Polynomial long division, synthetic division by a linear divisor, the remainder theorem linking division to evaluation, the factor theorem and its four equivalent statements, and using one known solution to finish a polynomial model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 5 — Polynomials and Polynomial Functions
Apply the Remainder and Factor Theorems
Objectives
Five outcomes. The third and fourth turn an arithmetic technique into a theory.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 362-367 — the lesson these objectives are drawn from
Warm-up
Lesson 5.2 built a synthetic tableau to evaluate a function, and its last line ended with an unexplained row of numbers.
Discussion prompt
For f of x equals x cubed plus 5x squared minus 7x plus 2, run a synthetic tableau at x equal to 2. Then multiply the quantity x squared plus 7x plus 7 by x minus 2 and add 16. What do you notice?
Hint: Compare the bottom row of the tableau with the coefficients you were asked to multiply.
Answer:
\[ (x^2 + 7x + 7)(x-2) + 16 = x^3 + 5x^2 - 7x + 2 \]
The tableau's bottom row was 1, 7, 7, 16 — the coefficients of that quadratic, followed by 16. So the tableau was not just evaluating the function; it was dividing it by x minus 2, and the value f of 2 turned out to be the remainder. This lesson explains why.
Concept
Dividing a polynomial by x minus k leaves a remainder that is exactly the value of the function at k. So a zero of the function is precisely a divisor that goes in exactly, and one known zero converts a high-degree polynomial into a lower-degree one.
synthetic division — A compact method for dividing a polynomial by a divisor of the form x minus k, using only the coefficients. Its bottom row gives the quotient's coefficients followed by the remainder.
\[ \frac{f(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)} \]
That link is what makes the whole chapter possible. Finding one zero of a cubic reduces it to a quadratic, which Chapter 4 can finish, and Lesson 5.6 supplies a way to find that first zero.
Figure (svg): The same value obtained twice: as a remainder from division and as a function value
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 362-364
Section
Section 1
Concept
Divide the highest-power term of what remains of the dividend by the first term of the divisor to get the next term of the quotient, multiply the divisor by it, subtract, and bring down the next term. The remainder must have degree less than the divisor.
polynomial long division — A division algorithm for polynomials that mirrors long division of numbers. It expresses the dividend as the divisor times a quotient, plus a remainder of lower degree than the divisor.
\[ \frac{f(x)}{d(x)} = q(x) + \frac{r(x)}{d(x)} \]
Write a zero coefficient for every missing power in the dividend. Without it the columns misalign and every later step is wrong, exactly as in the synthetic tableau of Lesson 5.2.
Figure (svg): A polynomial long division laid out step by step with a linear divisor
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 362-362 — Use polynomial long division
Picture it
Example 2: a cubic divided by a linear divisor.
Figure (svg): A polynomial long division laid out step by step with a linear divisor
The remainder 16 is written over the divisor, not on its own. The whole expression must equal the original quotient, and 16 by itself would not.
Worked example
Example 1. Note the zero for the missing term.
\[ \text{Divide } 3x^4 - 5x^3 + 4x - 6 \text{ by } x^2 - 3x + 5. \]
Write a zero for the missing power
Why: The dividend has no x squared term, so it is written as 3x to the fourth minus 5x cubed plus 0x squared plus 4x minus 6.
\[ \text{insert } 0 x ^{2} \]
Divide the leading terms
Why: Three x to the fourth over x squared is 3x squared, the first term of the quotient; multiplying and subtracting leaves 4x cubed minus 15x squared plus 4x.
\[ 3 x ^{2} \]
Repeat
Why: Four x cubed over x squared is 4x; multiplying and subtracting leaves negative 3x squared minus 16x minus 6.
\[ 4 x \]
Repeat again
Why: Negative 3x squared over x squared is negative 3; multiplying and subtracting leaves negative 25x plus 9.
\[ -3 \]
Stop when the degree drops below the divisor's
Why: The remainder is linear and the divisor is quadratic, so the division ends.
\[ \text{remainder } -25 x + 9 \]
Figure (svg): The solution to Worked example divide by a quadratic shown as a ladder of expressions, one row per algebraic move
\[ 3x^2 + 4x - 3 + \frac{-25x + 9}{x^2 - 3x + 5} \]
Verify: multiply the quotient by the divisor and add the remainder
Why: The quantity 3x squared plus 4x minus 3 times x squared minus 3x plus 5 expands to 3x to the fourth minus 5x cubed plus 0x squared plus 29x minus 15; adding negative 25x plus 9 gives 3x to the fourth minus 5x cubed plus 4x minus 6, the original dividend. This check is complete rather than partial, which is why the textbook shows it in full.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 362-362
Ranking
One pass of the algorithm.
Put in order
Why: Step one is the setup and the only place a whole division can go silently wrong. The stopping condition in step five is what defines a remainder: it must have lower degree than the divisor, exactly as a numerical remainder must be smaller than the divisor.
Worked example
Example 2 and Guided Practice 2.
\[ \text{Divide } x^3 + 5x^2 - 7x + 2 \text{ by } x - 2, \text{ and } x^3 - x^2 + 4x - 10 \text{ by } x + 2. \]
First: divide the leading terms
Why: X cubed over x is x squared; multiplying and subtracting leaves 7x squared minus 7x.
\[ x ^{2} \]
First: repeat twice more
Why: Seven x squared over x is 7x, then 7x over x is 7, leaving a constant remainder.
\[ x ^{2} + 7 x + 7,\text{ remainder } 16 \]
Second: the same process with x plus 2
Why: The quotient terms are x squared, then negative 3x, then 10.
\[ x ^{2} - 3 x + 10 \]
Second: read the remainder
Why: The last subtraction leaves negative 30.
\[ \text{remainder } -30 \]
Figure (svg): The solution to Worked example divide by a linear divisor shown as a ladder of expressions, one row per algebraic move
\[ x^2 + 7x + 7 + \tfrac{16}{x-2}; \quad x^2 - 3x + 10 - \tfrac{30}{x+2} \]
Verify: evaluate each dividend at the divisor's zero
Why: For the first, f of 2 is 8 plus 20 minus 14 plus 2, which is 16 — the remainder. For the second, f of negative 2 is negative 8 minus 4 minus 8 minus 10, which is negative 30 — again the remainder. That is not a coincidence, and it is the subject of the third idea.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 363-363
Trap
\[ \frac{x^3 + 5x^2 - 7x + 2}{x-2} \]
Report the quotient and add the remainder
Why: The remainder 16 is written as a separate term.
\[ x^2 + 7x + 7 + 16 \quad \text{(wrong)} \]
At x equal to 3 the left side is 45 over 1, or 45, while x squared plus 7x plus 7 plus 16 is 9 plus 21 plus 7 plus 16, which is 53.
\[ x^2 + 7x + 7 + \frac{16}{x-2} \]
Write the remainder over the divisor
Why: The remainder is what is left to be divided, so it keeps its denominator.
\[ \text{at } x = 3: \; 9 + 21 + 7 + \tfrac{16}{1} = 53 \;\text{... and } \tfrac{45}{1} = 45? \]
Checking at x equal to 3: the original is 27 plus 45 minus 21 plus 2, or 53, over 1, which is 53 — matching. The book flags this in an Avoid Errors note: what gets added is r of x over d of x, never r of x alone.
Fill the middle
Example 1, at the second stage.
Fill in the blanks
\frac4x___ = ___ \;\Longrightarrow\; \text___
Why: Four x cubed divided by x squared is 4x, by the quotient of powers rule from Lesson 5.1. Only the leading terms are used at each stage; the rest of the divisor comes back in when you multiply and subtract.
Prediction
Commit before reasoning.
Predict first
Dividing by a quadratic, what is the highest degree the remainder can have?
Correct: One — the remainder has degree less than the divisor.
\[ \deg r(x) < \deg d(x) \]
Why: If the remainder still had degree 2, the division could continue, since its leading term would be divisible by the divisor's. So the algorithm only stops when the degree has dropped below the divisor's, which for a quadratic divisor means degree 1 or 0. Example 1's remainder, negative 25x plus 9, is linear for exactly this reason, while a linear divisor always leaves a constant.
Comparison
Fill the blanks. The same algorithm twice.
Comparison matrix
| Question | Dividing numbers | Dividing polynomials |
|---|---|---|
| What you compare at each step | the leading digits | the leading terms |
| When you stop | when the remainder is less than the divisor | when its degree is less than the divisor's |
| How the answer is written | quotient plus remainder over divisor | quotient plus remainder over divisor |
| The check | divisor times quotient plus remainder | divisor times quotient plus remainder |
The algorithm is genuinely the same one, with degree playing the role that size plays for numbers.
Section
Section 2
Concept
When the divisor is x minus k, the whole long division collapses into a row of coefficients. The bottom row gives the quotient's coefficients, one degree lower than the dividend, followed by the remainder.
\[ \frac{2x^3 + x^2 - 8x + 5}{x+3} = 2x^2 - 5x + 7 - \frac{16}{x+3} \]
For a divisor x plus 3, the value used is negative 3, because x plus 3 is x minus negative 3. Getting that sign wrong is the commonest error in the method.
Figure (svg): A synthetic division tableau with the quotient coefficients and the remainder identified
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 363-363 — Use synthetic division
Picture it
Example 3: a cubic divided by x plus 3.
Figure (svg): A synthetic division tableau with the quotient coefficients and the remainder identified
Four numbers in the bottom row: three are the coefficients of a quadratic quotient and the last is the remainder. The count is automatic, since a cubic divided by a linear gives a quadratic.
Worked example
Example 3. Watch the sign of the divider value.
\[ \text{Divide } 2x^3 + x^2 - 8x + 5 \text{ by } x + 3 \text{ using synthetic division.} \]
Find the value to use
Why: The divisor x plus 3 equals x minus negative 3, so the value is negative 3.
\[ k = -3 \]
Write the coefficients
Why: Nothing is missing: 2, 1, negative 8, 5.
\[ 2, 1, -8, 5 \]
Run the row
Why: Bring down 2; 2 times negative 3 is negative 6, giving negative 5; negative 5 times negative 3 is 15, giving 7; 7 times negative 3 is negative 21, giving negative 16.
\[ 2, -5, 7, -16 \]
Read the result
Why: The first three numbers are the quotient's coefficients and the last is the remainder.
\[ 2 x ^{2} - 5 x + 7,\text{ remainder } -16 \]
Figure (svg): The solution to Worked example divide synthetically shown as a ladder of expressions, one row per algebraic move
\[ 2x^2 - 5x + 7 - \frac{16}{x+3} \]
Verify: multiply back
Why: The quantity 2x squared minus 5x plus 7 times x plus 3 is 2x cubed plus 6x squared minus 5x squared minus 15x plus 7x plus 21, which is 2x cubed plus x squared minus 8x plus 21; subtracting 16 gives the original. The check confirms both the quotient and the sign of the remainder.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 363-363
Matching
Write each as x minus k.
Match the pairs
Why: The value used is always the zero of the divisor: the number that makes it vanish. Asking what makes x plus 3 equal zero gives negative 3 immediately, which is the same habit that fixed the sign of h in Lesson 4.2 and the roots in Lesson 4.3.
Worked example
Guided Practice 3 and 4.
\[ \text{Divide } x^3 + 4x^2 - x - 1 \text{ by } x + 3, \text{ and } 4x^3 + x^2 - 3x + 7 \text{ by } x - 1. \]
First: use negative 3
Why: The coefficients are 1, 4, negative 1, negative 1.
\[ k = -3 \]
First: run the row
Why: Bring down 1; negative 3 gives 1; negative 3 gives negative 4; 12 gives 11.
\[ 1, 1, -4, 11 \]
Second: use positive 1
Why: The divisor is x minus 1, so k is 1, and the coefficients are 4, 1, negative 3, 7.
\[ k = 1 \]
Second: run the row
Why: Bring down 4; 4 gives 5; 5 gives 2; 2 gives 9.
\[ 4, 5, 2, 9 \]
Figure (svg): The solution to Worked example two more synthetic divisions shown as a ladder of expressions, one row per algebraic move
\[ x^2 + x - 4 + \tfrac{11}{x+3}; \quad 4x^2 + 5x + 2 + \tfrac{9}{x-1} \]
Verify: check each remainder by evaluating
Why: For the first, f of negative 3 is negative 27 plus 36 plus 3 minus 1, which is 11 — the remainder. For the second, f of 1 is 4 plus 1 minus 3 plus 7, which is 9. Both match, which is the remainder theorem again and a very fast way to confirm a tableau.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 364-364
Error analysis
A student divides by x plus 3 synthetically.
Annotate
On: \( x + 3 \;\Longrightarrow\; \text{use } 3 \text{ in the tableau} \)
Write the divisor in the form x minus k explicitly before starting. The sign flip takes a second and it is the only setup decision in the method.
Fill the middle
Example 3, at the bottom row.
Fill in the blanks
\text7 2, -5, 7, -16 \;\Longrightarrow\; \text___ 2x^2 - 5x + ___
Why: The first three numbers are the quotient's coefficients and the last is the remainder. A cubic divided by a linear divisor gives a quadratic, so exactly three coefficients are expected — counting them is a quick check that no column was dropped.
Sorting
Look at the divisor.
Sort into buckets
Sort each division by the better method.
Synthetic division is faster but narrower. Long division always works, which is why it is worth being fluent in both.
Prediction
Commit before reasoning.
Predict first
A degree-5 polynomial is divided by x minus k. What degree is the quotient?
Correct: Four — one less.
\[ \deg q = \deg f - \deg d \]
Why: The degrees subtract, because quotient times divisor must reproduce the dividend's leading term and the divisor contributes one degree. That gives a useful count: a degree-n polynomial divided by a linear divisor has n coefficients in its quotient plus one remainder, so the bottom row of the tableau has n plus 1 entries — the same as the top row.
Section
Section 3
Concept
If a polynomial f is divided by x minus k, the remainder equals f of k. So dividing and evaluating are two ways of computing the same number, and either can be used to do the other.
remainder theorem — If a polynomial f is divided by x minus k, then the remainder is f of k.
\[ f(x) = (x-k)q(x) + r \;\Longrightarrow\; f(k) = 0 \cdot q(k) + r = r \]
The proof is one line. Writing f as the divisor times the quotient plus the remainder and substituting x equal to k makes the first term vanish, leaving the remainder.
Figure (svg): The same value obtained twice: as a remainder from division and as a function value
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 363-363 — Remainder Theorem
Picture it
Example 2's division beside a direct evaluation.
Figure (svg): The same value obtained twice: as a remainder from division and as a function value
The two computations share no steps and produce the same number. That is the theorem, and it explains why Lesson 5.2's tableau worked as an evaluator.
Worked example
One substitution does all the work.
\[ \text{Show that dividing } f(x) \text{ by } x - k \text{ leaves the remainder } f(k). \]
Write what division means
Why: The dividend is the divisor times the quotient, plus the remainder.
\[ f(x) = (x - k) q(x) + r(x) \]
Note the degree of the remainder
Why: The divisor is linear, so the remainder has degree 0: it is a constant.
\[ r(x) = r \]
Substitute x equal to k
Why: The factor x minus k becomes zero, so the whole first term vanishes.
\[ f(k) = 0 \times q(k) + r \]
Read the conclusion
Why: Whatever the quotient is, it is multiplied by zero, leaving only the remainder.
\[ f(k) = r \]
Figure (svg): The solution to Worked example prove the theorem shown as a ladder of expressions, one row per algebraic move
\[ f(k) = r \]
Verify: test it on Example 2
Why: Dividing x cubed plus 5x squared minus 7x plus 2 by x minus 2 gave a remainder of 16, and f of 2 is 8 plus 20 minus 14 plus 2, which is 16. The proof explains why no computation could have made these differ: the quotient's value at k is irrelevant because it gets multiplied by zero.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 363-363
Fill the middle
Example 2.
Fill in the blanks
\frac16___: \; \text___ = f(2) = 8 + 20 - 14 + 2 = ___
Why: Eight plus 20 is 28, minus 14 is 14, plus 2 is 16 — the same 16 the long division produced. Evaluating is usually quicker than dividing when only the remainder is wanted, and the theorem is what licenses the swap.
Worked example
The theorem run backwards, on two functions.
\[ \text{Find } f(-3) \text{ for } x^3 + 4x^2 - x - 1 \text{ and } f(1) \text{ for } 4x^3 + x^2 - 3x + 7. \]
First: run the synthetic tableau at negative 3
Why: The bottom row is 1, 1, negative 4, 11.
\[ \text{remainder } 11 \]
First: read the value
Why: By the remainder theorem, f of negative 3 is 11.
\[ f(-3) = 11 \]
Second: run the tableau at 1
Why: The bottom row is 4, 5, 2, 9.
\[ \text{remainder } 9 \]
Second: read the value
Why: So f of 1 is 9.
\[ f(1) = 9 \]
Figure (svg): The solution to Worked example evaluate by dividing shown as a ladder of expressions, one row per algebraic move
\[ f(-3) = 11, \qquad f(1) = 9 \]
Verify: check one by direct substitution
Why: For the second, 4 plus 1 minus 3 plus 7 is 9, matching. The theorem means the tableau serves two purposes at once: run it to divide and you get an evaluation free, or run it to evaluate and you get a factorisation free — which is exactly what the next idea uses.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 364-364
Error analysis
A student uses a tableau to evaluate a function.
Annotate
On: \( \text{bottom row } 2, -5, 7, -16 \;\Longrightarrow\; f(-3) = 2 \)
The row reads left to right as quotient coefficients and then one remainder. Drawing a small box around the last entry before reading anything makes the split explicit.
Prediction
Commit before reasoning.
Predict first
In the proof, why is the quotient's value at k irrelevant?
Correct: It is multiplied by x minus k, which is zero at k.
\[ f(k) = (k-k)q(k) + r = 0 \cdot q(k) + r = r \]
Why: Whatever number the quotient takes at k, multiplying it by zero gives zero, so it contributes nothing. That is why the theorem holds for every polynomial regardless of how complicated the quotient turns out to be — the argument never looks at the quotient at all.
Two truths and a lie
All three are about the remainder theorem.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The remainder theorem is stated for a LINEAR divisor of the form x minus k, and it is that form which makes the divisor vanish at k. Dividing by a quadratic leaves a remainder that is generally linear, not a single value, and no single substitution recovers it.
Comparison
Fill the blanks. The same row of numbers, two readings.
Comparison matrix
| You want | You read | Because |
|---|---|---|
| The value f(k) | the last entry | the remainder equals f(k) |
| The quotient | all the entries except the last | they are its coefficients |
| Whether x - k is a factor | whether the last entry is 0 | a zero remainder means it divides exactly |
| A lower-degree polynomial to factor | the quotient | its degree is one less |
One computation, four different questions answered. That efficiency is why synthetic division is worth being fluent in.
Section
Section 4
Concept
A polynomial f has x minus k as a factor exactly when f of k is zero. Since the remainder is f of k, a zero remainder means the division came out exactly and the divisor is a genuine factor.
factor theorem — A polynomial f has a factor x minus k if and only if f of k equals zero.
\[ x - k \text{ is a factor of } f \;\Longleftrightarrow\; f(k) = 0 \]
The if-and-only-if runs both ways, so knowing any one of the four statements — zero of the function, factor, root of the equation, x-intercept of the graph — gives the other three.
Figure (svg): Four equivalent statements about a value k, connected in a cycle
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 364-364 — Factor Theorem
Picture it
The vocabulary from Lesson 4.3, now with the factor added.
Figure (svg): Four equivalent statements about a value k, connected in a cycle
Lesson 4.3 connected zeros, roots and intercepts. The factor theorem adds the fourth corner and makes every connection an equivalence rather than an implication.
Worked example
Example 4. One synthetic division does the reduction.
\[ \text{Factor } 3x^3 - 4x^2 - 28x - 16 \text{ completely, given that } x + 2 \text{ is a factor.} \]
Translate the given factor
Why: If x plus 2 is a factor then f of negative 2 is zero, by the factor theorem.
\[ f(-2) = 0 \]
Divide synthetically by negative 2
Why: The bottom row is 3, negative 10, negative 8, 0, and the zero confirms the factor.
\[ 3, -10, -8, 0 \]
Write as a product of two factors
Why: The quotient is 3x squared minus 10x minus 8.
\[ (x + 2) (3 x ^{2} - 10 x - 8) \]
Factor the trinomial
Why: With a leading coefficient of 3 and a constant of negative 8, the arrangement 3x plus 2 times x minus 4 gives a middle term of negative 10x.
\[ (x + 2) (3 x + 2) (x - 4) \]
Figure (svg): The solution to Worked example factor given one factor shown as a ladder of expressions, one row per algebraic move
\[ (x+2)(3x+2)(x-4) \]
Verify: check the zero remainder and one value
Why: The remainder was exactly zero, which confirms that x plus 2 divides the cubic. Substituting x equal to 4 into the original gives 192 minus 64 minus 112 minus 16, which is zero — so x minus 4 really is a factor as well. The constant term also checks: 2 times 2 times negative 4 is negative 16.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 364-364
Matching
Ask what makes each factor vanish.
Match the pairs
Why: The third is the instructive one: a factor with a leading coefficient other than 1 corresponds to a fractional zero, negative two thirds. The factor theorem is stated for x minus k, but any linear factor can be put in that form by dividing out its leading coefficient.
Worked example
Guided Practice 5 and 6.
\[ \text{Factor } x^3 - 6x^2 + 5x + 12 \text{ and } x^3 - x^2 - 22x + 40, \text{ given } x - 4 \text{ is a factor of each.} \]
First: divide by 4
Why: The bottom row is 1, negative 2, negative 3, 0.
\[ \text{quotient } x ^{2} - 2 x - 3 \]
First: factor the quotient
Why: A product of negative 3 and a sum of negative 2 give the pair negative 3 and 1.
\[ (x - 4) (x - 3) (x + 1) \]
Second: divide by 4
Why: The bottom row is 1, 3, negative 10, 0.
\[ \text{quotient } x ^{2} + 3 x - 10 \]
Second: factor the quotient
Why: A product of negative 10 and a sum of 3 give the pair 5 and negative 2.
\[ (x - 4) (x + 5) (x - 2) \]
Figure (svg): The solution to Worked example two more, given x minus 4 shown as a ladder of expressions, one row per algebraic move
\[ (x-4)(x-3)(x+1), \qquad (x-4)(x+5)(x-2) \]
Verify: check the constant terms
Why: For the first, negative 4 times negative 3 times 1 is 12, matching. For the second, negative 4 times 5 times negative 2 is 40, also matching. The product of the constants in the factors always equals the polynomial's constant term when the leading coefficient is 1, which makes this a fast check.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 364-364
Trap
\[ x + 2 \text{ is a factor of } f \]
Divide synthetically using 2
Why: The number appearing in the factor is used directly.
\[ \text{remainder } 3(8) - 4(4) - 56 - 16 = -64 \neq 0 \quad \text{(wrong)} \]
The non-zero remainder says x minus 2 is not a factor, which is true but not what was asked.
\[ x + 2 = x - (-2) \;\Longrightarrow\; f(-2) = 0 \]
Use the zero of the factor, not the number written in it
Why: A factor x minus k corresponds to the zero k, so x plus 2 corresponds to negative 2.
\[ \text{remainder } 0 \;\Longrightarrow\; f(x) = (x+2)(3x^2 - 10x - 8) \]
A non-zero remainder is itself informative: it proves the divisor is not a factor. If you were told it was one and get a non-zero remainder, check the sign first.
Fill the middle
Example 4.
Fill in the blanks
\text- 8 3, -10, -8, 0 \;\Longrightarrow\; f(x) = (x+2)(3x^2 - 10x ___)
Why: The first three entries are the quotient's coefficients, so the quadratic is 3x squared minus 10x minus 8, and the final zero confirms that the division was exact. That zero is the whole point: a non-zero last entry would mean x plus 2 was not a factor at all.
Ranking
Factoring completely when one factor is known.
Put in order
Why: Confirming the zero remainder is worth its own step: it verifies both the given information and your arithmetic before you build anything on top of it. From step four onward the problem is an ordinary quadratic factorisation.
Prediction
Commit before reasoning.
Predict first
The factor theorem says x minus k is a factor if and only if f of k is zero. What does the second direction let you do?
Correct: Turn a known zero into a factor, and so reduce the degree.
\[ f(k) = 0 \;\Longrightarrow\; f(x) = (x-k)q(x), \; \deg q = \deg f - 1 \]
Why: One direction says a factor gives a zero, which is the easy half. The other says a zero gives a factor, and that is the useful half: it converts a single number into a division that drops the degree by one. That is how Example 4 turns a cubic into a quadratic, and it is the engine of Lessons 5.6 and 5.7.
Section
Section 5
Concept
Given one zero of a polynomial, divide by the corresponding factor and solve the quotient. For a cubic this reduces the problem to a quadratic, which factoring or the quadratic formula finishes.
\[ f(x) = (x-k)q(x): \; \text{solve } q(x) = 0 \text{ for the rest} \]
The quotient need not factor over the integers. In Example 6 the quotient is a quadratic with irrational roots, and the quadratic formula from Lesson 4.8 supplies them.
Figure (svg): A cubic profit model with two production levels giving the same profit
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 365-365 — Examples 5 and 6
Picture it
Example 6: a cubic profit model with a known solution at 1 million shoes.
Figure (svg): A cubic profit model with two production levels giving the same profit
The curve meets the line at 25 million dollars twice in the sensible range. The known solution supplied the factor; the quadratic formula supplied the other one.
Worked example
Example 5 and Guided Practice 7.
\[ \text{Given } f(2)=0 \text{ for } x^3 - 5x^2 - 12x + 36, \text{ and } f(-2)=0 \text{ for } x^3 + 2x^2 - 9x - 18, \text{ find all zeros.} \]
First: divide by x minus 2
Why: The bottom row is 1, negative 3, negative 18, 0.
\[ \text{quotient } x ^{2} - 3 x - 18 \]
First: factor the quotient
Why: A product of negative 18 and a sum of negative 3 give the pair negative 6 and 3.
\[ (x - 2) (x + 3) (x - 6) \]
First: list the zeros
Why: The three factors vanish at 2, negative 3 and 6.
\[ 2, -3, 6 \]
Second: divide by x plus 2
Why: The bottom row is 1, 0, negative 9, 0, so the quotient is x squared minus 9.
\[ \text{quotient } x ^{2} - 9 \]
Second: finish
Why: A difference of squares gives x plus 3 times x minus 3.
\[ -2, 3, -3 \]
Figure (svg): The solution to Worked example find the other zeros shown as a ladder of expressions, one row per algebraic move
\[ \{2, -3, 6\}; \qquad \{-2, 3, -3\} \]
Verify: check one zero from each
Why: For the first, substituting 6 gives 216 minus 180 minus 72 plus 36, which is zero. For the second, substituting 3 gives 27 plus 18 minus 27 minus 18, also zero. Note the zero in the middle of the second quotient's coefficients: the tableau produced x squared plus 0x minus 9, and dropping that zero would have given a linear quotient and lost a zero entirely.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 365-365
Fill the middle
Example 6, at the rearrangement.
Fill in the blanks
25 = -21x^3 + 46x \;\Longrightarrow\; 0 = 21x^3 - 46x + 25
Why: Adding 21x cubed and subtracting 46x from both sides moves everything to the right, leaving a constant of 25. Standard form is needed before any division, exactly as it was before the quadratic formula in Lesson 4.8.
Worked example
Example 6. The quotient does not factor, and that is fine.
\[ \text{With } P = -21x^3 + 46x \text{ and } P = 25 \text{ at } x = 1, \text{ find the smaller production giving the same profit.} \]
Substitute the profit and rearrange
Why: Twenty-five equals negative 21x cubed plus 46x, so 21x cubed minus 46x plus 25 equals zero.
\[ 21 x ^{3} - 46 x + 25 = 0 \]
Use the known solution
Why: The company already produces 1 million shoes at this profit, so x equal to 1 is a solution and x minus 1 is a factor.
\[ x - 1\text{ is } a\text{ factor} \]
Divide synthetically by 1
Why: The coefficients are 21, 0, negative 46, 25, and the bottom row is 21, 21, negative 25, 0.
\[ \text{quotient } 21 x ^{2} + 21 x - 25 \]
Solve the quotient
Why: The quadratic does not factor, so the quadratic formula gives roots of about 0.70 and about negative 1.70.
\[ x\text{ about } 0.7 \]
Reject the negative
Why: A negative number of shoes is meaningless.
\[ \text{about } 700, 000\text{ shoes} \]
Figure (svg): The solution to Worked example the shoe manufacturer shown as a ladder of expressions, one row per algebraic move
\[ x \approx 0.7 \;\Longrightarrow\; 700{,}000 \text{ shoes} \]
Verify: evaluate the model at 0.7
Why: Negative 21 times 0.343 is about negative 7.2, and 46 times 0.7 is 32.2, giving about 25 million dollars — the same profit as at 1 million shoes. Note that the coefficient row needed a zero for the missing x squared term, without which the division would have been meaningless.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 365-365
Error analysis
A student divides the shoe model's cubic by x minus 1.
Annotate
On: \( 21x^3 - 46x + 25 \;\Longrightarrow\; \text{row } 21, -46, 25 \)
Count the entries before starting: a degree-n polynomial has n plus 1 coefficients. This is the same slip as in Lesson 5.2 and it is just as costly here.
Ranking
Using one known solution to find the others.
Put in order
Why: Step four is where the quotient may resist factoring, and that is not a problem: for a cubic the quotient is quadratic, and the formula always finishes it. Step five is the same modelling judgement as in every model since Lesson 4.3.
Two truths and a lie
All three are about finding zeros.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. An odd-degree polynomial has at least one real zero, as Lesson 5.2's end behaviour argument shows, but the quotient's quadratic may have a repeated root or no real roots at all. Example 6's cubic has two positive solutions and one negative one; a cubic such as x cubed plus x has only one real zero.
Prediction
Commit before reasoning.
Predict first
Every example here was handed one zero. What if none is given?
Correct: Lesson 5.6 gives a list of candidate rational zeros to test.
\[ \text{candidates} = \frac{\text{factors of } a_0}{\text{factors of } a_n} \quad \text{(Lesson 5.6)} \]
Why: The rational zero theorem narrows the search to a finite list built from the factors of the constant term and the leading coefficient, and each candidate is tested with one synthetic tableau. That is exactly why this lesson comes first: without division there would be no way to use a candidate once it was found, and without the factor theorem no reason to look for one.
Comparison
Fill the blanks. Each column is the same fact from a different chapter.
Comparison matrix
| Language | Statement about k | Where it came from |
|---|---|---|
| Functions | f(k) = 0, so k is a zero | Lesson 4.3 |
| Equations | k is a root of f(x) = 0 | Lesson 4.3 |
| Graphs | (k, 0) is an x-intercept | Lesson 4.2 |
| Division | x - k divides f exactly | this lesson |
The factor theorem is what makes all four equivalent rather than merely related, and the fourth is the one that lowers the degree.
Pattern
One routine for dividing, and one for using a known zero.
Write the remainder over the divisor, never on its own. The whole expression has to equal the original quotient.
OpenStax Algebra and Trigonometry 2e, §5.4 Dividing Polynomials §5.4
Check
Long division. Mind the missing power.
Check your understanding
Divide x^3 + 5x^2 - 7x + 2 by x - 2.
Answer: A
Why: Synthetic division with 2 gives the bottom row 1, 7, 7, 16.
Check
The remainder theorem.
Check your understanding
What is the remainder when 4x^3 + x^2 - 3x + 7 is divided by x - 1?
Answer: A
Why: By the remainder theorem the remainder is f(1) = 4 + 1 - 3 + 7 = 9.
Check
The factor theorem.
Check your understanding
Factor 3x^3 - 4x^2 - 28x - 16 completely, given that x + 2 is a factor.
Answer: A
Why: Dividing by -2 gives the quotient 3x^2 - 10x - 8, which factors as (3x + 2)(x - 4).
Real world
A storage tank's volume in cubic metres is modelled by V of x equals x cubed minus 2x squared minus 5x plus 6, where x is a design parameter in metres. The engineers know that x equal to 1 gives a volume of zero, which is the degenerate case.
Discussion prompt
Factor the model completely, find every value of x making the volume zero, and say which values of x give a positive volume.
Hint: The factor theorem turns the known zero into a divisor.
Answer:
\[ V(1) = 1 - 2 - 5 + 6 = 0 \;\Longrightarrow\; x - 1 \text{ is a factor} \]
\[ \text{synthetic division by } 1: \; 1, -1, -6, 0 \;\Longrightarrow\; V(x) = (x-1)(x^2 - x - 6) \]
\[ = (x-1)(x-3)(x+2) \;\Longrightarrow\; x = 1, \; 3, \; -2 \]
The volume is zero at x equal to negative 2, 1 and 3. Testing between the zeros, the volume is positive for 1 less than x less than 3 and for x greater than 3 — but only x greater than 3 gives a design that grows without bound, so the usable range is 1 less than x less than 3 or x greater than 3.
Two things are worth noticing. One known zero reduced a cubic to a quadratic in a single division, which is the whole point of the factor theorem. And deciding where the volume is positive is Lesson 4.9's interval test applied to a cubic — the critical values are the zeros, and one test per interval decides it.
Commit first
Answer, then rate your confidence honestly.
Predict first
You divide a polynomial by x minus 5 and get a remainder of 0. What follows?
Correct: 5 is a zero and x minus 5 is a factor.
\[ r = 0 \;\Longleftrightarrow\; f(x) = (x-5)q(x) \;\Longleftrightarrow\; f(5) = 0 \]
Why: A zero remainder means the division came out exactly, so the polynomial equals x minus 5 times the quotient — which is exactly what being a factor means. By the remainder theorem it also means f of 5 is 0, so 5 is a zero of the function and a root of the equation, and the point (5, 0) is on the graph. One zero remainder delivers all four statements at once, which is what makes this the most useful single fact in the chapter.
Explain it
They can factor quadratics and are stuck on a cubic that will not factor by inspection.
Discussion prompt
In four sentences or fewer, explain how knowing one zero helps, and what you would do with it.
Hint: Talk about lowering the degree.
Answer:
If you know that some number k makes the polynomial zero, then x minus k divides it exactly — that is the factor theorem. Dividing by x minus k, which synthetic division does in one row, leaves a polynomial of degree one lower.
So a cubic becomes a quadratic, and you already know how to handle those, by factoring or by the formula. The whole difficulty is finding that first zero, which is what the next lesson is about.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For long division, do one stage at a time and write every subtraction out rather than doing it mentally. For the sign, write the divisor as x minus k explicitly before starting. For missing powers, count that a degree-n polynomial has n plus 1 coefficients. For reading the row, box the last entry before reading anything else. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take f of x equals x cubed plus 5x squared minus 7x plus 2 and work it four ways. Top left: divide it by x minus 2 with full long division, showing every multiply and subtract. Top right: do the same division synthetically and mark which entries are the quotient and which is the remainder. Bottom left: evaluate f of 2 by direct substitution and draw an arrow to the remainder you found, writing the remainder theorem in one sentence beside it. Bottom right: draw the four-cornered diagram of zero, factor, root and x-intercept, and write beside it what each corner would say if the remainder had been zero instead of 16. In a margin, write the one-line proof that the remainder equals f of k.
If your two divisions disagree, check the synthetic one for a missing coefficient first — the long division makes a missing power visible, and the tableau does not.
Recap
Five things, and the last three are one idea seen three ways.
| If you see | Then |
|---|---|
| A linear divisor | Use synthetic division |
| A divisor of degree 2 or more | Use long division |
| A missing power | Write a zero coefficient |
| A remainder of zero | The divisor is a factor |
| A known zero k | Divide by x - k to lower the degree |
| A quotient that will not factor | Use the quadratic formula |
Every problem here handed you a zero to start from. Lesson 5.6 removes that crutch: the rational zero theorem produces a finite list of candidates, and one synthetic tableau tests each.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.5 Apply the Remainder and Factor Theorems §5.5, pp. 362-367 — everything on these slides traces back here
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