The Central Idea

Ordinary division can be written in the form

\[ \text{dividend} = \text{divisor}\cdot\text{quotient} + \text{remainder}. \]

Polynomial division uses exactly the same structure.

If \(P(x)\) is divided by \(D(x)\), then

\[ \boxed{ P(x)=D(x)Q(x)+R(x) } \]

where:

The remainder must have degree less than the degree of the divisor.

Division as the Reverse of Multiplication

Suppose

\[ (x+2)(x^2+3x+4) = x^3+5x^2+10x+8. \]

Then dividing

\[ x^3+5x^2+10x+8 \]

by

\[ x+2 \]

must return

\[ x^2+3x+4. \]

Polynomial division therefore reverses a known multiplication structure.

Exact Division

Division is exact when the remainder is zero.

For example,

\[ \frac{x^2-9}{x-3} \]

is exact because

\[ x^2-9=(x-3)(x+3). \]

Therefore,

\[ \boxed{ \frac{x^2-9}{x-3}=x+3 }. \]

Division with a Remainder

Not every polynomial divides evenly.

For example, dividing

\[ x^2+1 \]

by

\[ x-1 \]

produces quotient

\[ x+1 \]

and remainder

\[ 2. \]

Thus

\[ \boxed{ x^2+1=(x-1)(x+1)+2 }. \]

Quotient-Remainder Form

The result of polynomial division can be written in two equivalent ways.

If

\[ P(x)=D(x)Q(x)+R(x), \]

then dividing both sides by \(D(x)\) gives

\[ \boxed{ \frac{P(x)}{D(x)} = Q(x)+\frac{R(x)}{D(x)} }. \]

This form is especially useful when working with rational expressions.

Polynomial Long Division

Polynomial long division follows the same repeated pattern as long division with numbers.

At each stage:

  1. divide the leading term of the current dividend by the leading term of the divisor;
  2. place that result in the quotient;
  3. multiply the divisor by the new quotient term;
  4. subtract;
  5. repeat.

Example: Divide a Quadratic by a Linear Polynomial

Divide

\[ x^2+5x+6 \]

by

\[ x+2. \]

Start with the leading terms:

\[ \frac{x^2}{x}=x. \]

Therefore, the first quotient term is

\[ x. \]

Multiply and Subtract

Multiply the divisor by \(x\):

\[ x(x+2)=x^2+2x. \]

Subtract:

\[ (x^2+5x+6)-(x^2+2x). \]

This leaves

\[ 3x+6. \]

Repeat the Process

Divide the new leading term by the leading term of the divisor:

\[ \frac{3x}{x}=3. \]

Add \(3\) to the quotient.

Multiply:

\[ 3(x+2)=3x+6. \]

Subtract:

\[ (3x+6)-(3x+6)=0. \]

Therefore,

\[ \boxed{ \frac{x^2+5x+6}{x+2}=x+3 }. \]

Long Division Written Compactly

\[ \begin{array}{r|l} x+2 & x^2+5x+6 \\ \hline & x+3 \end{array} \]

The compact written algorithm is useful, but the meaning remains the same: divide, multiply, subtract, repeat.

Example with a Remainder

Divide

\[ x^2+4x+7 \]

by

\[ x+2. \]

First,

\[ \frac{x^2}{x}=x. \]

Multiply:

\[ x(x+2)=x^2+2x. \]

Subtract:

\[ 2x+7. \]

Next,

\[ \frac{2x}{x}=2. \]

Multiply:

\[ 2(x+2)=2x+4. \]

Subtract:

\[ 3. \]

Therefore,

\[ \boxed{ x^2+4x+7=(x+2)(x+2)+3 }. \]

Writing the Same Result as a Rational Expression

From

\[ x^2+4x+7=(x+2)(x+2)+3, \]

we obtain

\[ \boxed{ \frac{x^2+4x+7}{x+2} = x+2+\frac{3}{x+2} }. \]

Missing Powers Must Be Included

Polynomial long division depends on aligning like powers.

Consider

\[ x^3-8. \]

Written with missing powers included,

\[ x^3+0x^2+0x-8. \]

The zero terms act as placeholders and keep the columns aligned during division.

Example with Missing Powers

Divide

\[ x^3-8 \]

by

\[ x-2. \]

Write the dividend as

\[ x^3+0x^2+0x-8. \]

Begin:

\[ \frac{x^3}{x}=x^2. \]

Multiply:

\[ x^2(x-2)=x^3-2x^2. \]

Subtract:

\[ 2x^2+0x. \]

Continue the Missing-Power Example

Next,

\[ \frac{2x^2}{x}=2x. \]

Multiply:

\[ 2x(x-2)=2x^2-4x. \]

Subtract:

\[ 4x-8. \]

Then

\[ \frac{4x}{x}=4. \]

Multiply:

\[ 4(x-2)=4x-8. \]

Subtract:

\[ 0. \]

Therefore,

\[ \boxed{ \frac{x^3-8}{x-2}=x^2+2x+4 }. \]

Degree Predicts the Quotient Degree

When dividing nonzero polynomials,

\[ \deg(Q) = \deg(P)-\deg(D) \]

whenever the dividend degree is at least the divisor degree.

For example, dividing a degree-\(5\) polynomial by a degree-\(2\) polynomial produces a quotient of degree

\[ 5-2=3. \]

This gives a useful structural check before the division begins.

When the Divisor Has Higher Degree

If

\[ \deg(P)<\deg(D), \]

then the quotient is zero and the entire dividend is the remainder.

For example,

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

has

\[ Q(x)=0 \]

and

\[ R(x)=x+1. \]

This satisfies the requirement that

\[ \deg(R)<\deg(D). \]

Dividing by a Higher-Degree Polynomial

Long division works even when the divisor is not linear.

Divide

\[ x^4+3x^3+2x^2+5x+4 \]

by

\[ x^2+x+1. \]

Begin with

\[ \frac{x^4}{x^2}=x^2. \]

Multiply:

\[ x^2(x^2+x+1) = x^4+x^3+x^2. \]

Subtract:

\[ 2x^3+x^2+5x+4. \]

Continue the Higher-Degree Example

Next,

\[ \frac{2x^3}{x^2}=2x. \]

Multiply:

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

Subtract:

\[ -x^2+3x+4. \]

Then

\[ \frac{-x^2}{x^2}=-1. \]

Multiply:

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

Subtract:

\[ 4x+5. \]

Since

\[ \deg(4x+5)<2, \]

the division stops.

Result of the Higher-Degree Division

The quotient is

\[ x^2+2x-1 \]

and the remainder is

\[ 4x+5. \]

Therefore,

\[ \boxed{ x^4+3x^3+2x^2+5x+4 = (x^2+x+1)(x^2+2x-1)+(4x+5) }. \]

The Remainder Must Be Smaller Than the Divisor

Long division stops when the remainder has lower degree than the divisor.

If the divisor has degree \(2\), then the remainder may have degree \(1\) or \(0\), but not \(2\) or greater.

This is analogous to ordinary integer division, where the remainder must be smaller than the divisor.

Checking Long Division

Every division result can be checked using

\[ \boxed{ P=DQ+R }. \]

Suppose division gives

\[ Q(x)=x+2 \]

and

\[ R(x)=3 \]

when dividing \(x^2+4x+7\) by \(x+2\).

Check:

\[ (x+2)(x+2)+3 \] \[ = x^2+4x+4+3 \] \[ = x^2+4x+7. \]

The original dividend returns.

Synthetic Division

Synthetic division is a compact form of polynomial division that works especially well when the divisor has the form

\[ x-c. \]

Instead of repeatedly writing variables and powers, synthetic division works directly with coefficients.

Why Synthetic Division Is Possible

When dividing by

\[ x-c, \]

the leading coefficient of the divisor is \(1\), and the same sequence of multiply-and-add operations repeats at every stage.

Because the powers are already determined by position, we can suppress the written variables and operate on the coefficient list alone.

Use the Opposite Sign from the Divisor

For a divisor

\[ x-c, \]

synthetic division uses the number

\[ \boxed{c}. \]

Therefore:

This sign convention is one of the most common sources of synthetic division errors.

Example: Synthetic Division

Divide

\[ x^3+5x^2+2x-8 \]

by

\[ x-1. \]

The coefficients are

\[ 1,\quad5,\quad2,\quad-8. \]

Since the divisor is \(x-1\), use

\[ 1. \]

Synthetic Division Procedure

Start with

\[ \begin{array}{r|rrrr} 1 & 1 & 5 & 2 & -8 \end{array} \]

Bring down the first coefficient:

\[ 1. \]

Multiply by the synthetic number:

\[ 1\cdot1=1. \]

Add to the next coefficient:

\[ 5+1=6. \]

Continue the Synthetic Division

Multiply:

\[ 6\cdot1=6. \]

Add:

\[ 2+6=8. \]

Multiply:

\[ 8\cdot1=8. \]

Add:

\[ -8+8=0. \]

The synthetic row is

\[ \boxed{1,\ 6,\ 8,\ 0}. \]

Interpreting the Synthetic Row

The final entry is the remainder:

\[ R=0. \]

The preceding entries are the coefficients of the quotient.

Since the original dividend has degree \(3\), the quotient has degree \(2\):

\[ x^2+6x+8. \]

Therefore,

\[ \boxed{ \frac{x^3+5x^2+2x-8}{x-1} = x^2+6x+8 }. \]

Synthetic Division with a Remainder

Divide

\[ 2x^3-3x^2+4x+5 \]

by

\[ x-2. \]

Use \(2\) and the coefficients

\[ 2,\quad-3,\quad4,\quad5. \]

Bring down \(2\).

Multiply:

\[ 2\cdot2=4. \]

Add:

\[ -3+4=1. \]

Continue the Remainder Example

Multiply:

\[ 1\cdot2=2. \]

Add:

\[ 4+2=6. \]

Multiply:

\[ 6\cdot2=12. \]

Add:

\[ 5+12=17. \]

The synthetic result is

\[ \boxed{2,\ 1,\ 6,\ 17}. \]

Interpret the Quotient and Remainder

The quotient coefficients are

\[ 2,\quad1,\quad6, \]

so

\[ Q(x)=2x^2+x+6. \]

The remainder is

\[ 17. \]

Therefore,

\[ \boxed{ 2x^3-3x^2+4x+5 = (x-2)(2x^2+x+6)+17 }. \]

Synthetic Division with \(x+3\)

Divide

\[ x^3+2x^2-5x+4 \]

by

\[ x+3. \]

Rewrite the divisor as

\[ x-(-3). \]

Therefore, use

\[ \boxed{-3}. \]

The coefficients are

\[ 1,\quad2,\quad-5,\quad4. \]

Carry Out the \(x+3\) Division

Bring down \(1\).

Multiply:

\[ 1(-3)=-3. \]

Add:

\[ 2+(-3)=-1. \]

Multiply:

\[ (-1)(-3)=3. \]

Add:

\[ -5+3=-2. \]

Multiply:

\[ (-2)(-3)=6. \]

Add:

\[ 4+6=10. \]

Therefore,

\[ \boxed{ Q(x)=x^2-x-2,\qquad R=10 }. \]

Missing Powers in Synthetic Division

Missing powers must appear as zero coefficients.

Consider

\[ x^4-5x^2+8. \]

The complete coefficient list is

\[ 1,\quad0,\quad-5,\quad0,\quad8. \]

Omitting either zero would shift every later coefficient into the wrong power position.

Example with Missing Powers

Divide

\[ x^4-5x^2+8 \]

by

\[ x-2. \]

Use

\[ 2 \]

with coefficients

\[ 1,\quad0,\quad-5,\quad0,\quad8. \]

Bring down \(1\).

Multiply and add:

\[ 1(2)=2,\qquad0+2=2, \] \[ 2(2)=4,\qquad-5+4=-1, \] \[ (-1)(2)=-2,\qquad0-2=-2, \] \[ (-2)(2)=-4,\qquad8-4=4. \]

Interpret the Missing-Power Example

The synthetic row is

\[ 1,\quad2,\quad-1,\quad-2,\quad4. \]

Therefore,

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

and

\[ R=4. \]

So

\[ \boxed{ x^4-5x^2+8 = (x-2)(x^3+2x^2-x-2)+4 }. \]

Synthetic Division as Repeated Multiply-and-Add

The synthetic algorithm has a simple rhythm:

  1. bring down;
  2. multiply by \(c\);
  3. add;
  4. multiply by \(c\);
  5. add;
  6. continue until the final coefficient.

The final number is the remainder.

All earlier numbers form the quotient coefficients.

Why the Quotient Degree Drops by One

Synthetic division uses a linear divisor:

\[ x-c. \]

A linear divisor has degree \(1\).

Therefore, dividing a degree-\(n\) polynomial by \(x-c\) produces a quotient of degree

\[ n-1. \]

If the dividend is cubic, the quotient is quadratic. If the dividend is degree \(7\), the quotient is degree \(6\).

When Synthetic Division Applies

In its standard elementary form, synthetic division applies directly when the divisor is

\[ \boxed{x-c}. \]

Examples include

\[ x-4, \] \[ x+7=x-(-7), \]

and

\[ x-\frac12. \]

For general polynomial divisors, long division remains the universal method.

What About a Divisor Like \(2x-3\)?

The divisor

\[ 2x-3 \]

is linear, but it is not in the simple form \(x-c\) because its leading coefficient is \(2\).

Polynomial long division handles it directly.

More advanced synthetic methods can be adapted to divisors of the form \(ax-b\), but the standard synthetic algorithm in this course will use monic linear divisors \(x-c\).

Long Division and Synthetic Division Produce the Same Result

Synthetic division is not a different kind of division.

It is a compressed bookkeeping system for the same quotient-remainder process.

If both methods are applicable and performed correctly, they must produce the same quotient and remainder.

Comparing the Two Methods

Feature Long Division Synthetic Division
Works with general polynomial divisors Yes No, not in the standard elementary form
Uses full terms and variables Yes No
Uses coefficient lists Not primarily Yes
Especially efficient for \(x-c\) Works Yes
Makes quotient-remainder structure explicit Yes Yes

Choosing the Best Method

Use long division when:

Use synthetic division when:

Division and Factorization

If division by a polynomial produces remainder zero, then the divisor divides the polynomial exactly.

For example,

\[ \frac{x^3-8}{x-2}=x^2+2x+4. \]

Therefore,

\[ x^3-8=(x-2)(x^2+2x+4). \]

This connects division directly to factorization.

The next lesson will formalize this relationship through the Remainder Theorem and Factor Theorem.

Division and Function Evaluation

The final synthetic entry has an important relationship to evaluating the polynomial at the synthetic number.

For example, when dividing by

\[ x-c, \]

the final synthetic value matches the value obtained from

\[ P(c). \]

We will prove and formalize this fact in the next lesson.

Common Error: Forgetting Missing Terms

Suppose the dividend is

\[ x^4+3x-2. \]

The coefficient list is not

\[ 1,\ 3,\ -2. \]

The correct list is

\[ \boxed{1,\ 0,\ 0,\ 3,\ -2}. \]

The zeros represent the missing \(x^3\) and \(x^2\) terms.

Common Error: Using the Wrong Synthetic Number

For the divisor

\[ x+5, \]

the synthetic number is

\[ -5, \]

because

\[ x+5=x-(-5). \]

Using \(5\) would correspond to dividing by \(x-5\), which is a different polynomial.

Common Error: Forgetting the Remainder

Suppose division gives quotient

\[ x^2+1 \]

and remainder

\[ 4. \]

The result is not merely

\[ x^2+1. \]

The complete relationship is

\[ \boxed{ P(x)=D(x)(x^2+1)+4 }. \]

Or, in rational-expression form,

\[ \boxed{ \frac{P(x)}{D(x)} = x^2+1+\frac{4}{D(x)} }. \]

Common Error: Stopping Long Division Too Soon

Long division continues until the remainder has degree less than the divisor.

If the divisor is linear and the current remainder still contains an \(x\)-term, division must continue.

For a linear divisor, the final remainder must be a constant.

Common Error: Misreading the Synthetic Quotient

Suppose synthetic division of a cubic polynomial produces

\[ 2,\quad5,\quad-1,\quad7. \]

The last number is the remainder.

The quotient coefficients are

\[ 2,\quad5,\quad-1. \]

Since the quotient must be quadratic,

\[ \boxed{ Q(x)=2x^2+5x-1 }. \]

The coefficient positions determine the powers.

A Reliable Procedure for Polynomial Long Division

  1. Write the dividend and divisor in standard form.
  2. Insert zero terms for missing powers.
  3. Divide the leading term of the current dividend by the leading term of the divisor.
  4. Place the result in the quotient.
  5. Multiply the entire divisor by that quotient term.
  6. Subtract carefully.
  7. Bring down the next term or continue with the remaining polynomial.
  8. Repeat until the remainder degree is less than the divisor degree.
  9. Verify using \(P=DQ+R\).

A Reliable Procedure for Synthetic Division

  1. Confirm that the divisor is in the form \(x-c\).
  2. Identify the synthetic number \(c\).
  3. Write every coefficient, including zeros for missing powers.
  4. Bring down the first coefficient.
  5. Multiply by \(c\).
  6. Add to the next coefficient.
  7. Repeat multiply-and-add through the final coefficient.
  8. Interpret the final number as the remainder.
  9. Interpret the earlier numbers as quotient coefficients.
  10. Check using \(P=DQ+R\).

Example: Full Synthetic Workflow

Divide

\[ 2x^4-3x^3+0x^2+5x-6 \]

by

\[ x+2. \]

Use

\[ c=-2. \]

Coefficients:

\[ 2,\ -3,\ 0,\ 5,\ -6. \]

Bring down:

\[ 2. \]

Multiply and add:

\[ 2(-2)=-4,\qquad -3-4=-7, \] \[ (-7)(-2)=14,\qquad0+14=14, \] \[ 14(-2)=-28,\qquad5-28=-23, \] \[ (-23)(-2)=46,\qquad-6+46=40. \]

Therefore,

\[ \boxed{ Q(x)=2x^3-7x^2+14x-23,\qquad R=40 }. \]

Example: Full Long-Division Workflow

Divide

\[ 2x^3+3x^2-5x+4 \]

by

\[ x^2+x-1. \]

First:

\[ \frac{2x^3}{x^2}=2x. \]

Multiply:

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

Subtract:

\[ x^2-3x+4. \]

Next:

\[ \frac{x^2}{x^2}=1. \]

Multiply:

\[ x^2+x-1. \]

Subtract:

\[ -4x+5. \]

Since the remainder has degree \(1\), which is less than the divisor's degree \(2\), stop.

Therefore,

\[ \boxed{ 2x^3+3x^2-5x+4 = (x^2+x-1)(2x+1)+(-4x+5) }. \]

Practice 1: Exact Long Division

Divide:

\[ \frac{x^2+6x+8}{x+2}. \]

Practice 2: Long Division with Remainder

Divide:

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

Practice 3: Missing Powers

Divide:

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

Practice 4: Higher-Degree Divisor

Divide:

\[ \frac{x^4+2x^3+x^2+3x+2}{x^2+x+1}. \]

Practice 5: Synthetic Division

Use synthetic division:

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

Practice 6: Synthetic Division with \(x+2\)

Use synthetic division:

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

Practice 7: Missing Coefficients

Use synthetic division:

\[ \frac{x^4-4x+1}{x-1}. \]

Practice 8: Identify the Quotient Degree

Before dividing, determine the expected quotient degree when a degree-\(8\) polynomial is divided by a degree-\(3\) polynomial.

Practice 9: Identify the Synthetic Number

State the synthetic number used for each divisor:

  1. \(x-7\)
  2. \(x+3\)
  3. \(x-\frac12\)

Practice 10: Verify a Division Result

Verify that

\[ x^3+2x^2-5x+7 = (x-2)(x^2+4x+3)+13. \]

Practice Answers

Practice 1

\[ x^2+6x+8=(x+2)(x+4). \]

Therefore,

\[ \boxed{x+4}. \]

Practice 2

Divide:

\[ \frac{x^2}{x}=x. \]

Subtract \(x^2+x\), leaving

\[ 2x+5. \]

Then

\[ \frac{2x}{x}=2. \]

Subtract \(2x+2\), leaving

\[ 3. \]

Therefore,

\[ \boxed{ Q(x)=x+2,\qquad R=3 }. \]

Practice 3

Include the missing term:

\[ x^3+0x^2+0x+1. \]

The quotient is

\[ x^2-x+1 \]

with remainder \(0\).

Therefore,

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

Practice 4

The quotient is

\[ x^2+x-1 \]

and the remainder is

\[ 3x+3. \]

Therefore,

\[ \boxed{ x^4+2x^3+x^2+3x+2 = (x^2+x+1)(x^2+x-1)+(3x+3) }. \]

Practice 5

Use \(2\) with coefficients

\[ 1,\ -4,\ 1,\ 6. \]

Synthetic division gives

\[ 1,\ -2,\ -3,\ 0. \]

Therefore,

\[ \boxed{ Q(x)=x^2-2x-3,\qquad R=0 }. \]

Practice 6

Use \(-2\) with coefficients

\[ 2,\ 1,\ -5,\ 3. \]

Synthetic division gives

\[ 2,\ -3,\ 1,\ 1. \]

Therefore,

\[ \boxed{ Q(x)=2x^2-3x+1,\qquad R=1 }. \]

Practice 7

Use the complete coefficient list

\[ 1,\ 0,\ 0,\ -4,\ 1. \]

Using \(1\), synthetic division gives

\[ 1,\ 1,\ 1,\ -3,\ -2. \]

Therefore,

\[ \boxed{ Q(x)=x^3+x^2+x-3,\qquad R=-2 }. \]

Practice 8

\[ 8-3=\boxed{5}. \]

Practice 9

  1. \(x-7\): \(\boxed{7}\)
  2. \(x+3=x-(-3)\): \(\boxed{-3}\)
  3. \(x-\frac12\): \(\boxed{\frac12}\)

Practice 10

Multiply:

\[ (x-2)(x^2+4x+3). \]

Expand:

\[ x^3+4x^2+3x-2x^2-8x-6 \] \[ = x^3+2x^2-5x-6. \]

Add the remainder:

\[ x^3+2x^2-5x-6+13 \] \[ = x^3+2x^2-5x+7. \]

The original dividend is recovered.

What to Remember

Polynomial division follows the quotient-remainder relationship

\[ \boxed{ P(x)=D(x)Q(x)+R(x) }. \]

Polynomial long division is the general algorithm:

Synthetic division is a compressed coefficient method for divisors of the form

\[ \boxed{x-c}. \]

Missing powers must always be represented by zero coefficients.

A zero remainder means the division is exact and the divisor is a factor of the polynomial.

Where We Go Next

Synthetic division reveals a striking connection between division and function evaluation: the remainder produced by dividing \(P(x)\) by \(x-c\) is exactly \(P(c)\).

The next lesson, Remainder and Factor Theorems, proves and applies this relationship to evaluate polynomials efficiently, test possible factors, and connect zeros directly to linear factors.