}a\left(x\right):6x^4-x^3+2x^2-7x+2\\ b\left(x\right):2x+3\end{array}\]

Find the quotient polynomial and the remainder when \(a\left( x \right)\)is divided by \(b\left( x \right)\).

**Solution:** You are urged to first work out this problem on your own and then compare your solution with the one presented below, in which all the steps of the division algorithm have been combined. In this solution, *M*_{1}, *M*_{2}, etc, are the multipliers with which we multiply the divisor at each successive step of the algorithm. *M*_{1} is the first multiplier, *M*_{2} is the second multiplier, and so on:

\[2x+3\overset{\overbrace{3{{x}^{3}}}^{{{M}_{1}}}-\overbrace{5{{x}^{2}}}^{{{M}_{2}}}+\overbrace{\frac{17}{2}x}^{{{M}_{3}}}-\overbrace{\frac{65}{4}}^{{{M}_{4}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\overline{\left){\begin{align} 6{{x}^{4}}-{{x}^{3}}+2{{x}^{2}}-7x+2 \\ \underline{6{{x}^{4}}+9{{x}^{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\underline{\begin{align} -10{{x}^{3}}+2{{x}^{2}}-7x+2 \\ -10{{x}^{3}}-15{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ \end{align}} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,17{{x}^{2}}-7x+2 \\ \,\,\,\,\,\,\,\,\,\,\underline{17{{x}^{2}}+\frac{51}{2}x\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,-\frac{65}{2}x+2 \\ \,\,\,\,\underline{-\frac{65}{2}x-\frac{195}{4}\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{203}{4} \\ \end{align}}\right.}}\]

Thus, the quotient polynomial and the remainder are:

\[\begin{array}{l}q\left(x\right):3x^3-5x^2+\frac{17}2x-\frac{65}4\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;r=\frac{203}4\end{array}\]

**Example 2:** Consider the following two polynomials:

\[\begin{array}{l}a\left(x\right):x^3-x^2+x-1\\b\left(x\right):2x+1\end{array}\]

Find the quotient polynomial and the remainder when \(a\left( x \right)\) is divided by \(b\left( x \right)\).

**Solution:** The division is shown below:

\[2x+1\overset{\overbrace{\,\frac{1}{2}{{x}^{2}}}^{{{M}_{1}}}-\overbrace{\frac{3}{4}x}^{{{M}_{2}}}+\overbrace{\frac{7}{8}}^{{{M}_{3}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{\overline{\left){\begin{align} & {{x}^{3}}-{{x}^{2}}+x-1 \\ & \underline{{{x}^{3}}+\frac{1}{2}{{x}^{2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ & \,\,\,\,\,\,-\frac{3}{2}{{x}^{2}}+x-1 \\ & \,\,\,\,\,\,\underline{\,-\frac{3}{2}{{x}^{2}}-\frac{3}{4}x\,\,\,\,\,\,\,\,\,\,} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{7}{4}x-1 \\ & \,\,\,\,\,\,\,\underline{\frac{7}{4}x+\frac{7}{8}\,\,\,} \\ & -\frac{15}{8}\,\,\,\,\,\,\, \\ \end{align}}\right.}}\]

Thus, we have

\[\begin{array}{l}q\left(x\right):\frac12x^2-\frac34x+\frac78\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;r=\;-\frac{15}8\end{array}\]