The Central Idea
Suppose a polynomial \(P(x)\) is divided by the linear divisor
\[ x-c. \]Polynomial division gives
\[ P(x)=(x-c)Q(x)+R. \]Because the divisor has degree \(1\), the remainder must have degree less than \(1\). Therefore, the remainder is a constant.
Now substitute
\[ x=c. \]Then
\[ P(c)=(c-c)Q(c)+R. \]Since
\[ c-c=0, \]we obtain
\[ \boxed{P(c)=R}. \]This is the Remainder Theorem.
The Remainder Theorem
The Remainder Theorem states:
If a polynomial \(P(x)\) is divided by
\[ x-c, \]then the remainder is
\[ \boxed{P(c)}. \]This means polynomial evaluation and polynomial division are connected by the same number.
Why the Theorem Is Useful
Without performing full polynomial division, we can determine the remainder from
\[ P(c). \]For example, if
\[ P(x)=x^3-4x+7, \]then the remainder when dividing by
\[ x-2 \]is
\[ P(2). \]Compute:
\[ P(2)=2^3-4(2)+7. \] \[ P(2)=8-8+7. \] \[ \boxed{P(2)=7}. \]Therefore, the remainder is
\[ \boxed{7}. \]Example: Remainder from Direct Evaluation
Find the remainder when
\[ P(x)=2x^4-3x^2+5x-1 \]is divided by
\[ x+1. \]Rewrite
\[ x+1=x-(-1). \]Therefore, use
\[ c=-1. \]Evaluate:
\[ P(-1) = 2(-1)^4-3(-1)^2+5(-1)-1. \]Thus,
\[ 2-3-5-1=-7. \]Therefore, the remainder is
\[ \boxed{-7}. \]Remainders from Synthetic Division
Synthetic division produces the same remainder.
Suppose we divide
\[ P(x)=x^3+2x^2-5x+4 \]by
\[ x-3. \]Use the synthetic number
\[ 3 \]with coefficients
\[ 1,\quad2,\quad-5,\quad4. \]Synthetic division gives:
\[ 1,\quad5,\quad10,\quad34. \]The final number is the remainder:
\[ \boxed{34}. \]Check with Direct Evaluation
Evaluate
\[ P(3). \]We get
\[ 3^3+2(3)^2-5(3)+4. \] \[ 27+18-15+4. \] \[ \boxed{34}. \]The synthetic remainder and the direct evaluation agree.
Synthetic Division as Efficient Evaluation
Because the final synthetic value equals \(P(c)\), synthetic division can be used simply to evaluate a polynomial.
This is especially useful when the polynomial has high degree or many terms.
Instead of repeatedly computing powers such as
\[ c^5,\quad c^4,\quad c^3, \]synthetic evaluation uses repeated multiplication and addition.
Example: Efficient Evaluation
Evaluate
\[ P(2) \]for
\[ P(x)=3x^5-4x^4+2x^3-7x+6. \]Include the missing \(x^2\) coefficient:
\[ 3,\quad-4,\quad2,\quad0,\quad-7,\quad6. \]Use synthetic evaluation with \(2\).
Bring down \(3\).
Multiply and add:
\[ 3(2)=6,\qquad -4+6=2, \] \[ 2(2)=4,\qquad2+4=6, \] \[ 6(2)=12,\qquad0+12=12, \] \[ 12(2)=24,\qquad-7+24=17, \] \[ 17(2)=34,\qquad6+34=40. \]Therefore,
\[ \boxed{P(2)=40}. \]The Factor Theorem
The Remainder Theorem immediately leads to another important result.
If dividing \(P(x)\) by
\[ x-c \]produces remainder zero, then the division is exact.
That means
\[ x-c \]is a factor of \(P(x)\).
Since the remainder equals \(P(c)\), we obtain the Factor Theorem.
Factor Theorem Statement
The Factor Theorem states:
\[ \boxed{ x-c\text{ is a factor of }P(x) \iff P(c)=0 }. \]The symbol
\[ \iff \]means “if and only if.”
Thus, the relationship works in both directions.
From Factor to Zero
If
\[ x-c \]is a factor of \(P(x)\), then
\[ P(x)=(x-c)Q(x) \]for some polynomial \(Q(x)\).
Substitute \(x=c\):
\[ P(c)=(c-c)Q(c). \]Therefore,
\[ \boxed{P(c)=0}. \]From Zero to Factor
If
\[ P(c)=0, \]then the Remainder Theorem says that the remainder from dividing by
\[ x-c \]is zero.
Therefore, the division is exact and
\[ \boxed{x-c\text{ is a factor of }P(x)}. \]Example: Test a Possible Factor
Determine whether
\[ x-2 \]is a factor of
\[ P(x)=x^3-3x^2-4x+12. \]Evaluate
\[ P(2). \]We get
\[ 8-12-8+12. \] \[ \boxed{P(2)=0}. \]Therefore,
\[ \boxed{x-2\text{ is a factor of }P(x)}. \]Find the Remaining Factor
Once the Factor Theorem confirms that
\[ x-2 \]is a factor, synthetic division can find the quotient.
Use \(2\) with coefficients
\[ 1,\quad-3,\quad-4,\quad12. \]Synthetic division gives
\[ 1,\quad-1,\quad-6,\quad0. \]Therefore,
\[ Q(x)=x^2-x-6. \]So
\[ P(x)=(x-2)(x^2-x-6). \]Continue the Factorization
The quadratic factor
\[ x^2-x-6 \]factors as
\[ (x-3)(x+2). \]Therefore,
\[ \boxed{ x^3-3x^2-4x+12 = (x-2)(x-3)(x+2) }. \]The Factor Theorem can therefore become the first step in a larger factorization.
Testing a Nonfactor
Determine whether
\[ x-4 \]is a factor of
\[ P(x)=x^3-3x^2-4x+12. \]Evaluate
\[ P(4). \]Then
\[ 64-48-16+12=12. \]Since
\[ P(4)\neq0, \]we conclude
\[ \boxed{x-4\text{ is not a factor}}. \]The remainder from division by \(x-4\) is \(12\).
The Sign Connection
If the possible factor is
\[ x+5, \]write it as
\[ x-(-5). \]Therefore, test
\[ \boxed{P(-5)}. \]The sign in the factor and the number substituted into the polynomial are opposites.
Example with a Positive Binomial Factor
Determine whether
\[ x+3 \]is a factor of
\[ P(x)=x^3+4x^2-3x-18. \]Test
\[ P(-3). \]Compute:
\[ (-3)^3+4(-3)^2-3(-3)-18. \] \[ -27+36+9-18. \] \[ \boxed{0}. \]Therefore,
\[ \boxed{x+3\text{ is a factor}}. \]Factors and Zeros
If
\[ P(c)=0, \]then \(c\) is called a zero of the polynomial function.
By the Factor Theorem,
\[ P(c)=0 \]is equivalent to saying
\[ x-c \]is a factor.
Thus:
\[ \boxed{ c\text{ is a zero} \iff x-c\text{ is a factor} }. \]Zeros and Graphs
If
\[ P(c)=0, \]then the point
\[ (c,0) \]lies on the graph of
\[ y=P(x). \]Therefore, a real zero is an \(x\)-intercept of the polynomial graph.
This gives a three-way connection:
\[ \boxed{ P(c)=0 \iff x-c\text{ is a factor} \iff (c,0)\text{ lies on the graph} }. \]Example: Factor, Zero, and Intercept
Suppose
\[ P(x)=(x-4)(x+1). \]Then
\[ P(4)=0 \]and
\[ P(-1)=0. \]Therefore, the zeros are
\[ 4 \]and
\[ -1. \]The graph has \(x\)-intercepts at
\[ \boxed{(4,0)} \]and
\[ \boxed{(-1,0)}. \]Factor Testing with Synthetic Division
Suppose we want to test whether \(x-3\) is a factor of
\[ P(x)=2x^4-x^3-13x^2+4x+12. \]Use synthetic division with
\[ c=3. \]Coefficients:
\[ 2,\quad-1,\quad-13,\quad4,\quad12. \]Bring down \(2\).
Multiply and add:
\[ 2(3)=6,\qquad -1+6=5, \] \[ 5(3)=15,\qquad -13+15=2, \] \[ 2(3)=6,\qquad4+6=10, \] \[ 10(3)=30,\qquad12+30=42. \]The remainder is
\[ \boxed{42}. \]Therefore, \(x-3\) is not a factor.
Factor Testing Without Full Quotient Interpretation
If the only question is whether \(x-c\) is a factor, we care only about whether the synthetic remainder is zero.
We do not need to interpret the quotient unless the remainder is zero and we want to continue factoring.
This can save substantial work when testing several possible factors.
Using the Theorem to Find an Unknown Coefficient
The Factor Theorem can also determine unknown coefficients.
Suppose
\[ P(x)=x^3+kx^2-4x+8 \]and we are told that
\[ x-2 \]is a factor.
Then
\[ P(2)=0. \]Solve for the Unknown Coefficient
Substitute \(x=2\):
\[ 2^3+k(2)^2-4(2)+8=0. \]Simplify:
\[ 8+4k-8+8=0. \] \[ 4k+8=0. \] \[ 4k=-8. \] \[ \boxed{k=-2}. \]Thus the factor condition determines the missing coefficient.
Using a Known Remainder to Find an Unknown
Suppose
\[ P(x)=2x^3+ax^2-5x+1 \]leaves remainder \(7\) when divided by
\[ x-2. \]By the Remainder Theorem,
\[ P(2)=7. \]Therefore,
\[ 2(8)+4a-10+1=7. \] \[ 16+4a-9=7. \] \[ 4a+7=7. \] \[ \boxed{a=0}. \]Remainder Theorem as a Shortcut
Suppose the problem asks only:
“What is the remainder when \(P(x)\) is divided by \(x-4\)?”
Full long division is unnecessary.
Simply calculate
\[ \boxed{P(4)}. \]Choosing the method that matches the question is an important part of efficient algebra.
Factor Theorem as a Decision Test
Suppose the question asks:
“Is \(x+2\) a factor of \(P(x)\)?”
The most direct test is
\[ \boxed{P(-2)}. \]If the result is zero, the factor works.
If the result is not zero, the factor does not work.
Several Candidate Factors
Suppose we want to determine which of
\[ x-1,\qquad x-2,\qquad x+2 \]are factors of
\[ P(x)=x^3-x^2-4x+4. \]Test the corresponding values:
\[ P(1)=1-1-4+4=0, \] \[ P(2)=8-4-8+4=0, \] \[ P(-2)=-8-4+8+4=0. \]Therefore, all three are factors:
\[ \boxed{ P(x)=(x-1)(x-2)(x+2) }. \]Verification by Expansion
Multiply the last two factors:
\[ (x-2)(x+2)=x^2-4. \]Then
\[ (x-1)(x^2-4) \]gives
\[ x^3-x^2-4x+4. \]The original polynomial is recovered.
Known Factors Reduce the Polynomial Degree
If a degree-\(5\) polynomial has known factor
\[ x-c, \]synthetic division produces a degree-\(4\) quotient.
Finding one factor therefore reduces the remaining problem to a lower-degree polynomial.
Repeating this process can eventually reduce a higher-degree polynomial to quadratic or linear factors.
Example: Reducing Degree
Suppose
\[ P(x)=x^4-5x^3+5x^2+5x-6 \]and we know that \(x-1\) is a factor.
Synthetic division by \(x-1\) uses coefficients
\[ 1,\ -5,\ 5,\ 5,\ -6. \]The result is
\[ 1,\ -4,\ 1,\ 6,\ 0. \]Therefore,
\[ P(x)=(x-1)(x^3-4x^2+x+6). \]A degree-\(4\) problem has become a degree-\(3\) problem.
The Theorems and Polynomial Structure
The Remainder and Factor Theorems connect several ideas that were previously studied separately:
- function evaluation;
- polynomial division;
- synthetic division;
- factoring;
- zeros;
- graph intercepts.
This is an important example of mathematical structure: one relationship appears in several different representations.
Equivalent Statements
For a polynomial \(P(x)\) and number \(c\), the following statements are equivalent:
- \(P(c)=0\);
- the remainder from division by \(x-c\) is \(0\);
- \(x-c\) divides \(P(x)\) exactly;
- \(x-c\) is a factor of \(P(x)\);
- \(c\) is a zero of \(P\);
- if \(c\) is real, \((c,0)\) is an \(x\)-intercept of the graph.
Moving fluently among these statements is more powerful than memorizing them separately.
Common Error: Using the Wrong Sign
If the factor is
\[ x+4, \]the value to test is
\[ \boxed{-4}, \]not \(4\).
The factor must be written
\[ x-c. \]Since
\[ x+4=x-(-4), \]we have
\[ c=-4. \]Common Error: Confusing Zero with the Remainder Requirement
A nonzero remainder does not mean the division procedure failed.
It simply means the divisor is not a factor.
If
\[ P(c)=5, \]then the remainder from division by
\[ x-c \]is
\[ 5. \]The division is valid; it is just not exact.
Common Error: Forgetting Missing Coefficients
When using synthetic division to evaluate
\[ P(x)=x^4-3x+2, \]the coefficient list is
\[ \boxed{1,\ 0,\ 0,\ -3,\ 2}. \]Missing \(x^3\) and \(x^2\) terms require zero placeholders.
Common Error: Assuming One Successful Test Finds All Factors
Showing that
\[ x-2 \]is a factor identifies only one factor.
The quotient may contain additional factors.
Continue factoring or testing the reduced polynomial when the problem asks for complete factorization.
Common Error: Treating Every \(x\)-Intercept the Same Way
The Factor Theorem tells us that a real zero corresponds to a real \(x\)-intercept.
Later we will see that repeated factors can affect how the graph behaves at that intercept.
The graph may cross the axis or touch and turn depending on multiplicity.
That distinction belongs to the later lesson on Zeros, Roots, Factors, and Multiplicity.
A Reliable Procedure for Finding a Remainder
- Write the divisor in the form \(x-c\).
- Identify \(c\).
- Evaluate \(P(c)\) directly or synthetically.
- The result is the remainder.
A Reliable Procedure for Testing a Factor
- Write the possible factor in the form \(x-c\).
- Identify \(c\).
- Compute \(P(c)\).
- If \(P(c)=0\), then \(x-c\) is a factor.
- If \(P(c)\neq0\), then \(x-c\) is not a factor.
- If the factor works and further factorization is needed, use synthetic division to find the quotient.
Example: Complete Factor-Test Workflow
Determine whether
\[ x+2 \]is a factor of
\[ P(x)=x^3+x^2-4x-4. \]Since
\[ x+2=x-(-2), \]evaluate
\[ P(-2). \]We get
\[ -8+4+8-4=0. \]Therefore,
\[ x+2 \]is a factor.
Continue the Complete Workflow
Use synthetic division with \(-2\):
\[ 1,\quad1,\quad-4,\quad-4. \]The synthetic result is
\[ 1,\quad-1,\quad-2,\quad0. \]Therefore,
\[ P(x)=(x+2)(x^2-x-2). \]Factor the quadratic:
\[ x^2-x-2=(x-2)(x+1). \]Thus,
\[ \boxed{ P(x)=(x+2)(x-2)(x+1) }. \]Practice 1: Find a Remainder
Find the remainder when
\[ P(x)=x^3-2x^2+5x-1 \]is divided by
\[ x-3. \]Practice 2: Positive Binomial Divisor
Find the remainder when
\[ P(x)=2x^4+x^2-3 \]is divided by
\[ x+2. \]Practice 3: Test a Factor
Determine whether
\[ x-1 \]is a factor of
\[ P(x)=x^3-4x^2+x+2. \]Practice 4: Test Another Factor
Determine whether
\[ x+1 \]is a factor of
\[ P(x)=x^3-4x^2+x+2. \]Practice 5: Find an Unknown Coefficient
Find \(k\) if
\[ P(x)=x^3+kx^2-5x+3 \]has factor
\[ x-1. \]Practice 6: Known Remainder
Find \(a\) if
\[ P(x)=x^3+ax^2+2x-4 \]leaves remainder \(6\) when divided by
\[ x-2. \]Practice 7: Synthetic Factor Test
Use synthetic division to determine whether
\[ x-2 \]is a factor of
\[ P(x)=x^4-3x^3-3x^2+11x-6. \]Practice 8: Factor Completely
Given that
\[ x-3 \]is a factor of
\[ P(x)=x^3-2x^2-5x+6, \]use synthetic division and factor completely.
Practice 9: Identify the Zero
If
\[ x+7 \]is a factor of \(P(x)\), what zero of \(P\) does it identify?
Practice 10: Interpret the Remainder
Synthetic division of \(P(x)\) by \(x-4\) ends with the number \(-9\).
State:
- the remainder;
- the value \(P(4)\);
- whether \(x-4\) is a factor.
Practice Answers
Practice 1
By the Remainder Theorem, compute
\[ P(3)=27-18+15-1. \] \[ \boxed{23}. \]The remainder is \(23\).
Practice 2
Since
\[ x+2=x-(-2), \]compute
\[ P(-2)=2(16)+4-3. \] \[ 32+4-3=33. \]Therefore, the remainder is
\[ \boxed{33}. \]Practice 3
Evaluate
\[ P(1)=1-4+1+2=0. \]Therefore,
\[ \boxed{x-1\text{ is a factor}}. \]Practice 4
Evaluate
\[ P(-1)=-1-4-1+2=-4. \]Since the result is not zero,
\[ \boxed{x+1\text{ is not a factor}}. \]Practice 5
Because \(x-1\) is a factor,
\[ P(1)=0. \]Therefore,
\[ 1+k-5+3=0. \] \[ k-1=0. \] \[ \boxed{k=1}. \]Practice 6
Since the remainder is \(6\),
\[ P(2)=6. \]Thus,
\[ 8+4a+4-4=6. \] \[ 8+4a=6. \] \[ 4a=-2. \] \[ \boxed{a=-\frac12}. \]Practice 7
Use \(2\) with coefficients
\[ 1,\ -3,\ -3,\ 11,\ -6. \]Synthetic division gives
\[ 1,\ -1,\ -5,\ 1,\ -4. \]The remainder is
\[ -4. \]Therefore,
\[ \boxed{x-2\text{ is not a factor}}. \]Practice 8
Use \(3\) with coefficients
\[ 1,\ -2,\ -5,\ 6. \]Synthetic division gives
\[ 1,\ 1,\ -2,\ 0. \]Therefore,
\[ P(x)=(x-3)(x^2+x-2). \]Factor the quadratic:
\[ x^2+x-2=(x+2)(x-1). \]Thus,
\[ \boxed{ P(x)=(x-3)(x+2)(x-1) }. \]Practice 9
Since
\[ x+7=x-(-7), \]the zero is
\[ \boxed{-7}. \]Practice 10
- The remainder is \(\boxed{-9}\).
- \(P(4)=\boxed{-9}\).
- Since the remainder is not zero, \(x-4\) is not a factor.
What to Remember
The Remainder Theorem states:
\[ \boxed{ \text{remainder from dividing }P(x)\text{ by }x-c = P(c) }. \]The Factor Theorem states:
\[ \boxed{ x-c\text{ is a factor of }P(x) \iff P(c)=0 }. \]These results connect:
- evaluation;
- division;
- synthetic division;
- factors;
- zeros;
- graph intercepts.
A nonzero value \(P(c)\) is the remainder. A zero value means exact divisibility by \(x-c\).
Where We Go Next
The Remainder and Factor Theorems tell us how linear factors correspond to zeros. The next step is to study the simpler building blocks from which polynomial behavior grows.
The next lesson, Power Functions, examines functions of the form \(f(x)=ax^n\), including how the exponent and coefficient control shape, symmetry, growth, and end behavior.