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Dividing Polynomials

Rational Expression

A rational expression is a fraction in which both the numerator and the denominator are polynomials. The numerator represents the dividend, and the denominator represents the divisor. The degree of the polynomial in the numerator can be less than, equal to, or greater than the degree of the polynomial in the denominator. Rational expressions are analogous to rational numbers, where the numerator and denominator are integers.

Here are four examples of rational expressions:

  1. \( \frac{x^2 - 4}{x - 2} \)

  2. \( \frac{3x^3 + 2x - 7}{x^2 + x + 1} \)

  3. \( \frac{5x - 1}{3x^2 - x + 4} \)

  4. \( \frac{x^4 - 2x^2 + x - 6}{x^3 - 3x + 2} \)

Proper and Improper Rational Expressions

A rational expression is classified as proper if the degree of the numerator polynomial is less than the degree of the denominator polynomial. Conversely, it is termed improper if the degree of the numerator is equal to or greater than the degree of the denominator.

Examples of each include:

Proper rational expression:

\[ \frac{2x + 1}{x^2 + 3x + 2} \]

(The degree of the numerator is 1, and the degree of the denominator is 2.)

Improper rational expression:

\[ \frac{x^3 + 4x^2 - x + 7}{x^2 - 5x + 6} \]

(The degree of the numerator is 3, and the degree of the denominator is 2.)

Euclidian Division of Polynomials

Before we understand what the algorithm is, let us try to solve the following problem:

Problem:

Given two polynomials, \( p(x) = x^3 + 2x^2 - 3x + 1 \) and \( d(x) = x^2 + x - 1 \), find polynomials \( q(x) \) and \( r(x) \) such that \( p(x) \) can be expressed in the form \( p(x) = q(x)d(x) + r(x) \), where the degree of \( r(x) \) is less than the degree of \( d(x) \). The polynomial \( q(x) \) will be the quotient resulting from the division of \( p(x) \) by \( d(x) \), and \( r(x) \) will be the remainder of this division. The goal is to determine \( q(x) \) and \( r(x) \) such that the degree of \( r(x) \) is minimized.

Solution:

Given the polynomials:

  • \( p(x) = x^3 + 2x^2 - 3x + 1 \)
  • \( d(x) = x^2 + x - 1 \)

We aim to find \( q(x) \) and \( r(x) \) such that \( p(x) = q(x)d(x) + r(x) \).

Step 1: Matching the Highest Degree Term

We choose \( q(x) = x \) because multiplying \( d(x) \) by \( x \) gives us \( x^3 \), which matches the highest degree term of \( p(x) \).

So we start with:

\[ x^3 + 2x^2 - 3x + 1 = x(x^2 + x - 1) + r(x) \]

Step 2: Calculating the Initial Remainder \( r(x) \)

Multiplying out and subtracting from \( p(x) \), we get:

\[ x^3 + 2x^2 - 3x + 1 = x^3 + x^2 - x + r(x) \]
\[ \Rightarrow r(x) = x^2 - 2x + 1 \]

Step 3: Further Division of the Remainder

We see that the remainder \( r(x) \) is still divisible by \( d(x) \), so we perform another division:

\[ r(x) = x^2 - 2x + 1 = 1(x^2 + x - 1) + r'(x) \]

Multiplying out and finding the new remainder, we have:

\[ x^2 - 2x + 1 = x^2 + x - 1 + r'(x) \]
\[ \Rightarrow r'(x) = -3x + 2 \]

Step 4: Final Expression

Finally, we compile our results into the original polynomial \( p(x) \):

\[ p(x) = x(x^2 + x - 1) + (x^2 - 2x + 1) \]
\[ p(x) = x(x^2 + x - 1) + 1(x^2 + x - 1) + (-3x + 2) \]
\[ p(x) = (x + 1)(x^2 + x - 1) + (-3x + 2) \]

So we have found the quotient \( q(x) = x + 1 \) and the remainder \( r'(x) = -3x + 2 \) of minimum degree that satisfy \(p(x)=q(x)d(x)+r(x)\). You should verify whether both sides are equal or not.

In the context of polynomial division, the expression \( p(x) = q(x)d(x) + r(x) \) can be reformulated as \( \frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)} \). This formulation closely parallels the division of integers. For example, dividing 13 by 3 in integer arithmetic yields 4 with a remainder of 1, which can be represented as \( \frac{13}{3} = 4 + \frac{1}{3} \). In this analogy:

  • 4 represents the quotient, which is the integer part of the division.
  • \( \frac{1}{3} \) represents the remainder over the divisor, indicating what's left after the division.

Translating this concept to polynomial division:

  • \( p(x) \) is the dividend, the polynomial being divided.
  • \( d(x) \) is the divisor, the polynomial by which \( p(x) \) is divided.
  • \( q(x) \) is the quotient polynomial, analogous to the integer quotient in numerical division. It represents how many times \( d(x) \) can be "contained" within \( p(x) \).
  • \( r(x) \) is the remainder polynomial, similar to the numerical remainder. It's what remains from \( p(x) \) after subtracting the product of \( q(x) \) and \( d(x) \).

This division demonstrates that any polynomial \( p(x) \) divided by a non-zero polynomial \( d(x) \) can be expressed as a sum of a quotient polynomial \( q(x) \) and a fraction \( \frac{r(x)}{d(x)} \), where \( r(x) \) is a polynomial of lesser degree than \( d(x) \). If \( r(x) = 0 \), then \( p(x) \) is exactly divisible by \( d(x) \), and \( p(x)/d(x) = q(x) \).

This division process is foundational in understanding how polynomials can be broken down, analyzed, and simplified, providing insights into the structure of polynomial expressions and facilitating operations like factorization, finding roots, and simplifying complex rational expressions.

The Euclidean Division of Polynomials is an algorithmic process akin to the familiar division of integers. It states that for any two polynomials \( p(x) \) and \( d(x) \), where \( d(x) \) is not the zero polynomial and its degree is less than or equal to the degree of \( p(x) \), there exist unique polynomials \( q(x) \) and \( r(x) \) such that:

\[ p(x) = q(x)d(x) + r(x) \]

Here, \( q(x) \) is known as the quotient polynomial and \( r(x) \) as the remainder polynomial, with the degree of \( r(x) \) strictly less than the degree of \( d(x) \). This relationship can also be written as a rational expression:

\[ \frac{p(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)} \]

This resembles the division of integers, where you divide a number to get a quotient and a remainder.

In the context of polynomials: - \( p(x) \) is referred to as the dividend. - \( d(x) \) is the divisor. - The outcome of the division \( q(x) \) is the quotient. - \( r(x) \) is the remainder.

The degree of the quotient polynomial \( q(x) \) is the difference between the degrees of \( p(x) \) and \( d(x) \). Specifically, if the degree of \( p(x) \) is \( p \) and the degree of \( d(x) \) is \( d \), then:

\[ \delta(q) = p - d \]
\[ \delta(r) \in \{0, 1, 2, \ldots, d - 1\} \]

The division can be performed using long division, which is methodical and mirrors integer division, or synthetic division, which is a shortcut applicable when the divisor \( d(x) \) is a monic linear polynomial. The Euclidean Division algorithm is particularly useful for finding roots, factoring polynomials, and simplifying rational expressions.

Long Division of Polynomials

The long division of polynomials is a systematic method that mirrors the traditional long division of integers. Here's the step-by-step algorithm:

  1. Arrange Polynomials: Start by writing both the dividend \( p(x) \) and the divisor \( d(x) \) in descending order of their degrees. If there's any missing term in the polynomial sequence, introduce that term with a zero coefficient to maintain order.

  2. Divide Leading Terms: Take the leading term (the term with the highest power of \( x \)) of the dividend and divide it by the leading term of the divisor. This gives you the first term of the quotient.

  3. Multiply and Subtract: Multiply the entire divisor by this first term of the quotient. Then, subtract this result from the dividend, which will give you a new polynomial, often referred to as the remainder.

  4. New Dividend: This remainder becomes the new dividend. Now, take its leading term and divide it by the leading term of the original divisor to get the next term of the quotient.

  5. Iterate the Process: Repeat the multiplication and subtraction steps with the new dividend and the divisor. Continue this process, each time appending the next term to the quotient, until you obtain a remainder with a degree that is less than the degree of the divisor.

  6. Conclude with Remainder: The process stops when you can no longer divide because the remainder's degree is lower than the divisor's. This remainder is then written over the divisor to complete the expression of the result.

Let us understand with a few examples how it works.

Example

Problem: Divide the polynomial \(3{{x}^{3}}-2x+1\) by \({{x}^{2}}-x+1\) .

Solution:

Let the dividend be \(p(x)=3{{x}^{3}}+0{{x}^{2}}-2x+1\) (insert the missing term having coefficient 0)

Long Division process

Thus, we get,

Long Division Process

Example

Problem: Find the remainder when \(x^5+1\) is divided by \(x^3-1\).

Solution:

Long division process

Here's a breakdown of the steps:

  1. Division: The first step is to divide \( x^5 \), the leading term of the dividend, by \( x^3 \), the leading term of the divisor, which gives us \( x^2 \) as the first term of the quotient.

  2. Multiplication: Multiply the entire divisor \( x^3 - 1 \) by \( x^2 \) to get \( x^5 - x^2 \).

  3. Subtraction: Subtract \( x^5 - x^2 \) from \( x^5 + 1 \). Since there are no other terms with \( x^5 \), this just eliminates the \( x^5 \) term, leaving \( x^2 + 1 \).

  4. Bring Down Next Term: Since there are no \( x^4 \) or \( x^3 \) terms in the original dividend, we treat them as zero and continue with the subtraction result, \( x^2 + 1 \).

The result \( x^2 + 1 \) has a lower degree than the divisor \( x^3 - 1 \), so this concludes the division.

Thus, the quotient is \( x^2 \) and the remainder is \( x^2 + 1 \). The division process shows that:

\[ \frac{x^5 + 1}{x^3 - 1} = x^2 + \frac{x^2 + 1}{x^3 - 1} \]

Synthetic Method of division of Polynomials:

This is faster and requires lesser numerical calculations than long division method. Also instead of subtractions we are only dealing with the addition, so synthetic division is less prone to sign errors.

We can have two cases based on whether the divisor is a monic polynomial or not.

  1. Divisor is a monic polynomial.

    Let us explain this method with an example. Suppose we want to divide \(2x^5 + 4x^3 + 3x - 1\) by \(x^2 - 4x + 6\).

    Step 1: Put the terms of divisor and dividend in descending order of degrees. The missing terms are replaced by terms having zero coefficients. Multiply the divisor with -1 (Negate all coefficients of divisor)

    \[ -1(x^2 - 4x + 6) = -x^2 + 4x - 6 \]

    Step 2: Neglect the leading coefficient and put the rest of coefficient in diagonal (let us call this diagonal the "divisor diagonal") as shown in the example. Also put the coefficients of dividend in a row (let us call this row as "dividend row") as shown in the example.

    Synthetic Division explantion of terms

    Step 3: Leave the space of number of rows equal to the number of elements in divisor diagonal(which is equal to degree of divisor) Copy down the first element of dividend row directly to the final result row.

    step 3

    Step 4: Now multiply the first element of final result row with the elements of divisor diagonal row. You will obtain a diagonal whose lower left most element is in the second column of dividend coefficients.

    Step 4

    Step 5: Add the elements of second column and put the sum in result row.

    Step 5

    Step 6: Multiply the second element of the final result row to the divisor diagonal and put the obtained diagonal below dividend row that lower left most element now is in the third column.

    Step 6

    Step 7: Now add the elements of the third column.

    Step 7

    Step 8: Now continue with this process in Step 6 and Step 7. Stop as soon as the last column is filled with upper right most element of diagonal and add the last column elements to get the last element of the final result row.

    Step 8

    Step 9: If the degree of the divisor is \(n\) , then the last \(n\) elements of the final result row are the coefficients of the remainder and the remaining elements are the coefficients of the quotient.

    Step 9

    So in the example the remainder is finally \( 51x + 287 \) and the quotient is \( 2x^3 + 8x^2 + 24x + 48 \). Thus,

    \[ \frac{2x^5 + 4x^3 + 3x - 1}{x^2 - 4x + 6} = 2x^3 + 8x^2 + 24x + 48 + \frac{51x + 287}{x^2 - 4x + 6} \]

    Let us take more examples which will make process clearer.

    Example

    Divide \( x^7 + 2x + 1 \) by \( x^3 - x^2 + 2 \)

    Solution: Multiply the divisor by -1. We get \( -x^3 + x^2 - 2 = -x^3 + 0.x^2 + x - 2 \)

    Fill the table:

    Synthetics division

    Therefore remainder\( = -5x^2 + 4x + 7 \) and quotient\( = x^4 + x^3 + x^2 - x - 3 \)

    Thus,

    \[ \frac{x^7 + 2x + 1}{x^3 - x^2 + 2} = x^4 + x^3 + x^2 - x - 3 + \frac{-5x^2 + 4x + 7}{x^3 - x^2 + 2} \]

  2. Divisor is a non-monic polynomial

    When the divisor is a non-monic polynomial the synthetic division is modified a little bit. We first take the leading coefficient of divisor, let’s say \( k \), and what is left is a monic polynomial. We divide the dividend with this monic polynomial with the usual method. We obtain the quotient \( q(x) \) and \( r(x) \).

    Then \( \frac{q(x)}{k} \) is the required quotient and \( r(x) \) is the remainder (remainder is the same).

    Why?

    Suppose we want to divide polynomial \( p(x) \) by non-monic polynomial \( d(x) \). Let us make \( d(x) \) monic by taking the leading coefficient \( k \) common. Let’s say that the new monic divisor is \( d'(x) = \frac{d(x)}{k} \). Then by Euclid’s Division Lemma, when we divide \( p(x) \) by \( d'(x) \), there exists polynomials \( q(x) \) and \( r(x) \) such that

    \[ p(x) = q(x)d'(x) + r(x) \quad .....[1] \]

    Suppose we rewrite [1] as

    \[ p(x) = \left( \frac{q(x)}{k} \right)(kd'(x)) + r(x) \quad .....[2] \]

    Which is same as \( p(x) = \left( \frac{q(x)}{k} \right)d(x) + r(x) \) (Since, \( d(x) = kd'(x) \)) .....[3]

    Now [3] says that if we divide \( p(x) \) by \( d(x) \) then the quotient is \( \frac{q(x)}{k} \) and the remainder is \( r(x) \).

    We may thus write the following theorem:

    "When a polynomial \( P(x) \) is divided by a non-zero polynomial \( d(x) \) if \( q(x) \) is the quotient and \( r(x) \) is the remainder, then when \( P(x) \) is divided by \( kd(x) \), the remainder remains \( r(x) \) but the quotient becomes \( \frac{q(x)}{k} \)."