# 5.4: Dividing Polynomials

Learning Objectives

• Use long division to divide polynomials.
• Use synthetic division to divide polynomials.

The exterior of the Lincoln Memorial in Washington, D.C., is a large rectangular solid with length 61.5 meters (m), width 40 m, and height 30 m.1 We can easily find the volume using elementary geometry.

[egin{align*} V&=l ; {cdot} ; w ; {cdot} ; h &=61.5 ; {cdot} ; 40 ; {cdot} ; 30 &=73,800 end{align*}]

So the volume is 73,800 cubic meters ((m^3)).

Suppose we knew the volume, length, and width. We could divide to find the height.

[egin{align*} h&=dfrac{V}{l{cdot}w} &=dfrac{73,800}{61.5{cdot}40} &=30 end{align*}]

As we can confirm from the dimensions above, the height is 30 m. We can use similar methods to find any of the missing dimensions. We can also use the same method if any or all of the measurements contain variable expressions. For example, suppose the volume of a rectangular solid is given by the polynomial (3x^4−3x^3−33x^2+54x). The length of the solid is given by (3x); the width is given by (x−2).

To find the height of the solid, we can use polynomial division, which is the focus of this section.

## Using Long Division to Divide Polynomials

We are familiar with the long division algorithm for ordinary arithmetic. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division.

Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.

[egin{align*} ext{dividend}&=( ext{divisor}{cdot} ext{quotient})+ ext{remainder} 178&=(3{cdot}59)+1 &=177+1 &=178end{align*}]

We call this the Division Algorithm and will discuss it more formally after looking at an example.

Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide (2x^3−3x^2+4x+5) by (x+2) using the long division algorithm, it would look like this:

We have found

[dfrac{2x^3−3x^2+4x+5}{x+2}=2x^2−7x+18−dfrac{31}{x+2}]

or

[2x^3−3x^2+4x+5=(x+2)(2x^2−7x+18)−31]

We can identify the dividend, the divisor, the quotient, and the remainder.

Writing the result in this manner illustrates the Division Algorithm.

The Division Algorithm

The Division Algorithm states that, given a polynomial dividend (f(x)) and a non-zero polynomial divisor (d(x)) where the degree of (d(x)) is less than or equal to the degree of (f(x)), there exist unique polynomials (q(x)) and (r(x)) such that

[f(x)=d(x)q(x)+r(x)]

(q(x)) is the quotient and (r(x)) is the remainder. The remainder is either equal to zero or has degree strictly less than (d(x)).

If (r(x)=0), then (d(x)) divides evenly into (f(x)). This means that, in this case, both (d(x)) and (q(x)) are factors of (f(x)).

Given a polynomial and a binomial, use long division to divide the polynomial by the binomial

1. Set up the division problem.
2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
3. Multiply the answer by the divisor and write it below the like terms of the dividend.
4. Subtract the bottom binomial from the top binomial.
5. Bring down the next term of the dividend.
6. Repeat steps 2–5 until reaching the last term of the dividend.
7. If the remainder is non-zero, express as a fraction using the divisor as the denominator.

Example (PageIndex{1}): Using Long Division to Divide a Second-Degree Polynomial

Divide (5x^2+3x−2) by (x+1).

Solution

The quotient is (5x−2). The remainder is 0. We write the result as

[dfrac{5x^2+3x−2}{x+1}=5x−2]

or

[5x^2+3x−2=(x+1)(5x−2)]

Analysis

This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend.

Example (PageIndex{2}): Using Long Division to Divide a Third-Degree Polynomial

Divide (6x^3+11x^2−31x+15) by (3x−2).

Solution

There is a remainder of 1. We can express the result as:

[dfrac{6x^3+11x^2−31x+15}{3x−2}=2x^2+5x−7+dfrac{1}{3x−2}]

Analysis

We can check our work by using the Division Algorithm to rewrite the solution. Then multiply.

[(3x−2)(2x^2+5x−7)+1=6x^3+11x^2−31x+15]

Notice, as we write our result,

• the dividend is (6x^3+11x^2−31x+15)
• the divisor is (3x−2)
• the quotient is (2x^2+5x−7)
• the remainder is (1)

(PageIndex{2})

Divide (16x^3−12x^2+20x−3) by (4x+5).

Solution

(4x^2−8x+15−dfrac{78}{4x+5})

## Using Synthetic Division to Divide Polynomials

As we’ve seen, long division of polynomials can involve many steps and be quite cumbersome. Synthetic division is a shorthand method of dividing polynomials for the special case of dividing by a linear factor whose leading coefficient is 1.

To illustrate the process, recall the example at the beginning of the section.

Divide (2x^3−3x^2+4x+5) by (x+2) using the long division algorithm.

The final form of the process looked like this:

There is a lot of repetition in the table. If we don’t write the variables but, instead, line up their coefficients in columns under the division sign and also eliminate the partial products, we already have a simpler version of the entire problem.

Synthetic division carries this simplification even a few more steps. Collapse the table by moving each of the rows up to fill any vacant spots. Also, instead of dividing by 2, as we would in division of whole numbers, then multiplying and subtracting the middle product, we change the sign of the “divisor” to –2, multiply and add. The process starts by bringing down the leading coefficient.

We then multiply it by the “divisor” and add, repeating this process column by column, until there are no entries left. The bottom row represents the coefficients of the quotient; the last entry of the bottom row is the remainder. In this case, the quotient is (2x^2–7x+18) and the remainder is –31.The process will be made more clear in Example (PageIndex{3}).

Synthetic Division

Synthetic division is a shortcut that can be used when the divisor is a binomial in the form (x−k). In synthetic division, only the coefficients are used in the division process.

Given two polynomials, use synthetic division to divide

1. Write (k) for the divisor.
2. Write the coefficients of the dividend.
3. Bring the lead coefficient down.
4. Multiply the lead coefficient by (k). Write the product in the next column.
5. Add the terms of the second column.
6. Multiply the result by (k). Write the product in the next column.
7. Repeat steps 5 and 6 for the remaining columns.
8. Use the bottom numbers to write the quotient. The number in the last column is the remainder and has degree 0, the next number from the right has degree 1, the next number from the right has degree 2, and so on.

Example (PageIndex{3}): Using Synthetic Division to Divide a Second-Degree Polynomial

Use synthetic division to divide (5x^2−3x−36) by (x−3).

Solution

Begin by setting up the synthetic division. Write (k) and the coefficients.

Continue by adding the numbers in the second column. Multiply the resulting number by (k).Write the result in the next column. Then add the numbers in the third column.

The result is (5x+12). So (x−3) is a factor of the original polynomial.

Analysis

Just as with long division, we can check our work by multiplying the quotient by the divisor and adding the remainder.

[(x−3)(5x+12)+0=5x^2−3x−36]

Example (PageIndex{4}): Using Synthetic Division to Divide a Third-Degree Polynomial

Use synthetic division to divide (4x^3+10x^2−6x−20) by (x+2).

Solution

The binomial divisor is (x+2) so (k=−2). Add each column, multiply the result by –2, and repeat until the last column is reached.

The result is (4x^2+2x−10).

The remainder is 0. Thus, (x+2) is a factor of (4x^3+10x^2−6x−20).

Analysis

The graph of the polynomial function (f(x)=4x^3+10x^2−6x−20) in Figure (PageIndex{2}) shows a zero at (x=k=−2). This confirms that (x+2) is a factor of (4x^3+10x^2−6x−20).

Example (PageIndex{5}): Using Synthetic Division to Divide a Fourth-Degree Polynomial

Use synthetic division to divide (−9x^4+10x^3+7x^2−6) by (x−1).

Solution

Notice there is no x-term. We will use a zero as the coefficient for that term.

The result is (−9x^3+x^2+8x+8+frac{2}{x−1}).

(PageIndex{5})

Use synthetic division to divide (3x^4+18x^3−3x+40) by (x+7).

Solution

(3x^3−3x^2+21x−150+frac{1,090}{x+7})

## Using Polynomial Division to Solve Application Problems

Polynomial division can be used to solve a variety of application problems involving expressions for area and volume. We looked at an application at the beginning of this section. Now we will solve that problem in the following example.

Example (PageIndex{6}): Using Polynomial Division in an Application Problem

The volume of a rectangular solid is given by the polynomial (3x^4−3x^3−33x^2+54x). The length of the solid is given by (3x) and the width is given by (x−2). Find the height of the solid.

Solution

There are a few ways to approach this problem. We need to divide the expression for the volume of the solid by the expressions for the length and width. Let us create a sketch as in Figure (PageIndex{3}).

We can now write an equation by substituting the known values into the formula for the volume of a rectangular solid.

[egin{align*} V&=l{cdot}w{cdot}h 3x^4−3x^3−33x^2+54x&=3x{cdot}(x−2){cdot}h end{align*}]

To solve for (h), first divide both sides by (3x).

[dfrac{3x{cdot}(x−2){cdot}h}{3x}=dfrac{3x^4−3x^3−33x^2+54x}{3x}]

[(x-2)h=dfrac{x^3-x^2-11x+18}{x-2}]

Now solve for (h) using synthetic division.

[h=dfrac{x^3−x^2−11x+18}{x−2}]

The quotient is (x^2+x−9) and the remainder is 0. The height of the solid is (x^2+x−9).

(PageIndex{6})

The area of a rectangle is given by (3x^3+14x^2−23x+6). The width of the rectangle is given by (x+6). Find an expression for the length of the rectangle.

Solution

(3x^2−4x+1)

## Key Equations

Division Algorithm (f(x)=d(x)q(x)+r(x)) where (q(x){ eq}0)

## Key Concepts

• Polynomial long division can be used to divide a polynomial by any polynomial with equal or lower degree.
• The Division Algorithm tells us that a polynomial dividend can be written as the product of the divisor and the quotient added to the remainder.
• Synthetic division is a shortcut that can be used to divide a polynomial by a binomial in the form x−k.
• Polynomial division can be used to solve application problems, including area and volume.

1 National Park Service. "Lincoln Memorial Building Statistics." www.nps.gov/linc/historycultu...statistics.htm. Accessed 4/3/2014

## Glossary

Division Algorithm

given a polynomial dividend (f(x)) and a non-zero polynomial divisor (d(x)) where the degree of (d(x)) is less than or equal to the degree of (f(x)), there exist unique polynomials (q(x)) and (r(x)) such that (f(x)=d(x)q(x)+r(x)) where (q(x)) is the quotient and (r(x)) is the remainder. The remainder is either equal to zero or has degree strictly less than (d(x)).

synthetic division

a shortcut method that can be used to divide a polynomial by a binomial of the form (x−k)

## 5.4: Dividing Polynomials

Recall the quotient rule for exponents: if x is nonzero and m and n are positive integers, then

In other words, when dividing two expressions with the same base, subtract the exponents. This rule applies when dividing a monomial by a monomial. In this section, we will assume that all variables in the denominator are nonzero.

Example 1: Divide: 28 y 3 7 y .

Solution: Divide the coefficients and subtract the exponents of the variable y.

Example 2: Divide: 24 x 7 y 5 8 x 3 y 2 .

Solution: Divide the coefficients and apply the quotient rule by subtracting the exponents of the like bases.

When dividing a polynomial by a monomial, we may treat the monomial as a common denominator and break up the fraction using the following property:

Applying this property results in terms that can be treated as quotients of monomials.

Example 3: Divide: − 5 x 4 + 25 x 3 − 15 x 2 5 x 2 .

Solution: Break up the fraction by dividing each term in the numerator by the monomial in the denominator and then simplify each term.

Check your division by multiplying the answer, the quotient The result after dividing. , by the monomial in the denominator, the divisor The denominator of a quotient. , to see if you obtain the original numerator, the dividend The numerator of a quotient. .

Example 4: Divide: 9 a 4 b − 7 a 3 b 2 + 3 a 2 b − 3 a 2 b .

Answer: − 3 a 2 + 7 3 a b − 1 . The check is optional and is left to the reader.

Try this! Divide: ( 16 x 5 − 8 x 4 + 5 x 3 + 2 x 2 ) ÷ ( 2 x 2 ) .

Answer: 8 x 3 − 4 x 2 + 5 2 x + 1

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## Using Long Division to Divide Polynomials

We are familiar with the algorithm for ordinary arithmetic. We begin by dividing into the digits of the dividend that have the greatest place value. We divide, multiply, subtract, include the digit in the next place value position, and repeat. For example, let’s divide 178 by 3 using long division.

Another way to look at the solution is as a sum of parts. This should look familiar, since it is the same method used to check division in elementary arithmetic.

We call this the Division Algorithm and will discuss it more formally after looking at an example.

Division of polynomials that contain more than one term has similarities to long division of whole numbers. We can write a polynomial dividend as the product of the divisor and the quotient added to the remainder. The terms of the polynomial division correspond to the digits (and place values) of the whole number division. This method allows us to divide two polynomials. For example, if we were to divide 2 x 3 − 3 x 2 + 4 x + 5 2 x 3 − 3 x 2 + 4 x + 5 by x + 2 x + 2 using the long division algorithm, it would look like this:

We can identify the , the , the , and the .

Writing the result in this manner illustrates the Division Algorithm.

### Note: The Division Algorithm:

The states that, given a polynomial dividend f ( x ) f ( x ) and a non-zero polynomial divisor d ( x ) d ( x ) where the degree of d ( x ) d ( x ) is less than or equal to the degree of f ( x ) , f ( x ) , there exist unique polynomials q ( x ) q ( x ) and r ( x ) r ( x ) such that

q ( x ) q ( x ) is the quotient and r ( x ) r ( x ) is the remainder. The remainder is either equal to zero or has degree strictly less than d ( x ) . d ( x ) .

If r ( x ) = 0 , r ( x ) = 0 , then d ( x ) d ( x ) divides evenly into f ( x ) . f ( x ) . This means that, in this case, both d ( x ) d ( x ) and q ( x ) q ( x ) are factors of f ( x ) . f ( x ) .

### How To:

1. Set up the division problem.
2. Determine the first term of the quotient by dividing the leading term of the dividend by the leading term of the divisor.
3. Multiply the answer by the divisor and write it below the like terms of the dividend.
4. Subtract the bottom from the top binomial.
5. Bring down the next term of the dividend.
6. Repeat steps 2–5 until reaching the last term of the dividend.
7. If the remainder is non-zero, express as a fraction using the divisor as the denominator.

### Example 1

#### Problem 1

##### Using Long Division to Divide a Second-Degree Polynomial

Divide 5 x 2 + 3 x − 2 5 x 2 + 3 x − 2 by x + 1. x + 1.

##### Solution

The quotient is 5 x − 2. 5 x − 2. The remainder is 0. We write the result as

##### Analysis

This division problem had a remainder of 0. This tells us that the dividend is divided evenly by the divisor, and that the divisor is a factor of the dividend.

### Example 2

#### Problem 1

##### Using Long Division to Divide a Third-Degree Polynomial

Divide 6 x 3 + 11 x 2 − 31 x + 15 6 x 3 + 11 x 2 − 31 x + 15 by 3 x − 2. 3 x − 2.

##### Solution

There is a remainder of 1. We can express the result as:

##### Analysis

We can check our work by using the Division Algorithm to rewrite the solution. Then multiply.

Notice, as we write our result,

• the dividend is 6 x 3 + 11 x 2 − 31 x + 15 6 x 3 + 11 x 2 − 31 x + 15
• the divisor is 3 x − 2 3 x − 2
• the quotient is 2 x 2 + 5 x − 7 2 x 2 + 5 x − 7
• the remainder is 1 1

### Try It:

#### Exercise 1

Divide 16 x 3 − 12 x 2 + 20 x − 3 16 x 3 − 12 x 2 + 20 x − 3 by 4 x + 5. 4 x + 5.

##### Solution

4 x 2 − 8 x + 15 − 78 4 x + 5 4 x 2 − 8 x + 15 − 78 4 x + 5

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## Binomial

An algebraic expression that contains two non zero terms is known as a binomial. It is expressed in the form ax m + bx n where a and b number, x is variable, m and n are nonnegative distinct integers.

• g + 3m is a binomial in two variables g and m.
• 3a 4 – 5b 2 is a binomial in two variables a and b.
• -4x 2 – 9y is a binomial in two variables x and y.
• a 2 /4 + b/2 is a binomial in two variables a and b.

### Binomial Equation

Any equation that contains one or more binomials is known as a binomial equation.

### Operations on Binomials

• Factorization
• Subtraction
• Multiplication
• Raising to the nth power
• Converting to lower-order binomials

Factorization:

A binomial can be expressed as the product of the other two.

Two binomials can be added if both contain the same variable and the same exponent.

(2a 2 + 3b) + (4a 2 + 5b) = 6a 2 + 8b

Subtraction:

It is similar to addition, two binomials should contain the same variable and exponent.

Multiplication:

When we multiply two binomials distributive property is used and it ends up with four terms. In this method, multiplication is carried by multiplying each term of the first factor to the second factor.

(ax + b) (mx + n) can be expressed as amx 2 + (an + mb) x + bn

Raising to nth Power:

A binomial can be raised to the nth power and expressed in the form of (x + y) n

Converting to Lower order binomials:

Higher-order binomials can be factored to lower-order binomials such as cubes can be factored to products of squares and another monomial.

a 3 + b 3 can be expressed as (a + b) (a 2 – ab + b 2 ).

## 5.4: Dividing Polynomials

The Improving Mathematics Education in Schools (TIMES) Project

Number and Algebra : Module 39Years : 9-10

• Familiarity with coordinate geometry, and elementary graphs, including the graph
of y = x 2 .

Polynomials represent the next level of algebraic complexity after quadratics. Indeed a quadratic is a polynomial of degree 2. We can factor quadratic expressions, solve quadratic equations and graph quadratic functions, the obvious question arises as to
how these things might be performed with algebraic expressions of higher degree.

The quadratic x 2 &minus 5 x + 6 factors as ( x &minus 2)( x &minus 3). Hence the equation x 2 &minus 5 x + 6 = 0
has solutions x = 2 and x = 3.

Similarly we can factor the cubic x 3 &minus 6 x 2 + 11 x &minus 6 as ( x &minus 1)( x &minus 2)( x &minus 3), which enables us to show that the solutions of x 3 &minus 6 x 2 + 11 x &minus 6 = 0 are x = 1, x = 2 or x = 3. In this module we will see how to arrive at this factorisation.

Polynomials in many respects behave like whole numbers or the integers. We can add, subtract and multiply two or more polynomials together to obtain another polynomial. Just as we can divide one whole number by another, producing a quotient and remainder, we can divide one polynomial by another and obtain a quotient and remainder, which are also polynomials.

A quadratic equation of the form ax 2 + bx + c has either 0, 1 or 2 solutions, depending on whether the discriminant is negative, zero or positive. The number of solutions of the this equation assisted us in drawing the graph of the quadratic function y = ax 2 + bx + c . Similarly, information about the roots of a polynomial equation enables us to give a rough sketch of the corresponding polynomial function.

As well as being intrinsically interesting objects, polynomials have important applications in the real world. One such application to error-correcting codes is discussed in the Appendix to this module.

A polynomial is as expression such as x 5 &minus 2 x 3 + 8 x + 3 or x 4 &minus x 2 + 1. There may be any number of terms, but each term must be a multiple of a whole number power of x . Thus 2 x 3 &minus x is not a polynomial.

The term with the highest power of is called the leading term and its coefficient is called the leading coefficicent . If the leading coefficient is 1, then the polynomial is called monic. The index of the leading term is called the degree of the polynomial. The term independent of is called the constant term .

Thus x 5 &minus 2 x 3 + 8 x + 3 is a monic polynomial of degree 5 with constant term 3, while
x 4 &minus x 2 + 1 is a non-monic polynomials of degree 4 with leading coefficient and constant term 1.

In the first polynomial, the coefficients are all integer while the second polynomials has an irrational coefficient. For the most part, we will consider only polynomials of the first type, but much of what follows applies equally well to the second.

To name polynomials, we will use the function notation such as p ( x ) or q ( x ). Thus we can write p ( x ) = x 5 &minus 2 x 3 + 8 x + 3, or q ( x ) = x 4 &minus x 2 + 1. This enables us to conveniently substitute values of x when required.

The general polynomial has the form

p ( x ) = a n x n &minus a n &minus 1 x n &minus 1 + . + a 1 x + a 0 ,

where a n ≠ 0 and n is a whole number. The coefficients are, in general, real numbers.

Write down the leading term, the leading coefficient, the degree and the constant term in the general polynomial given above.

For small degree polynomials, we use the following names.

• a polynomial of degree 1 is called linear
• a polynomial of degree 2 is called a quadratic
• a polynomial of degree 3 is called a cubic
• a polynomial of degree 4 is called a quartic
• a polynomial of degree 5 is called a quintic

A polynomial that consists only of a non-zero constant, is called a constant polynomial and has degree 0. The polynomial p ( x ) = 0 is called the zero polynomial. It has no terms and so there is no leading term. It is best not to define the degree of the zero polynomial. Some books write its degree as &minus1 or &minus &infin .

To add or subtract two polynomials, we collect the like terms.

Let p ( x ) = 3 x 4 &minus 2 x 2 + x &minus 1, q ( x ) = 7 x 5 + 2 x 2 .

Find a p ( x ) + q ( x ) b p ( x ) &minus q ( x ) c 2 p ( x ) &minus 3 q ( x ) .

a p ( x ) + q ( x ) = 7 x 5 + 3 x 4 + x &minus 1

b p ( x ) &minus q ( x ) = &minus7 x 5 + 3 x 4 &minus 4 x 2 &minus 1

c 2 p ( x ) &minus 3 q ( x ) = 14 x 5 &minus 9 x 4 + 10 x 2 &minus 3 x + 3.

Notice that we usually write the terms in the polynomial from largest to smallest degree. This is sometimes called the standard form of the polynomial.

To multiply two polynomials, we multiply each term in the first polynomial by second polynomial and collect like terms.

The polynomials P ( x ), Q ( x ) and R ( x ) are given by P ( x ) = x 3 &minus 2 x 2 + x &minus 1, Q ( x ) = 3 x 3 &minus 2 x 2 and R ( x ) = &minus x 4 + 2 x 3 &minus 3 x 2 . Find:

a P ( x ) Q ( x ) b Q ( x ) R ( x )

 a P ( x ) Q ( x ) = ( x 3 &minus x 2 + x &minus 1)(3 x 3 &minus 2 x 2 ) = x 3 (3 x 2 &minus 2 x 2 ) &minus x 2 (3 x 3 &minus 2 x 2 ) + x (3 x 3 &minus 2 x 2 ) &minus (3 x 3 &minus 2 x 2 ) = 3 x 6 &minus 2 x 5 &minus 3 x 5 + 2 x 4 + 3 x 4 &minus 2 x 3 &minus 3 x 3 + 2 x 2 = 3 x 6 &minus 5 x 5 + 5 x 4 &minus 5 x 3 + 2 x 2 b Q ( x ) R ( x ) = (3 x 3 &minus 2 x 2 )(&minus x 4 + 2 x 3 &minus 3 x 2 ) = 3 x 3 (&minus x 4 + 2 x 3 &minus 3 x 2 ) &minus 2 x 2 (&minus x 4 + 2 x 3 &minus 3 x 2 ) = &minus3 x 7 + 6 x 6 &minus 9 x 5 + 2 x 6 &minus 4 x 5 + 6 x 4 = &minus3 x 7 + 8 x 6 &minus 13 x 5 + 6 x 4

a Square the polynomial p ( x ) = 2 x 3 &minus x + 1.

b Complete the statements:

/>i />When two non-zero polynomials are multiplied, the degree of the product is .

/>ii />The constant term in the product of two polynomials is .

When one whole number is divided into another a quotient and remainder is formed. Thus gives 7 remainder 2. There are various ways to write this result.

In the statement 37 ÷ 5 equals 7 remainder 2

• the number 5 that we are dividing by is called the divisor
• the number 37 that we are dividing into is called the dividend
• the number 7 is called the quotient
• the number 2 is called the remainder

The key point about the remainder is that it is non-negative but strictly less than the divisor. Thus, using the third representation, when we divide two whole numbers p and d > 0 we can write

p = dq + r , where 0 ≤ r ≤ d .

If the remainder is zero, then we cay that d is a factor of p .

These basic statements regarding arithmetic have analogues when we come to divide one polynomial by another. We will use the same terminology when discussing polynomial division. The process of doing this is modelled on long division.

The division of the polynomial p ( x ) by the polynomial d ( x ) also produces a quotient q ( x ) and a remainder r ( x ) and so we can write

The key idea in performing the division is to keep working with the leading terms, as the following example shows.

We will divide the polynomial p ( x ) = 5 x 4 &minus7 x 3 + 2 x &minus 4 by the polynomial d ( x ) = x &minus 2, and then express the division in the form p ( x ) = d ( x ) q ( x ) + r ( x ).

 5 x 3 + 3 x 2 + 6 x + 14 x &minus 2 5 x 4 &minus 7 x 3 + 2 x &minus 4 (Divide x into 5 x 4 , giving 5 x 3 .) 5 x 4 &minus 10 x 3 (Multiply x &minus 2 by 5 x 3 and then subtract.) 3 x 3 + 2 x &minus 4 (Divide x into 3 x 3 , giving 3 x 2 .) 3 x 3 &minus 6 x 2 (Multiply x &minus 2 by 3 x 2 and then subtract.) 6 x 2 + 2 x &minus 4 (Divide x into 6 x 2 , giving 6 x .) 6 x 2 &minus 12 x (Multiply x &minus 2 by 6 x and then subtract.) 14 x &minus 4 (Divide x into 14 x , giving 14.) 14 x &minus 28 (Multiply x &minus 2 by 14 and then subtract.) 24 (This is the final remainder.)

Hence 5 x 4 &minus 7 x 3 + 2 x &minus 4 = ( x &minus 2)(5 x 3 + 3 x 2 + 6 x + 14) + 24. (1)

In this example, we see that the quotient is q ( x ) = 5 x 3 + 3 x 3 + 6 x &minus 14 and the remainder is r ( x ) = 24.

We can perform a partial check by substituting x = 2 into the last line.

It can be seen from the above example that the degree of the remainder is less than the degree of the divisor, since otherwise, we could continue the division. Thus, in the case when is a linear factor, the remainder will be a constant and so we can write it as .

In general, we can now write p ( x ) = d ( x ) q ( x ) + r ( x ), where r ( x ) = 0, or, 0 ≤ degree (r ( x )) < degree ( d ( x )).

Divide p ( x ) = 5 x 4 &minus 7 x 3 + 2 x &minus 4 by d ( x ) = x 2 &minus 2. E x press the result in the form
p ( x ) = d ( x ) q ( x ) + r ( x ) where the degree of r ( x ) is less than the degree of d ( x ).

Note that in this case, since the divisor has degree 2, the remainder will either be 0 or have degree at most 1.

Long division of polynomials is a cumbersome process and in some instances we are only interested in the remainder. This does not appear until the end of the computation. When we divide a polynomial p ( x ) by a linear factor ( x &minus a ), we can find the remainder quite easily.

Since the divisor is linear, p ( x ) = ( x &minus a ) q ( x ) + r where r is a constant. Substituting x = a into both sides, we have r = p ( a ).

Thus the remainder is equal to the polynomial evaluated at x = a .

This surprising result is called the remainder theorem . We should keep in mind that it says nothing at all about the quotient q ( x ) and only works when we are dividing by a linear factor ( x &minus a ).

Find the remainder when p( x ) = x 4 &minus 3 x 2 &minus 10 x + 2 is divided by

a ( x &minus 3) b ( x + 4) c (2 x &minus 1)

a The remainder is p (3) = 81 &minus 27 &minus 30 + 2 = 26
b Since ( x + 4) = ( x &minus (&minus4)), the remainder is p (&minus4) = 256 &minus 48 + 40 + 2 = 250
c Write p ( x ) = (2 x &minus 1) q ( x ) + r , so we can obtain r by substituting x = in both sides. Hence the remainder is r = &minus &minus 5 + 2 = &minus .

The polynomial p ( x ) = x 5 &minus 7 x 3 + ax + 1 has remainder 13 after division by x &minus 1. Find the value of the coefficient a .

Factoring quadratics is an important technique which we used to solve quadratic equations. In a similar way, we would like to be able to develop some techniques to factor polynomials.

If the linear polynomial ( x &minus a ) is a factor of a polynomial p ( x ) then we can write
p ( x ) = q ( x )( x &minus a ) and so the remainder when p ( x ) is divided by ( x &minus a ) is equal to 0.
Using the remainder theorem, we have now proven:

( x &minus a ) is a factor of the polynomial p ( x ) if p(a) = 0. If p ( a ) ≠ 0, then

( x &minus a ) is not a factor of p ( x ).

The number a which gives p ( a ) = 0 is called a zero of the polynomial.

Which of the following are factors of p ( x ) = x 3 &minus 6 x 2 + 11 x &minus 6?

a ( x &minus 2) b ( x + 1) c ( x &minus 1).

a p (2) = 8 &minus 24 + 22 &minus 6 = 0 and so ( x &minus 2) is a factor of p ( x ).
b p (&minus1) = &minus 1 &minus 6 &minus 11 &minus 6 ≠ 0 and so ( x + 1) is not a factor of p ( x ).
c p (1) = 1 &minus 6 + 11 &minus 6 = 0 and so ( x &minus 1) is a factor of p ( x ).

Notice that since ( x &minus 2) and ( x &minus 1) are both factors of p ( x ) then so is their product
( x &minus 2)( x &minus 1) = x 2 &minus 3 x + 2. We could thus find the third factor by long division.
Dividing p ( x ) by x 2 &minus 3 x + 2 gives ( x &minus 3) as the third factor and so we have the
following factorisation,

p ( x ) = x 3 &minus 6 x 2 + 11 x &minus 6 = ( x &minus 1)( x &minus 2)( x &minus 3).

Since p (2) = 0 and p (1) = 0, and p ( x ) has degree 3, we can write p ( x ) = ( x &minus 2)( x &minus 1) ( x &minus a )

Where a is a number to be determined. Since p (0) = &minus6 we have &minus2 a = &minus6, so a = 3.

The polynomial p ( x ) = 3 x 6 &minus 5 x 3 + ax 2 + bx + 10 is divisible by x + 1 and x &minus 2. Find the values of the coefficients a and b .

Our aim is to take a polynomial with integer coefficients and write it as a product of polynomials of smaller degree which also has integer coefficients. This process is called factoring over the integers.

The factor theorem allows us to check if a polynomial p ( x ) has a linear factor ( x &minus a ). If it does, then we can use long division to find a polynomial q ( x ) such that p ( x ) = ( x &minus a ) q ( x ) and q ( x ) has degree one less than the degree of p ( x ). Thus we may be able to repeat the process on q ( x ) and so on as often as possible to give a complete factorisation of p ( x ).

We begin with polynomial p ( x ) = x 3 + 4 x 2 &minus 7 x &minus 10.

• We search systematically for one linear factor, by checking the numbers a = 1, &minus 1, 2, &minus2, … until we find an integer value of a such that p ( a ) = 0.

p (1) = &minus12 ≠ 0, p (&minus1) = 0 so ( x + 1) is factor.

Now the quadratic q( x ) = x 2 + 3 x &minus 10 can be factored, using our knowledge of quadratics, as ( x + 5)( x &minus 2) and so the complete factorisation of p ( x ) is

To assist us in finding an integer zero of the polynomial we use the following result.

p ( x ) = a n x n + a n &minus1 x n &minus1 + … + a 1 x + a 0

has an integer zero a , then a is a factor of the constant term a 0 .

Thus, in the e x ample above, the only possible integer zeroes are ±1, ±2, ±5 or ±10.

E x plain why the above theorem is true.

Factorise the polynomial p ( x ) = x 4 &minus 2 x 3 &minus 8 x + 16.

• We only need to test the positive and negative factors of 16.
P (1) = 1 &minus 2 &minus 8 + 16 ≠ 0, so x &minus 1 is not a factor of p ( x ).
P (&minus1) = 1 + 2 + 8 + 16 ≠ 0, so x + 1 is not a factor of p ( x ).
P (2) = 16 &minus 16 &minus 16 + 16 = 0, so x &minus 2 is a factor.
After long division of p ( x ) by x &minus 2, p ( x ) = ( x &minus 2)( x 3 &minus 8)
• Let Q ( x ) = x 3 &minus 8.
x &minus 1 and x + 1 are not factors of Q ( x ) since P ( x ) = ( x &minus 2) Q ( x ), and they are not factors of P ( x ).
However, Q (2) = 8 &minus 8 = 0, so x &minus 2 is a factor of Q ( x ) as well.
After long division of Q ( x ) by x &minus 2, Q ( x ) = ( x &minus 2) ( x 2 +2 x + 4).
• The quadratic x 2 + 2 x + 4 cannot be factorised, see below.
Hence P ( x ) = ( x &minus 2) 2 ( x 2 + 2 x + 4) is the complete factorisation of P ( x ).

Note that the quadratic x 2 + 2 x + 4 = ( x + 1) 2 + 3 which is always greater or equal to 3, hence the quadratic has no factors.

By firstly removing the obvious common factor, factorise the polynomial
p ( x ) = 2 x 5 &minus 22 x 4 + 78 x 3 &minus 90 x 2 .

One of the main methods of solving quadratic equations was the method of factoring. Similarly, one of the main applications of factoring polynomials is to solve polynomial equations.

Solve x 3 + 4 x 2 &minus 7 x &minus 10 = 0

In an earlier e x ample, we factored the polynomial
p ( x ) = x 3 + 4 x 2 &minus 7 x &minus 10 = 0 as ( x + 1)( x + 5)( x &minus 2).

Thus, the equation x 3 + 4 x 2 &minus 7 x &minus 10 = 0 becomes ( x + 1)( x + 5)( x &minus 2) = 0

Since the product of the three factors is zero, we can equate each factor to zero to find the solutions. Thus, x + 1 = 0 or x &minus 2 = 0 or x + 5 = 0 giving x = &minus1, x = 2 or x = &minus5.

Note : The polynomial p ( x ) = ( x &minus 1)( x &minus 2) 2 ( x &minus 4) 3 is of degree 6, but the polynomial equation ( x &minus 1)( x &minus 2) 2 ( x &minus 4) 3 = 0 has only 3 (distinct) solutions x = 1 or x = 2 or x = 4. Thus the number of (distinct) solutions may be less than the degree, but it can never e x ceed the degree.

Use the factorisation of p ( x ) = 2 x 5 &minus 22 x 4 + 78 x 3 &minus90 x 2 from the above e x ercise to solve the equation 2 x 5 &minus 22 x 4 + 78 x 3 &minus 90 x 2 = 0.

In some situations, the factorisation results in a quadratic equation with either no real solutions or irrational solutions. In this case, we may need to complete the problem by using the quadratic formula.

Solve x 4 + 7 x 3 &minus 2 x 2 &minus 7 x + 1 = 0

The polynomial x 4 + 7 x 3 &minus 2 x 2 &minus 7 x + 1 has factorisation ( x &minus 1)( x + 1)( x 2 + 7 x &minus 1).

Hence the equation becomes ( x &minus 1)( x + 1)( x 2 + 7 x &minus 1) = 0.

Thus the solutions are x = 1, x = &minus1 and the solutions to x 2 + 7 x &minus 1 = 0.

Using the quadratic formula, b 2 &minus 4 ac = 49 + 4 = 53, so the quadratic has solutions
x = and x = .

Hence the quartic has four solutions x = 1, &minus1, and x = .

Note that there are polynomial equations with irrational roots that cannot be solved using the procedure above. For e x ample, the polynomial p ( x ) = x 5 &minus 3 x 3 &minus 2 x 2 + 6 factors as
( x 2 &minus 3)( x 3 &minus 2) and so the equation x 5 &minus 3 x 3 &minus 2 x 2 + 6 = 0 has solutions, x = , x = &minus, and x = .

In general, factoring polynomials over the integers is a difficult problem. The polynomial
x 3 &minus 2, for e x ample, cannot be factored over the integers, but it does have one real solution, x = .

Sketching polynomial functions

A polynomial function is a function of the form y = p ( x ), where p ( x ) is a polynomial.

In the module, Quadratic Functions we saw how to sketch the graph of a quadratic
by locating

The verte x is an e x ample of a turning point.

For polynomials of degree greater than 2, finding turning points is not an elementary procedure and usually requires the use of calculus, however:

• To find the y -intercept, we put x = 0.
• To find the x -intercepts, we put y = 0 and solve the corresponding polynomial equation, if possible.

Let us take the polynomial y = x 3 + x 2 &minus 6 x = x ( x &minus 2)( x &minus 3).

The y -intercept is 0 and the x -intercepts occur when x ( x &minus 2)( x &minus 3) = 0 that is, when x = 0, 2 and 3.

To get a picture of the overall shape of the curve, we can substitute some test points.

We can represent the sign of y using a sign diagram:

With this information, we can begin to give a sketch of the graph of y = x ( x &minus 2)( x + 3).

The sign diagram tells us that the graph cuts the x -a x is at the points x = &minus3, 0 and 2 and also whether the graph is above or below the a -a x is on each side of these points. It does not tell us the ma x imum and minimum values of y between the zeroes.

Notice that if x is a large positive number, then p ( x ) is also large and positive. For e x ample, if x = 10, then y = 1040. If x is a large negative number, then p ( x ) is also a large negative number. For e x ample, if x = &minus10, then y = &minus840.

Sketch the graph of y = ( x + 2)( x + 1)( x &minus 1)( x &minus 2).

Graphs of polynomials with repeated factors

Polynomial functions such as p ( x ) = 3( x &minus 1) 2 ( x + 3) 5 ( x &minus 4) which contain repeated factors require a deal of care.

If we e x amine, for e x ample, the size of x 4 for various values of x , we notice

• x 4 is positive for both positive and negative values of x
• for values of x between &minus1 and 1, the size of x 4 is smaller than the value of x .

Graphically this tells us that the graph of y = x 4 has a minimum at x = 0 and that near x = 0 the graph is quite flat, but starts to increase sharply for x > 1 and for x < &minus1.

The graphs of y = x 2 and y = x 4 are shown for comparison.

Both of these graphs have a minimum at x = 0. In the case of the parabola, we call this a verte x but we do not generally use this word for polynomials of higher degree. Instead we talk of a turning point and further classify it as a ma x imum or minimum.

The graphs of y = &minus x 2 and y = &minus x 4 each has a ma x imum at x = 0.

The same phenomenon occurs for all positive even powers of x , and for even powers of ( x &minus a ). Hence near a zero of a polynomial that arises from a factor with an even power, the graph has a minimum or a ma x imum and is ‘flat’ near that zero.

The graph of y = ( x &minus 2) cuts the x -a x is at x = 2, but does not have a ma x imum or minimum there. Since it is a straight line, the graph is not flat at this point, indeed is has a gradient of 45° On the other hand the graph of y = ( x &minus 2) 3 behaves slightly differently at
x = 2. In the following we will consider odd powers greater or equal to 3.

Since an odd power of a negative number is negative, a sign diagram shows that the
y values of the graph of y = x 3 move from negative to positive as the x values move from &minus1 to 1. As above, the graph is flat near the origin.

Hence the graph of y = x 3 looks like:

At the origin we have neither a ma x imum nor a minimum. The point x = 0 is called an infection point of the curve.

Plot the graph of y = 2 x 3 ( x &minus 2) 2 .

The graph passes through the origin and cuts the x -a x is at x = 0 and x = 2. At x = 0 the graph has an inflection point and at x = 2 it has a minimum. The sign diagram is.

Sketch the graph of the polynomial function p ( x ) = ( x + 3) 3 ( x &minus 1) 3 . (You should find that the graph is symmetric about x = &minus1 can you see why?)

Fundamental theorem of algebra

The zeroes of a polynomial are also called the roots of the corresponding polynomial equation.

The polynomial equation x 2 &minus 4 = 0, has two integer roots, x = 2, x = &minus2, while the equation x 2 &minus 2 = 0, has two real roots, x = , x = &minus. On the other hand, the equation x 2 + 2 = 0 has no real roots. Furthermore, the equation x 3 &minus 1 = 0, which factors as ( x &minus 1)( x 2 + x + 1 = 0), has only one real root since the quadratic x 2 + x + 1 = 0 has no solutions.

To properly understand how many solutions a polynomial equation may have, we need to introduce the comple x numbers. A complex number is a number of the form a + ib where a , b are real numbers and i 2 = &minus 1. The comple x number i is often referred to as an imaginary number. Notice that if we put b = 0, we obtain a real number and so the comple x numbers contains the set of all real numbers.

Thus, although the equation x 2 + 1 = 0 has no real solutions, it does have two comple x solutions, x = i , and x = &minus i , and the polynomial x 2 + 1 can be factored as ( x &minus i )( x + i ).

The great mathematician Gauss (1777&minus1855) gave the first proof of the following amazing result, which came to be known as the Fundamental theorem of algebra.

Every polynomial equation of degree greater than 0, has at least one comple x solution.

Given this result, it is not hard to show that:

Every polynomial equation of degree n , greater than 0, has e x actly n solutions, counting multiplicity, over the comple x numbers.

E x plain how the corollary may be deduced from the theorem.

Hence, every polynomial of degree n , greater than 0, can be factored into n linear factors using comple x numbers.

Note that the e x pression counting multiplicity means that given the polynomial equation ( x &minus 2) 3 ( x + 1) 2 = 0 , for e x ample, we say that the roots are x = 2, 2, 2, &minus1, &minus1. Thus we say that x = 2 is a root of multiplicity 3 and x = &minus1 is a root of multiplicity 2. However, the equation has only two (distinct) roots.

The verte x of a parabola is an e x ample of a turning point. The x -coordinate of the turning point of the parabola y = ax 2 + bx + c is given by x = &minus . The x -coordinates of the turning points of a polynomial are not so easy to find and require the use of differential calculus which is studied in senior mathematics.

Suppose that we can factor the monic quadratic x 2 + bx + c as ( x &minus &alpha )( x &minus &beta ). By e x panding out we can see that the sum of the roots, &alpha + &beta equals &minus b and the product of the roots, ab equals c .

We can perform a similar e x ercise on monic cubics. That is, if the roots of the cubic
x 3 + bx 2 + cx + d are &alpha , &beta , &gamma then it can be shown that

These identities give relationships between the roots of a polynomial and its coefficients.

Derive the formulas above.

The study of equations of degree greater than two goes back to Arabic mathematics. Omar Khayyam (1048&minus1141) spent much of his life trying to solve various cases of the cubic equation. It was not until the Renaissance that the general solution of the cubic was obtained. The precise details are sketchy, but the Italian mathematician Cardano (1501&minus1576) managed to prise the secret of solving the cubic out his compatriot Tartaglia and included it in his work the Ars Magna (The Great Art ), published in 1545.

Cardano showed how to reduce any cubic equation to the form x 3 + px + c = 0 and then, making the substitution x = u &minus v , reduce the problem to solving a quadratic. In practice it is easier to put x = u + v .

Put x = y &minus , in the equation x 3 + ax 2 + bx + c = 0 to show that the resulting equation contains no terms of degree 2.

Solve x 3 + 3 x &minus 1 = 0 using the substitution x = u + v .

Rearranging the equation and substituting, we have

( u + v ) 3 = &minus3( u + v ) + 1.

We now e x pand the left-hand side and factor 3uv from two of the terms to give

u 3 + v 3 + 3 uv ( u + v ) = 1 &minus3( u + v ).

Equating the coefficients of ( u + v ) on both sides and equating the remaining terms,
we have

u 3 + v 3 = 1 , 3 uv = &minus3.

Cubing the second equation produces

u 3 + v 3 = 1 , u 3 v 3 = &minus1 ,

At this stage, we have two numbers u 3 , v 3 whose sum and product we know. Hence
they will satisfy the quadratic equation z 2 &minus z &minus 1 = 0, which has solutions z = or
z = . These solutions represent the numbers u 3 + v 3 in either order, so taking cube roots we have the following solution to the original equation

x = u + v = + .

This is the only real solution to the equation.

Use your calculator to e x press this in decimal form and check that it satisfies the
original equation.

The solution of the general quartic equation was found soon afterwards by Cardano’s student and protégé Ferrari. He discovered a method to reduce the problem of solving
a quartic to that of solving a cubic.

In both cases it is possible to e x press the solution of the given equation using square and higher roots and the usual operations of arithmetic (addition, subtraction, multiplication and division). Such a solution is often called a solution using radicals . The solutions to the quadratic equation ax 2 + bx + c = 0 are x = and x = &minus , and so the quadratic equation can also be solved using radicals.

It was to take several hundred years before it was realised, by Abel (1802&minus1829) and Galois (1811&minus1832), that it was not possible to find the solution to the general quintic, or general higher degree equation, using radicals. Obviously, certain higher degree equations can be solved using radicals, for e x ample x 5 &minus 32 = 0, but this is not possible for the general case.

While Cardano’s work was a major breakthrough, there were still many unanswered questions regarding polynomials. In the 17 th century Descartes found a test, known as Descartes’ rule of signs , for determining the number of positive real roots of a polynomial, while Newton discovered the so-called Newton identities that find and relate formulas for the sum of the kth powers of the roots of a polynomial. In the 18 th and 19 th centuries, the great mathematicians, Euler, Lagrange, Eisenstein and Gauss further e x tended our understanding of polynomials and polynomial equations. This led to the development of what is nowdays called modern algebra which is concerned with the study of algebraic structures.

We have seen above that when we study a polynomial, we need to specify what kind of solutions/factors we are looking for. In particular, suppose p ( x ) is a polynomial with degree greater than 0, and real coefficients,

• over the comple x numbers p ( x ) factors into linear factors
• over the real numbers p ( x ) has all its factors either linear and/or quadratics
• over the rational numbers, it is possible to find polynomials of arbitrarily large
degree that are irreducible, that is, they cannot be e x pressed as the product of
two polynomials with rational coefficients, each of smaller degree.

The fundamental theorem of algebra is used to show the first of these statements. To obtain the second, we need to know the fact that when we have a polynomial with real coefficients, any comple x roots will occur in pairs, known as conjugate pairs. That is, if a + ib is a root, then so is a &minus ib . This fact can be used to prove the second statement.

(This requires a little knowledge of comple x numbers.)

Suppose that the polynomial p ( x ) = a n x n &minus a n &minus 1 x n &minus 1 + . + a 1 x + a 0 has all its coefficients real.

a If &alpha = a + ib is a solution to the polynomial equation, p ( x ) = a n x n &minus a n &minus 1 x n &minus 1 + . + a 1 x + a 0 = 0 , show that = a &minus ib is also a solution.
b Show that if x &minus &alpha is a factor of p ( x ), where &alpha = a + ib is a comple x number, then the quadratic ( x &minus &alpha )( x &minus ) is also a factor of p ( x ) and that ( x &minus &alpha )( x &minus ) has real coefficients.
c Deduce that every polynomial with real coefficients can, in theory, be factored as a product of linear and/or quadratic factors, with real coefficients. (Note that in practice this can be a very difficult task to perform.)

Eisenstein (c. 1850) came up with the following ingenious test for irreducibility of polynomials over the rationals.

Consider the polynomial p ( x ) = a n x n &minus a n &minus 1 x n &minus 1 + . + a 1 x + a 0 , where all the coefficients are integers. Suppose that we can find a prime number p that does not divide the leading coefficient an, but which does divide all of the other coefficients. The test says that if the square of that same prime does not divide the constant term, that is, p 2 + a 0 then p ( x ) is irreducible over the rational numbers.

The polynomial p ( x ) = 5 x 7 + 6 x 6 &minus 15 x 4 + 12 x &minus 21 satisfies Eisenstein’s criteria with the prime p = 3 and so p ( x ) cannot be e x pressed as the product of two polynomials each of smaller degree with integer coefficients. That is, p ( x ) is irreducible.

E x plain how to construct a polynomial of arbitrarily large degree that cannot be factored over the rationals.

In the 17 th and 18 th centuries, mathematicians made the remarkable discovery that functions such as y = sin x and y = cos x could be e x pressed using ‘infinite polynomials’, that is, polynomials whose powers of x continue indefinitely. These are called power series . Thus, for e x ample,

sin x = x &minus + &minus + …,

where the obvious pattern continues forever. The notation n! (read as n factorial ) is an abbreviation for n ( n &minus 1)( n &minus 2) … 3.2.1. Thus 5! = 5 × 4 × 3 × 2 × 1 = 120. These infinite series are often referred to as Maclaurin series and have very wide application in both mathematics and physics.

There still remain today unsolved problems related to polynomials. The appendi x below discusses in broad outline a remarkable application of polynomials to modern telecommunications.

An application of polynomials to error-correcting codes

Information is usually digitized by converting it to a sequence of 0’s and 1’s. For e x ample, the ASCII code for the digit 1 and letter A are ‘1’ />1000110, and ‘A’ />1000001. We will assume here that all messages under consideration are finite sequences of 0’s and 1’s.

When your mobile phone sends or receives messages, or data is sent from satellites deep in space, information may be lost or corrupted along the way to its destination.

Since the information is sent as sequences of 0’s and 1’s, a ‘corrupted’ 0 becomes a 1 and vice versa.

A simple check for errors is to add in a check digit s o that the string has an even number of 1’s, and hence the sum of the digits is even.

Thus we encode 1 as 10001101 and A as 10000010.

Now if a byte is transferred and one of the bits is corrupted, then the number of the 1’s becomes odd, and so the receiver can ask for a retransmission.

This code can detect one error, but cannot correct it.

A polynomial modulo 2 is a polynomial whose coefficients are either 0 or 1. Arithmetic is then performed modulo 2 so that 0 + 1 = 1 + 0 = 1 and 1 + 1 = 2= 0.

If p ( t ) = t 3 + 1 + 1, q ( t ) = t 4 + t 3 + t 2 + 1

Then p ( t ) + q ( t ) = t 4 + 2 t 3 + t 2 + 2 t + 2 = t 4 + t 2 .

Multiplication is done similarly.

p( t ) = t 3 + 1 + 1, q ( t ) = t + 1, then

p( t ) × q ( t ) = ( t 3 + t + 1)( t + 1) = t 4 + t 3 + t 2 + 2 t + 1 = t 4 + t 3 + t 2 + 1

We now fi x a polynomial m 1 ( t ) = t 3 + t + 1.

This polynomial cannot be factored modulo 2 since the only possible roots are 0 and 1 and neither work.

We now suppose that the polynomial m 1 ( t ) has a root a. This is, a has the property that
&alpha 3 + &alpha + 1 = 0 or &alpha 3 = &alpha + 1 (Recall that modulo 2, &minus1 = 1)

This number &alpha is very interesting and using the equation above, we can make up a table of its powers.

 &alpha | &alpha &alpha 2 | &alpha 2 &alpha 3 | &alpha + 1 &alpha 4 | &alpha 2 + &alpha &alpha 5 | &alpha 2 + &alpha + 1 &alpha 6 | &alpha 2 + 1 &alpha 7 | 1

For example, to obtain &alpha 5 we multiply &alpha 4 = &alpha 2 + &alpha by a giving &alpha 5 = &alpha 3 + &alpha 2 and replace &alpha 3 with &alpha + 1 to obtain &alpha 5 = &alpha + 1 + &alpha 2 .

Thus, we can write all the powers of &alpha as combinations of 1, &alpha and &alpha 2 (!!)

We start with a message ( a , b , c , d ) in binary of length 4 add in 3 check digits to obtain
( a , b , c , d , x , y , z ). We use these as the coefficients of the polynomial

p ( t ) = at 6 + bt 5 + ct 4 + dt 3 + xt 2 + yt + z .

The digits x , y , z are chosen so that p ( t ) is divisible by the polynomial m 1 ( t ) = t 3 + t + 1

Since a is a root of m 1 ( t ) it is also a root of p ( t ) and so p ( a ) = 0.

Take the message (1, 0, 0, 1) and encode it as (1, 0, 0, 1, x , y , z ). Converting to a polynomial, we have

 p ( t ) = t 6 + t 3 + xt 2 + yt + z . Substituting t = a and using the table to simplify we have p ( a ) = a 6 + a 3 + xa 2 + ya + z . = ( a 2 + 1) + ( a + 1) + xa 2 + ya + z = a 2 ( x + 1) + a ( y + 1) + z .

Now if we take x = 1, y = 1, z = 0 then p ( a ) will be zero. Hence we encode the message
(1, 0, 0, 1) as (1, 0, 0, 1, 1, 1, 0). We will call the polynomial corresponding to the sent message C ( t ) and so C ( a ) = 0.

Suppose that one error occurs in the fifth number from the left so we receive the message (1, 0, 1 , 1, 1, 1, 0).

Thus the t 4 coefficient is incorrect and to the polynomial for the received message is

R ( t ) = C ( t ) + t 4

Substituting t = &alpha we obtain

R ( &alpha ) = C ( &alpha ) + &alpha 4 = &alpha 4

and so we know where the error is.

In general, if there were e x actly one error in the th digit, we would receive

R ( t ) = C ( t ) + t i .

Then R ( &alpha ) = C ( &alpha ) + &alpha i = &alpha i and so the calculation of R ( a ) gives the position of the incorrect digit.

If R ( &alpha ) = 0 then there were no errors. This process, of course, assumes that at most 1
error occurred.

Assuming at most one error, correct and decode the message (1, 0, 0, 1, 0, 0, 1).

Converting to polynomials, we have

R ( t ) = t 6 + t 3 + 1

R ( &alpha ) = &alpha 6 + &alpha 3 + 1 = &alpha 2 + 1 + &alpha + 1 + 1 = &alpha 2 + &alpha + 1 = &alpha 5

So the corrected message was (1, 1 , 0, 1, 0, 0, 1) which decodes as (1, 1, 0, 1).

Correcting more than one error

The coding and correcting procedure outlined above only works if at most one error occurred in the transmission.

The polynomial method can be e x tended to detect and correct more than one error.
The details of these codes is slightly more complicated, but uses the same ideas developed above.

The codes used to perform multiple error correcting are called BCH codes and were discovered (independently) by Bose and Chaudhuri (in 1960) and Hocquenghem (in 1959) -hence the name BCH.

While the arithmetic now becomes very hard, it is easily performed by a computer, and codes can be constructed that can detect and correct an arbitrary number of errors.

The modern error correcting codes used in mobile phone technology are more sophisticated again, but essentially use the kinds of machinery which I have outlined above.

More recently, Reed-Solomon codes, which are a type of BCH code, have been used in applications such as satellite communications, compact disk players, DVDs and disk drives.

a 4 x 6 &minus 4 x 4 + 4 x 3 + x 2 &minus 2 x + 1

b />i />…. the sum of the degrees of the polynomials

ii… the product of the constants

5 x 4 &minus 7 x 3 + 2 x &minus 4 = ( x 2 &minus 2)(5 x 2 &minus 7 x + 10) + (&minus12 x + 16)

If P ( a ) = 0 then a 0 = &minus a n a n &minus a n &minus1 a n &minus1 ….&minus a 1 a . Hence a divides a 0 .

For the polynomial equation P ( x ) = 0 there is a solution x = a by the theorem. Hence x &minus a is a factor of P ( x ) and P ( x ) = ( x &minus a ) Q ( x ). The equation Q ( x ) = 0 also has a solution and so a new linear factor can be found by the theorem.

y 3 &minus + &minus + d

a Take conjugates of both sides. The conjugate of a real is the same real.
b An immediate consequence of a and the factor theorem and the fact that both the sum and product of a comple x number and its conjugate are real.
c The fundamental theorem and b gives the result

The Improving Mathematics Education in Schools (TIMES) Project 2009-2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations.

The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations.

## 5.4: Dividing Polynomials

Before studying this topic you may wish to review common fractions and the division of numbers. all the topics related to division available in this e-book.) -->

Suppose that A and B are any two expressions. Dividing A by B means setting up the quotient as

(This quotient is called an algebraic fraction .) Then you actually carry out the division.

Note that in the first form, , (which is the preferred form), no brackets are shown but they are implied . The reason that we put brackets around A and B is that they are expressions , not just numbers, and the division is supposed to apply to whatever A and B may contain. (We want the division to be at the end in the order of operations.)

How the division is actually carried out and what the result is, depends on whether A and B are monomials, multinomials or polynomials:

Case I: Numerator A and denominator B are both monomials . We merely reduce the algebraic fraction to lowest terms. The result of the division is a monomial.

• Get the sign using the rules for signs.
• Reduce the coefficient to lowest terms.
• Cancel identical factors that appear in both the numerator and denominator.
• Combine exponentials having the same base using the division property of exponentials.
 After carrying out all the simplifications, the denominator equals 1, so we don&rsquot have to display it. Thus the result is an ordinary expression, not an algebraic fraction.

Case II: Dividing a multinomial by a monomial

In this case each term of the multinomial is divided by the monomial like this:

and then each of the resulting terms is simplified as in Case I above.

This method is actually a consequence of replacing a division by a multiplication by a reciprocal and then using the distributive law to distribute the reciprocal onto each term of the multinomial, like this:

Note that the Algebra Coach does not divide a multinomial by a monomial when you click the Simplify button. You must use the Distribute button to do that. The reason is that the single fraction form of the expression is considered to be simpler than the multiple fraction form.

--> Example: Divide A = &minus2 x 4 z 2 &minus 3 x z by B = 6 x 3 z

Divide each term of the multinomial by the monomial.

Divide each term of the multinomial by the monomial. Notice how the signs are reversed (just like when distributing a negative).

Case III: Dividing a polynomial by a polynomial

An algebraic fraction whose numerator and denominator are both polynomials in the same variable is called a rational algebraic fraction . If the degree of the numerator polynomial is higher than the degree of the denominator polynomial then it is called an improper rational algebraic fraction . Just like long division can be used to convert an improper fraction to a mixed fraction (click here to review that topic), so long division can be used to convert an improper rational algebraic fraction into a polynomial plus a proper rational algebraic fraction (the analog of a mixed fraction).

Example: Divide 2 x 2 + 2 x &minus 3 by x &minus 2.

Set up the algebraic fraction in long division format, namely .

If you found this page in a web search you won&rsquot see the

## DIVISION OF POLYNOMIALS

The quotient law of exponents i.e. finds great use in the division of algebraic expressions.

### DIVISION OF A MONOMIAL BY A MONOMIAL:

Quotient of two monomials = (Quotient of their numerical coefficients) (Quotient of their variable parts)

### Divide :

question 1. 24a 2 bc 3 by &ndash 6abc 2 (ii) &ndash 56 xyz 3 by &ndash 6x 3 y 4 z

### DIVISION OF A POLYNOMIAL BY A MONOMIAL:

For dividing a polynomial by a monomial, we divide each term of the polynomial be the monomial.

question Divide :

question (i) 4x 5 &ndash 14x 4 + 6x 3 &ndash 2x 2 by 2x 2 (ii) 20x 3 y + 10xy 2 &ndash 15x 2 y by 5xy

question (i) (4x 5 &ndash 14x 4 + 6x 3 &ndash 2x 2 ) 5x 2

question (ii) (20x 3 y + 10xy 2 &ndash 15x 2 y) 5xy

### Sign convertions

Division of a monomial by a monomial

Quotient can be found by subtracting smaller power of a letter from greater power of the same letter.

To divide a multinomial by a monomial we have to divide each term of the dividend and take the sum of those partial quotients for the complete quotient

Ex: Divide 18x 8 + 24x 6 + 12x 4 6x 2

### Division of a multinomial by another multinomial

• The dividend and the divisor both stand arranged according to descending powers of a common letter, namely, a
• Divide the first term of the dividend by the first term of the divisor and write down the result as the first term of the quotient. Multiply the divisor by the quantity thus found and subtract the product from the dividend.
• (Regard the remainder as a new dividend and see if it is arranged according to the descending powers of the common letter. Divide its first terms by the first term of the divisor and write down the result as the next term of the quotient. Multiply the divisor by this term and subtract the product from the new dividend. Then goes similarly with the successive remainders until there is no remainder.

e.g.: x 4 &ndash 4x 2 + 12x &ndash 9 x 2 &ndash 2x + 3

Divided = x 4 &ndash 4x 2 + 12x &ndash 9

Dividend = Divisor × Quotient + Remainder

question 1. Divide x + 6x 2 &ndash 15 by 2x &ndash 3.

Solution: Arranging the terms of the dividend and the divisor in descending order of powers of x and then dividing, we get:

question 2. Find the quotient and remainder when (x 5 + 3x 4 &ndash 5x 3 + 14x 2 + 36x - 13) is divided by (x 2 + 4x &ndash 2).

Solution: On dividing, we get :

x 2 + 4x &ndash 2 x 5 + 3x 4 &ndash 5x 3 + 14x 2 + 36 x &ndash 13 x 3 &ndash x 2 + x + 8

Quotient = x 3 &ndash x 2 + x + 8, Remainder = 6x + 3.

### IDENTITY:

An identity is an equality which is true for all values of the variable(s).

Identity 1 : (a + b) 2 = a 2 + 2ab + b 2

Identity 2 : (a &ndash b) 2 = a 2 &ndash 2ab + b 2

Identity 3 : (a + b) (a &ndash b) = a 2 &ndash b 2

Identity 4 : (x + a) (x + b) = x 2 + (a + b) x + ab

Identity 5 : (a + b + c) 2 = (a 2 + b 2 + c 2 ) + 2(ab + bc + ca)

Identity 6 : (a + b &ndash c) 2 = (a 2 + b 2 + c 2 ) + 2(ab &ndash bc - ca)

Identity 7 : (a + b) 3 = a 3 + b 3 + 3ab (a + b)

Identity 8 : (a - b) 3 = a 3 &ndash b 3 &ndash 3ab (a &ndash b)

General expressed in symbols is called a formula. Some of the formulae are listed below.

In addition to the answer from Matt L., I want to emphasize the fact why the comment mentions the polynomial division. Consider to finite-length sequences $h[n]$ and $x[n]$, with length $N$ (one sequence can be zero-padded to length $N$).

Now, look at the convolution of both:

As an example, consider the two sequences $h[n]=[1, 2, 3]$ and $x[n]=[4, 6, 5]$. Their convolution is (in Matlab)

Now, consider the product of the two polynomials $h(x)=1+2x+3x^2$ and $x(x)=4+6x+5x^2$ (in Mathematica):

As you can see, the coefficients of the polynomial of the product are equal to the values of the convolution of both sequences.

So, now to get $x[n]$ from $y[n]$, you need to perform the polynomial division of $y(x)/h(x)$:

Hence, the coefficients of the polynomial yield the original sequence.

This technique can be extended to infinite sequences.

To get the connection to the Z-Transform here: The Z-Transform of the finite sequences $x[n]$ and $h[n]$ are given by

And the Z-Transform of the output $y[n]$ is given by

Here, you again have the polynomial multiplication. Also, here it becomes even easier to spot, why it also works with infinite sequences: Then both $H(z)$ and $X(z)$ become fractions of polynomials, but the principle of the polynomial multipication remains.

For finite length sequences that's indeed easily possible. The convolution of two sequences $x[n]$ and $h[n]$

corresponds to multiplication of their $mathcal$-transforms:

follows. Consequently, $x[n]$ can be obtained by computing the impulse response of a recursive filter with numerator polynomial $Y(z)$ and denominator polynomial $H(z)$. Note that this recursive filter is actually an FIR filter (due to pole/zero cancellation), because, obviously, its impulse response $x[n]$ is of finite length.

This little Matlab/Octave example should help clarify this:

If you "invert a deconvolution", you get a convolution. Looking a Matlab convolution conv you can read the one-liner:

Convolution and polynomial multiplication

So they are (almost) the same.

To get the idea, let us start from the standard multiplication, just forgetting about the carry for the moment. If you multiply 123 by 456.

To get the "units" (carry-free, remember), your only option is to multiply units ($3 imes 6$). To get tens, you can now multiply tens by units two options: tens from 123, units from 456, and units from 123, tens from 456. To get hundreds, you can multiply hundreds by units (two options) and tens by tens (one option). At each step, the sum of couples of affected indices is constant, and this sum grows by one unit as you move to the left.

Polynomial multiplication works like a carryless multiplication: to get the $x^D$ term, you have to combine the products of terms $x^d$ and $x^$, as long as $dge 0$ and $D-dge 0$. Here you see that the sum of the powers $d$ and $D-d$ is constant for a given $D$, and grows with $D$. For (infinite) Laurent or power series, you get a similar grouping of terms as in the polynomial product (known as a Cauchy products).

Now, take a sequence or signal $(ldots,a_<-1>,a_<0>,a_<1>,a_<2>,ldots,a_,ldots)$. You can convert it into a simple object, easier to manipulate, by encoding the index $n$ into a power $z^<-n>$, because "powers don't mix": one cannot confuse two different powers. You then have a bijection between the space of sequences and the different power series: $cdots+a_<-1>z^<1>+a_<0>z^<0>+a_<1>z^<-1>+a_<2>z^<2>+cdots+a_z^<-n>+cdots$. Since you have this analogy between indices and powers, the Cauchy product naturally encodes the convolution (an isomorphism).