Polynomial Division

I. Introduction

Let us understand what is a polynomial division. In normal integer division, we obtain a quotient and a remainder, such that $dividend = quotient \times divisor + remainder$. Similarly, in polynomial division, when we divided $P(x)$ by $G(x)$ we obtain two polynomials $Q(x)$ and $R(x)$ as the quotient and remainder polynomials such that $P(x) = Q(x) \times G(x) + R(x)$.

II. Method

We begin by arranging each polynomial such that the terms are in descending order of its degree, that is, the highest degree term is at the leftmost position and the lowest degree is on the rightmost position.

At each step we eliminate the term with the highest degree by subtracting a multiple of the divisor, then write down the remaining dividend.

We continue to do this, till we are left with a remainder, which has an order less than the order of the divisor. This is our final remainder.

We will take a couple of examples to understand the process. We will start with a relatively small example to begin with:

Let us divide polynomial

$P(x) = 2x^2 + x - 15$ by polynomial

$G(x) = x+3$ and understand the steps that are required.

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The long division goes as follows:

$\begin{array}{rl}\phantom{000000}2x-5 \\[-5pt]x+3 \enclose{longdiv}{2x^2 + x - 22} \\ \underline{-(2x^2+6x)}\phantom{000} & \Leftarrow & \hbox{Step 1} \\ -5x-22 \\ \underline{-(-5x-15)} & \Leftarrow & \hbox{Step 2}\\ -7\end{array}$

Let us look at the details of each step now.

$\underline{Step\ 1}$:

In the first step we are eliminating the term $2x^2$ from the dividend, which is the highest power of $x$ in the dividend. We do so by multiplying the divisor $(x+3)$ by $2x$ and subtracting it from the dividend.

$2x^2 + x - 22 - 2x(x+3)$

$= 2x^2 + x - 22 - 2x^2 -6x$

$= -5x -22$

So, we are left with $-5x - 22$ as the dividend and our quotient, so far, is $2x\ldots$

$\underline{Step\ 2}$:

Next, we eliminate the term $-5x$ from the dividend, by multiplying the divisor by $-5$ and subtracting it from the dividend.

$-5x - 22 - (-5)(x+3)$

$= -5x - 22 + 5x + 15$

$= -7$

Here, we see that the order of the remainder is 0, which is less than the order of the divisor, which is 1 and our division is complete. Therefore, our final quotient is $2x - 5$ and remainder is $-7$

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Now, let us try this with slightly larger polynomials. We will use the following polynomials for our next example:

$P(x) = x^5 - 3x^4 - 28x^3 + 47x^2 + 49x - 45$, and divide this by the polynomial using the long division method,

$G(x) = x^2 + 4x - 3$, also, note that the order of $G(x)$ is $2$.

Like before, we will first perform the long division and then look at what is happening in each step.

$\begin{array}{rl} \phantom{0000000000000}x^3-7x^2+\ \ 3x+14 \\[-5pt]x^2 + 4x - 3 \enclose{longdiv} {x^5 - 3x^4 - 28x^3 + 47x^2 + 49x - 45} \\ \underline{-(x^5+4x^4-3x^3)}\phantom{0000000000000000} & \Leftarrow & \hbox{Step 1} \\ -7x^4-25x^3+47x^2+49x-45 \\ \underline{-(-7x^4-28x^3+21x^2)}\phantom{000000000} & \Leftarrow & \hbox{Step 2}\\ 3x^3+26x^2+49x-45 \\ \underline{-(3x^3+12x^2-9x)}\phantom{0000} & \Leftarrow & \hbox{Step 3} \\ 14x^2+58x-45 \\ \underline{-(14x^2+56x-42)} & \Leftarrow & \hbox{Step 4} \\ 2x-3\end{array}$

$\underline{Step\ 1}$:

We started by eliminating the highest power of $x$ in the dividend, which is $x^5$. We did so by multiplying $G(x)$ by $x^3$, and subtracting from the divisor.

$(x^5 - 3x^4 - 28x^3 + 47x^2 + 49x - 45) - x^3(x^2 +4x -3)$

$= x^5 -3x^4 - 28x^3 + 47x^2 + 49x - 45 - x^5 -4x^4 + 3x^3$

$= -7x^4 - 25x^3 + 47x^2 + 49x - 45$, and we got the first term of our quotient.

Thus, we got the remaining polynomial as $-7x^4 - 25x^3 + 47x^2 + 49x - 45$ and the quotient, as $x^3\ldots$

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$\underline{Step\ 2}$:

Next we eliminated the highest power of $x$ in the remaining dividend, which is $-7x^4$, so by multiplying the divisor by $-7x^2$, and subtracting:

$-7x^4 - 25x^3 + 47x^2 + 49x - 45 - (-7x^2)(x^2 + 4x - 3)$

$= -7x^4 - 25x^3 + 47x^2 + 49x - 45 + 7x^4 + 28x^3 - 21x^2$

$= 3x^3 + 26x^2 + 49x - 45$, which is our remaining dividend,

and, our quotient, till now, is $x^3 - 7x^2\ldots$

$\underline{Step\ 3}$:

We eliminated the next highest power of $x$, that is, $3x^3$, by multiplying the divisor by $3x$.

$3x^3 + 26x^2 + 49x - 45 - 3x(x^2 + 4x - 3)$

$= 3x^3 + 26x^2 + 49x - 45 - 3x^3 -12x^2 + 9x$

$= 14x^2 + 58x -45$, which is our remaining dividend, and our quotient, till now is

$x^3 - 7x^2 + 3x\ldots$

$\underline{Step\ 4}$:

We eliminated the next highest power of $x$ in the remaining dividend, which was $14x^2$, by multiplying the divisor by $14$, and we got:

$14x^2 + 58x - 45 - 14(x^2 + 4x - 3)$

$= 14x^2 + 58x - 45 - 14x^2 - 56x + 42$

$= 2x - 3$

Now, we see that the highest power in the remaining dividend is $1$ which is less than the order of the divisor which is $2$ and our division is complete.

So, our final quotient is $x^3 - 7x^2 + 3x + 14$ and the remainder is $2x - 3$.

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Now let us try the following problem:

--------- Reference to question: eb48d9a7-6543-42c3-aae3-9943ec06ae7c ---------

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III. Points To Observe:

The remainder is always of a lower order than the divisor. An easy way to remember this is that in integer division the remainder is always less than the divisor.
If you divide a polynomial of a lower order by a polynomial of higher order, then the quotient will be $0$ and the remainder is same as the dividend. Analogy to integer division is that, when you divide an integer by a larger than it, the quotient is $0$ and the remainder is the dividend itself.
For a single variable polynomial, the degree of the quotient is equal to the degree of the divisor subtracted from the degree of the dividend.

Polynomial Multiplication