Division Algorithm for General Divisors

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Let us now discuss polynomial division in the case of general divisors, that is, the degree of the divisor can be any positive integer less than that of the dividend. Consider the following example:

\[\begin{align}&a\left( x \right): \quad {x^4} - 3{x^3} + 2{x^2} - 7x + 4\\&b\left( x \right): \quad {x^2} + x + 2\end{align}\]

To divide these polynomials, we follow an approach exactly analogous to the case of linear divisors. At each step, we pick the appropriate multiplier for the divisor, do the subtraction process, and create a new dividend. In the following, we have broken down the division process into a number of steps:

Step-1

We start by figuring out the first multiplier:

\[{x^2} \times \left( ? \right) = {x^4}\]

Clearly, this multiplier will be \({x^2}\). We multiply this multiplier with the divisor, and carry out subtraction as follows:

\[{x^2} + x + 2\mathop{\left){\vphantom{1\begin{array}{l}{x^4} - 3{x^3} + 2{x^2} - 7x + 4\\{x^4} + {x^3} + 2{x^2}\\\;\;\;\;\overline {- 4{x^3} + 0{x^2} - 7x}\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{x^4} - 3{x^3} + 2{x^2} - 7x + 4\\{x^4} + {x^3} + 2{x^2}\\\;\;\;\;\overline { - 4{x^3} + 0{x^2} - 7x}\end{array}}}}\limits^{\displaystyle {{x^2}}}\]

The expression at the bottom is our new dividend.

Step-2

Now, we have to figure out our next multiplier:

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

Thus, we have:

\[{x^2} + x + 2\mathop{\left){\vphantom{1\begin{array}{l}{x^4} - 3{x^3} + 2{x^2} - 7x + 4\\\underline {{x^4} + {x^3} + 2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\, - 4{x^3} + 0{x^2} - 7x\\\;\;\;\;\underline { - 4{x^3} - 4{x^2} - 8x\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\,4{x^2} + x + 4\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{x^4} - 3{x^3} + 2{x^2} - 7x + 4\\\underline {{x^4} + {x^3} + 2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\, - 4{x^3} + 0{x^2} - 7x\\\;\;\;\;\underline { - 4{x^3} - 4{x^2} - 8x\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\,4{x^2} + x + 4\end{array}}}}\limits^{\displaystyle\,\,\, {{x^2} - 4x}}\]

Step-3

The next multiplier is similarly obtained:

\[{x^2} \times \left( ? \right) = 4{x^2}\]

Thus,

\[{x^2} + x + 2\mathop{\left){\vphantom{1\begin{array}{l}{x^4} - 3{x^3} + 2{x^2} - 7x + 4\\\underline {{x^4} + {x^3} + 2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\, - 4{x^3} + 0{x^2} - 7x\\\;\;\;\;\underline { - 4{x^3} - 4{x^2} - 8x\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\,4{x^2} + x + 4\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\underline {4{x^2} + 4x + 8\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \;\;\;\;\;\,\,\,\;\;\;\;\;\; - 3x - 4\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{x^4} - 3{x^3} + 2{x^2} - 7x + 4\\\underline {{x^4} + {x^3} + 2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\, - 4{x^3} + 0{x^2} - 7x\\\;\;\;\;\underline { - 4{x^3} - 4{x^2} - 8x\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\,4{x^2} + x + 4\\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\underline {4{x^2} + 4x + 8\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \;\;\;\;\;\,\,\,\;\;\;\;\;\; - 3x - 4\end{array}}}}\limits^{\displaystyle\,\,\, {{x^2} - 4x + 4}}\]

The degree of the last expression is less than that of the divisor, which means that the division process has to stop now. Thus, the quotient and the remainder polynomials are (respectively):

\[\begin{align}& q\left( x \right): \quad {x^2} - 4x + 4\\& r\left( x \right):  \quad - 3x - 4 \end{align}\]

Note how we have obtained the remainder as a linear polynomial in this case, whereas in the case of a linear divisor, the remainder must always be a constant.

Example 1: Divide the following polynomials:

\[\begin{align}&a\left( x \right): \quad2{x^5} - {x^3} + 1\\&b\left( x \right): \quad{x^3} - 2\end{align}\]

Solution: You should give this a try on your own before proceeding. We will combine all the steps into a single step:

\[{x^3} - 2\mathop{\left){\vphantom{1\begin{array}{l}2{x^5} - {x^3} + 1\\\underline {2{x^5}\, - 4{x^2}\,\,\,\,\,\,\,\,\,\,} \\ - {x^3}\,\, + 4{x^2} + 1\\\underline { - {x^3}\;\,\,\,\,\,\,\,\,\,\,\,\, + 2\,\,} \\\;\;\;\;\;\;\;\,\,\,\,4{x^2} - 1\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}2{x^5} - {x^3} + 1\\\underline {2{x^5}\, - 4{x^2}\,\,\,\,\,\,\,\,\,\,} \\ - {x^3}\,\, + 4{x^2} + 1\\\underline { - {x^3}\;\,\,\,\,\,\,\,\,\,\,\,\, + 2\,\,} \\\;\;\;\;\;\;\;\,\,\,\,4{x^2} - 1\end{array}}}}\limits^{\displaystyle\,\,\, {2{x^2} - 1\,\,}}\]

Thus, the quotient and remainder polynomials are:

\[\begin{align}&q\left( x \right):\quad 2{x^2} - 1\\&r\left( x \right):\quad 4{x^2} - 1\end{align}\]

Example 2: Divide the following polynomials:

\[\begin{align}&a\left( x \right): \quad {x^4} + 1\\&b\left( x \right): \quad 2{x^2} + 1\end{align}\]

Solution: We have:

\[2{x^2} + 1\mathop{\left){\vphantom{1\begin{array}{l}{x^4} + 1\\\underline {{x^4} + \frac{1}{2}{x^2}\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,- \frac{1}{2}{x^2} + 1\\\underline {\,\,\,\,\,\,\,\,\,\,\,\, - \frac{1}{2}{x^2} - \frac{1}{4}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{5}{4}\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}{x^4} + 1\\\underline {{x^4} + \frac{1}{2}{x^2}\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\, - \frac{1}{2}{x^2} + 1\\\underline {\,\,\,\,\,\,\,\,\,\,\, - \frac{1}{2}{x^2} - \frac{1}{4}} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{5}{4}\end{array}}}}\limits^{\displaystyle\,\,\, {\frac{1}{2}{x^2} - \frac{1}{4}}}\]

Thus, the quotient and remainder polynomials are:

\[\begin{align}&q\left( x \right):\quad \frac{1}{2}{x^2} - \frac{1}{4}\\&r\left( x \right): \quad\frac{5}{4}\end{align}\]

Note that the remainder turned out to be a constant in this case, even though the divisor was a quadratic polynomial. This should not surprise you. The remainder can have any degree from zero to one less than the degree of the divisor.

Example 3: Divide the following polynomials:

\[\begin{align}&a\left( x \right): \quad 2{x^8} - 1\\&b\left( x \right): \quad {x^3} - x + 2\end{align}\]

Solution: In this case, many terms are missing from the dividend. Note carefully how we handle this issue. We have:

\[{x^3} - x + 2\mathop{\left){\vphantom{1\begin{array}{l}2{x^8} - 1\\\underline {2{x^8} - 2{x^6} + 4{x^5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\2{x^6} - 4{x^5} - 1\\\underline {2{x^6}\,\,\,\,\,\,\,\,\,\,\,\, - 2{x^4} + 4{x^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 4{x^5} + 2{x^4} - 4{x^3} - 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,  \underline { - 4{x^5}\;\;\;\;\;\;\;\;\; + 4{x^3}\,\,\,\,\,\,\,\,\, - 8{x^2}\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{x^4}\,\,\,\,\, - 8{x^3} + 8{x^2} - 1\\\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {2{x^4} \;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2{x^2}\;\;\;\;\;\,\, + 4x\,\,} \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,- 8{x^3} + 10{x^2} - 4x - 1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 8{x^3}\;\;\;\;\,\,\,\,\,\,\,\,\,\, + 8x - 16\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{x^2}-12x+15\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}2{x^8} - 1\\\underline {2{x^8} - 2{x^6} + 4{x^5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\2{x^6}-4{x^5}-1\\\underline{2{x^6}\,\,\,\,\,\,\,\,\,\,\,\, - 2{x^4} + 4{x^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 4{x^5} + 2{x^4} - 4{x^3} - 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \underline { - 4{x^5}\;\;\;\;\;\;\;\;\; + 4{x^3}\,\,\,\,\,\,\,\,\, - 8{x^2}\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2{x^4}\,\,\,\,\,-8{x^3}+8{x^2} - 1\\\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {2{x^4}  \;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2{x^2}\;\;\;\;\;\,\, + 4x\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 8{x^3} + 10{x^2} - 4x - 1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 8{x^3}\;\;\;\;\,\,\,\,\,\,\,\,\,\, + 8x - 16\,} \\  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{x^2}-12x+15\end{array}}}}\limits^{\displaystyle\,\,\, {2{x^5} + 2{x^3} - 4{x^2} + 2x - 8  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}\]

Thus, the quotient and remainder polynomials are:

\[\begin{align}&q\left( x \right): \quad 2{x^5} + 2{x^3} - 4{x^2} + 2x - 8\\&r\left( x \right): \quad10{x^2} - 12x + 15\end{align}\]

Note that at each step, we brought down all the remaining terms from the previous dividend, so that we do not miss out any terms. Also, if the degrees of two terms do not match, we do not align them vertically. And when we do the subtraction at any stage, we rearrange the terms from the highest to the lowest degree. Observe the working more than once if you have to, but make sure you are clear about the process followed.

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