# Tracking Only the Coefficients

Tracking Only the Coefficients

The division algorithm might strike you as lengthy and time consuming, but with a little thinking, we can use a technique which can make it significantly faster. This technique can be called the technique of tracking only the coefficients. Basically, the idea is to not write the variable at any stage, but only write the coefficients, since only the coefficients matter. Let us see how. Suppose that you have to divide the following polynomials:

\begin{align}&a\left( x \right)\;:\quad{x^4} - 2{x^3} + {x^2} + 4x - 5\\&b\left( x \right)\;:\quad{x^2} + 2x + 2\end{align}

Now, let us carry out the division process step-by-step. In each of the following steps, we have on the left our familiar division step, but on the right, we omit the variables, and keep only the coefficients. The two sides are exactly the same – the only difference is that on the right, we are trying to save some effort by not writing the variables, since the variables are not giving us any new information – only the coefficients are:

### Step-1

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

On the right side above, in the quotient area, we have written only 1. It should be understood that this corresponds to 1x2, that is, the degree of this term is 2, since the difference between the degrees of the dividend and the divisor is 2. Now, to the next step:

### Step-2

${x^2} + 2x + 2\mathop{\left){\vphantom{1\begin{array}{l} {x^4} - 2{x^3} + {x^2} + 4x - 5\\ \underline {{x^4} + 2{x^3} + 2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\, - 4{x^3}\,\,\, - {x^2} + 4x\\ \;\underline {\,\,\,\,\, - 4{x^3} - 8{x^2} - 8x\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7{x^2} + 12x - 5 \end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l} {x^4} - 2{x^3} + {x^2} + 4x - 5\\ \underline {{x^4} + 2{x^3} + 2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\, - 4{x^3}\,\,\, - {x^2} + 4x\\ \;\underline {\,\,\,\,\, - 4{x^3} - 8{x^2} - 8x\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7{x^2} + 12x - 5 \end{array}}}} \limits^{\displaystyle\,\,\, {{x^2} - 4x \,\,\,\,\,\,\,\,}}\,\,\,\,\,\,\,\,\left| \,\,\,\,\,\,\,\,{1\;\;\;2\;\;\;2\;\;\mathop{\left){\vphantom{1\begin{array}{l} 1\;\; - 2\;\;\,\;\,1\;\;\,\,\,\,4\;\; - 5\\ \underline {1\;\;\,\,\,\,2\;\;\,\,\;2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\, - 4\;\;\, - 1\;\;\,\,\,4\\ \,\,\,\,\,\underline { - 4\,\,\,\, - 8\,\,\, - 8\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,7\,\,\,\,\,\,12\,\,\, - 5 \end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l} 1\;\; - 2\;\;\,\;\,1\;\;\,\,\,\,4\;\; - 5\\ \underline {1\;\;\,\,\,\,2\;\;\,\,\;2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\, - 4\;\;\, - 1\;\;\,\,\,4\\ \,\,\,\,\,\underline { - 4\,\,\,\, - 8\,\,\, - 8\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,7\,\,\,\,\,\,12\,\,\, - 5 \end{array}}}} \limits^{\displaystyle\\\\ {1\;\; - 4 \,\,\;\;\; \,\,\,\,\,\,}}} \right.$

Again, on the right side, in the quotient area, the second term is written as –4, which should be understood as –4x.

### Step-3

${x^2} + 2x + 2\mathop{\left){\vphantom{1\begin{array}{l} {x^4} - 2{x^3} + {x^2} + 4x - 5\\ \underline {{x^4} + 2{x^3} + 2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\, - 4{x^3}\,\, - {x^2} + 4x\\ \,\,\,\,\,\,\,\underline { - 4{x^3} - 8{x^2} - 8x\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7{x^2} + 12x - 5\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {7{x^2} + 14x + 14\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\; - 2x\,\,\,\, - 19 \end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l} {x^4} - 2{x^3} + {x^2} + 4x - 5\\ \underline {{x^4} + 2{x^3} + 2{x^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\, - 4{x^3}\,\, - {x^2} + 4x\\ \,\,\,\,\,\,\,\underline { - 4{x^3} - 8{x^2} - 8x\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7{x^2} + 12x - 5\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {7{x^2} + 14x + 14\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\; \;\;\,\,\,\,\,\, - 2x\,\,\,\, - 19 \end{array}}}} \limits^{\displaystyle\\\\ {{x^2} - 4x + 7 \,\,\,}}\,\,\,\,\,\,\,\,\left| \,\,\,\,\,\,\,\,{1\;2\;2\;\mathop{\left){\vphantom{1\begin{array}{l} 1\;\;\, - 2\;\;\,\;\,1\;\;\,\,4\;\;\, - 5\\ \underline {1\;\;\,\,\,\,2\;\;\,\,\;2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\, - 4\;\;\, - 1\;\;\,\,4\\ \,\,\,\,\,\underline { - 4\,\,\,\, - 8\,\,\, - 8\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7\,\,\,\,\,12\,\,\, - 5\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,\,\,\,7\,\,\,\,\,14\,\,\,\,\,14\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\; - 2\,\, - 19 \end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l} 1\;\;\, - 2\;\;\,\;\,1\;\;\,\,4\;\;\, - 5\\ \underline {1\;\;\,\,\,\,2\;\;\,\,\;2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\, - 4\;\;\, - 1\;\;\,\,4\\ \,\,\,\,\,\underline { - 4\,\,\,\, - 8\,\,\, - 8\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,7\,\,\,\,\,12\,\,\, - 5\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,\,\,\,7\,\,\,\,\,14\,\,\,\,\,14\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \;\; - 2\,\, - 19 \end{array}}}} \limits^{\displaystyle\\\\ {1\;\;\; - 4\,\,\,\,\,\,7 \,\,\,\,\,\,\,\,\,}}} \right.$

Observe the right side carefully. The division process must stop now, since the last expression has a degree less than that of the divisor. Thus, the quotient and remainder polynomials can now be written from their coefficients:

\begin{align}&q\left( x \right) \equiv 1, - 4,\;7 \equiv {x^2} - 4x + 7\\&r\left( x \right) \equiv - 2,\; - 19 \equiv - 2x - 19\end{align}

You may say, in this case we were carrying out the complete division process on the left, and so it was easy to tell when to stop, and easy to figure out how to construct the quotient and remainder polynomials from only the coefficients on the right side, since we already had our answer on the left side. Well, your concern is genuine. If you were tracking only the coefficients (doing the stuff on the right side), if might not be obvious when to stop and how to reconstruct the quotient and remainder polynomials. But if you think carefully, it’s not difficult. For example, once you reach the last step where you obtain (–2, –19), it should be evident that –19 corresponds to the constant term in the remainder, and hence –2 must be the linear term. And since there are no more terms, (–2, –19) corresponds to a linear polynomial, and hence the division must stop at this stage.

Similarly, look carefully at the quotient we obtained (on the right side):

$q\left( x \right) \equiv 1,\; - 4,\;7$

How do we reconstruct the actual quotient polynomial from this? We already observed earlier that the first number (which is 1) corresponds to $${x^2}$$, which means that the next one, that is –4, must be the coefficient of x, and the third number, which is 7, must be the constant term. Thus, the reconstruction will be:

\begin{align}q\left( x \right) &= \left( {1 \times {x^2}} \right) + \left( { - 4 \times x} \right) + \left( 7 \right)\\& = {x^2} - 4x + 7\end{align}

Example 1: Divide the following polynomials by tracking only the coefficients:

\begin{align}&a\left( x \right)\;:\quad 2{x^5} - {x^4} + 2{x^3} - {x^2} + x + 2\\&b\left( x \right)\;:\quad{x^2} + 2x + 3\end{align}

Solution: We have presented the solution by breaking it down into steps:

### Step-1

$1\;\;\,2\,\;\;3\mathop{\left){\vphantom{1\begin{array}{l}2\;\;\;\; - 1\;\;\;\;2\;\;\;\; - 1\;\;\;\;1\;\;\;\;2\\\underline {2\;\;\;\;\,\,\,4\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\, - 5\,\,\,\, - 4\;\;\,\, - 1\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}2\;\;\;\; - 1\;\;\;\;2\;\;\;\; - 1\;\;\;\;1\;\;\;\;2\\\underline {2\;\;\;\;\,\,\,4\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\, - 5\,\,\,\, - 4\;\;\,\, - 1\end{array}}}}\limits^{\displaystyle\,\,\, {2\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}$

### Step-2

$1\;\,\;2\;\,\;3\mathop{\left){\vphantom{1\begin{array}{l}2\;\;\;\; - 1\;\;\;\;2\;\;\;\; - 1\;\;\;\;\,\,1\;\;\;\;2\\\underline {2\;\;\;\;\,\,\,4\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\, - 5\,\,\,\, - 4\;\;\,\, - 1\\\;\;\;\;\;\underline { - 5\;\;\; - 10\;\; - 15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,14\,\,\,\,\,1\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}2\;\;\;\; - 1\;\;\;\;2\;\;\;\; - 1\;\;\;\;\,\,1\;\;\;\;2\\\underline {2\;\;\;\;\,\,\,4\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\, - 5\,\,\,\, - 4\;\;\,\, - 1\\\;\;\;\;\;\underline { - 5\;\;\; - 10\;\; - 15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,14\,\,\,\,\,1\end{array}}}}\limits^{\displaystyle\,\,\, {2\;\;\;\; - 5 \; \,\,\,\,}}$

### Step-3

$1\;\;\,2\,\;\;3\mathop{\left){\vphantom{1\begin{array}{l}2\;\;\;\; - 1\;\;\;\;2\;\;\;\; - 1\;\;\;\;\,\,\,\,\,1\,\,\,\;\;\;\;2\\\underline{2\;\;\;\;\,\,\,4\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\, - 5\,\,\,\, - 4\;\;\,\, - 1\\\;\;\;\;\;\underline { - 5\;\;\; - 10\;\; - 15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,14\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\underline {\,\,\,\,\,6\,\,\,\,\,\,\,\,12\,\,\,\,\,\,\,\,18\,\,\,\,\,\,} \\ \,\,\,\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \,\, \,\,\,\,\,\,\,\,\,\,2\,\,\, - 17\,\,\,\,\,2\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}2\;\;\;\; - 1\;\;\;\;2\;\;\;\; - 1\;\;\;\;\,\,\,\,\,1\,\,\,\;\;\;\;2\\\underline {2\;\;\;\;\,\,\,4\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\, - 5\,\,\,\, - 4\;\;\,\, - 1\\\;\;\;\;\;\underline { - 5\;\;\; - 10\;\; - 15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,14\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,\,\,\,\,6\,\,\,\,\,\,\,\,12\,\,\,\,\,\,\,\,18\,\,\,\,\,\,} \\ \,\,\,\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \,\, \,\,\,\,\,\,\,\,\,\,2\,\,\, - 17\,\,\,\,\,2\end{array}}}}\limits^{\displaystyle\,\,\, {2\;\;\;\; - 5\;\;\;\;\;6 \,\,\,\,\,\,\,\,\,\,\,}}$

### Step-4

$1\;\;\,2\,\;\;3\mathop{\left){\vphantom{1\begin{array}{l}2\;\;\;\; - 1\;\;\;\;\,2\;\;\;\; - 1\;\;\;\;\,\,\,\,\,\,1\;\;\;\;\,\,2\\\underline {2\;\;\;\;\,\,\,4\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\, - 5\,\,\,\, - 4\;\;\,\, - 1\\\;\;\;\;\;\underline { - 5\;\;\; - 10\;\; - 15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,14\,\,\,\,\,\,\,\,1\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \underline {\,\,\,\,\,6\,\,\,\,\,\,\,\,12\,\,\,\,\,\,\,\,18\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,2\,\,\,\,\,\, - 17\,\,\,\,\,2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,\,\,2\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,6\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\; - 21\,\,\,\, - 4\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}2\;\;\;\; - 1\;\;\;\;\,2\;\;\;\; - 1\;\;\;\;\,\,\,\,\,\,1\;\;\;\;\,\,2\\\underline {2\;\;\;\;\,\,\,4\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\, - 5\,\,\,\, - 4\;\;\,\, - 1\\\;\;\;\;\;\underline { - 5\;\;\; - 10\;\; - 15\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,14\,\,\,\,\,\,\,\,1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \underline {\,\,\,\,\,6\,\,\,\,\,\,\,\,12\,\,\,\,\,\,\,\,18\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,2\,\,\,\,\,\, - 17\,\,\,\,\,2\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \underline {\,\,\,2\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,6\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\; - 21\,\,\,\, - 4\end{array}}}}\limits^{\displaystyle\,\,\, {2\;\;\;\; - 5\;\;\;\;\;6\,\,\,\,\,\,\,2\; \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}$

Thus, the quotient and remainder polynomials (in terms of their coefficients) are:

\begin{align}&q\left( x \right) \equiv 2,\; - 5,\;6,\;2\\&r\left( x \right) \equiv - 21,\; - 4\end{align}

How do we reconstruct them? Easy. The first number in the quotient polynomial, which is 2, must be the coefficient of $${x^3}$$, since that must be the highest degree term in the quotient. The next number, which is –5, will then be the coefficient of $${x^2}$$, and so on. Thus,

\begin{align}q\left( x \right) &= \left( {2 \times {x^3}} \right) + \left( { - 5 \times {x^2}} \right) + \left( {6 \times x} \right) + 2\\& = 2{x^3} - 5{x^2} + 6x + 2\end{align}

To reconstruct the remainder, note that the second number in the remainder is –4, and this must be the constant term, and therefore, the first number, which is –21, must be the coefficient of x. Thus,

$r\left( x \right) = - 21x - 4$

Example 2: Divide the following polynomials:

\begin{align}&a\left( x \right)\,:\; 2{x^4} - 3{x^2} + 2\\&b\left( x \right)\,: \; {x^2} + 3\end{align}

Solution: Note carefully that in both the dividend and the divisor polynomials, terms of some degrees are missing. While tracking the coefficients, we must keep track of terms of every degree, otherwise we risk mismatching terms while subtracting. Therefore, we will write the dividend polynomial as

$a\left( x \right) \equiv 2,\;0,\; - 3,\;0,\;2$

and the remainder polynomial as

$b\left( x \right) \equiv 1,\;0,\;3$

Now, we proceed with the division process:

$1\;\;0\;\;3\mathop{\left){\vphantom{1\begin{array}{l} 2\;\;\;\;0\;\;\;\;\,\, - 3\;\;\;\;\,\,0\;\;\;\,\,\,\,\,\,2\\ \underline {2\;\;\;\;0\,\,\,\,\,\,\,\,\,\,\,6\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,- 9\;\;\,\,\,\,\,0\\ \;\;\;\;\;\;\;\underline {\,0\;\;\;\,\,\,\,\,\,0\;\;\,\,\,\,\,0\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\; - 9\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,2\\ \,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\underline { - 9\,\,\,\,\,\,\,\,0\,\,\,\,\, \,\, - 27\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;29 \end{array}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l} 2\;\;\;\;0\;\;\;\;\,\, - 3\;\;\;\;\,\,0\;\;\;\,\, \,\,\,\,2\\ \underline {2\;\;\;\;0\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\fbox{0}- 9\;\;\,\,\, \,\,0\\ \;\;\;\;\;\;\underline {\,0\;\;\;\,\,\,\,\,\,0\;\;\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\; - 9\,\,\,\,\,\,\,\,0\, \,\,\,\,\,\,\,\,\,\,2\\ \,\,\;\;\;\;\;\;\;\;\;\;\;\;\underline { - 9\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\, - 27\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;29 \end{array}}}} \limits^{\displaystyle\,\,\, {2\;\;\; \;0\;\;\;\;\; - 9 \,\,\,\,\,}}$

The quotient polynomial is:

\begin{align}&q\left( x \right) \equiv 2,\;0,\; - 9\\&\Rightarrow \;\;q\left( x \right) = \left( {2 \times {x^2}} \right) + \left( {0 \times x} \right) + \left( { - 9} \right)\\& = 2{x^2} - 9\end{align}

The remainder turns out to be a constant:

\begin{align}&r\left( x \right) \equiv 29\\&\Rightarrow \;\;\;r\left( x \right) = 29\end{align}

Note carefully that the multiplier in the second step was 0, since the number in the boxed position turned out to be 0. A possible mistake could have been to ignore this 0, and take the second multiplier as –9. But that is plain wrong! Because then, while reconstructing the quotient polynomial, we would have taken –9 to be the coefficient of x, whereas actually, –9 is the constant term in the quotient polynomial. Thus, a word of advice while tracking only the coefficients: do not ignore zeroes anywhere.

Here is another example.

Example 3: Divide the following polynomials:

\begin{align}&a\left( x \right)\,: \; 2{x^7} - {x^5} + 5{x^3} - {x^2} + 6\\&b\left( x \right)\,:\; {x^3} + 3{x^2} - 1\end{align}

Solution: We have:

$1\;3\;0\; - 1\mathop{\left){\vphantom{1\begin{array}{l}2\;\;\;\;\;0\;\;\;\; - 1\;\;\;\;\;\,\,\,\,0\;\;\;\;\;\;\,\,\,\,\,5\;\;\;\;\;\,\,\, - 1\;\;\;\;\;\,0\;\;\;\;\;\;\,\,\,6\\\underline {2\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\, - 6\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,5\\\,\,\,\,\,\,\underline { - 6\,\,\,\,\, - 18\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,17\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\, - 1\\ \,\,\,\,\,\,\,\,\,\underline {\,17\,\,\,\,\,\,\,\,\,\,\,\,\,5\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,17\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\, - 49\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,16\,\,\,\,\,\,\,\,\,0\\ \,\,\,\,\,\,\,\,\,\,\,\,\underline { - 49\,\,\,\,\,\, - 147\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,49\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,146\,\,\,\,\,\,\,\,\,16\,\,\,\, - 49\,\,\,\,\,\,\,\,\,6\\ \,\,\,\,\,\,\,\,\,\underline {146\,\,\,\,\,\,\,\,438\,\,\,\,\,\,\,0\,\,\,\,\,\, - 146} \\ \;\;\;\;{\rm{END:}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 422\,\,\,\,\, - 49\,\,\,\,152\end{array}}}\right.\!\!\!\!\overline{\,\,\,\vphantom 1{\begin{array}{l}2\;\;\;\;\;0\;\;\;\; - 1\;\;\;\;\;\,\,\,\,0\;\;\;\;\;\;\,\,\,\,\,5\;\;\;\;\;\,\,\, - 1\;\;\;\;\;\,0\;\;\;\;\;\;\,\,\,6\\\underline {2\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\\,\,\,\,\,\, - 6\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,5\\\,\,\,\,\,\,\underline { - 6\,\,\,\,\, - 18\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,6\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,17\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\, - 1\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {\,17\,\,\,\,\,\,\,\,\,\,\,\,\,51\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\, - 17\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,-49\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,16\,\,\,\,\,\,\,\,\,0\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline { - 49\,\,\,\,\,\, - 147\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,49\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;146\,\,\,\,\,\,\,\,\,16\,\,\,\,49\,\,\,\,\,\,\,\,\,6\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\;\;\underline {146\,\,\,\,\,\,\,\,438\,\,\,\,\,\,\,0\,\,\,\,\,\, - 146} \\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{END:}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 422\,\,\,\,\, - 49\,\,\,\,152\end{array}}}}\limits^{\displaystyle\,\,\, {\,2\;\;\;\; - 6\;\;\;\;17\;\;\;\, - 49\;\;\;\;\;146\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}}$

The quotient and remainder polynomials are:

\begin{align}&q\left( x \right) \equiv 2,\; - 6,\;17,\; - 49,\;146\\ &\Rightarrow q\left( x \right) = 2{x^4} - 6{x^3} + 17{x^2} - 49x + 146\\&r\left( x \right) \equiv - 422,\; - 49,\;152\\& \Rightarrow r\left( x \right) = - 422{x^2} - 49x + 152\end{align}

Once you get a hang of this technique, division of polynomials becomes extremely fast to carry out.

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More Important Topics
Numbers
Algebra
Geometry
Measurement
Money
Data
Trigonometry
Calculus