HSA.APR.B.2: The Remainder Theorem and the Factor Theorem
In plain English: HSA.APR.B.2 is the Common Core algebra standard that asks students to know and apply the Remainder Theorem: when a polynomial p(x) is divided by x - a, the remainder is p(a). So p(a) = 0 exactly when x - a is a factor of p(x), and students can find remainders and test factors by evaluating instead of dividing. It is usually taught in Algebra II.
Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x - a is p(a), so p(a) = 0 if and only if (x - a) is a factor of p(x).
Common Core State Standards for Mathematics · Domain: Arithmetic with Polynomials and Rational Expressions (APR) · Cluster: Understand the relationship between zeros and factors of polynomials Also written as HSA-APR.B.2 or A-APR.2 · Official standard
Students divide polynomials by linear divisors of the form x - a, notice that the remainder always matches p(a), and then explain why: every division can be written as p(x) = (x - a)q(x) + r, and substituting x = a wipes out the first term. With the theorem in hand, students find remainders by evaluating instead of dividing, and use the special case p(a) = 0 to decide whether x - a is a factor.
The lesson treats both directions of the "if and only if": a zero value gives a factor, and a factor gives a zero value. Students also run the theorem backward to find an unknown coefficient. Graphing is used only to show what p(a) means; sketching graphs from zeros is the next standard, HSA.APR.B.3.
Learning Objectives
By the end of this lesson, students will be able to:
Divide a polynomial by x - a using long or synthetic division and name the quotient and the remainder
Explain why the remainder on division by x - a equals p(a), using p(x) = (x - a)q(x) + r
Find the remainder on division by x - a by evaluating p(a) instead of dividing
Decide whether x - a is a factor of p(x) by checking whether p(a) = 0, and explain both directions of the "if and only if"
Use the Remainder Theorem to find an unknown coefficient in a polynomial
Prior Knowledge Required
Students should already be comfortable with:
Adding, subtracting and multiplying polynomials HSA.APR.A.1
Evaluating functions written in function notation, including at negative inputs HSF.IF.A.2
Post both parts of the prompt at once. Give students 4 minutes to work alone, then 2 minutes to compare with a partner.
Warm-Up Prompt
"(1) Find the quotient and remainder when 247 is divided by 12, and write 247 = 12 × ___ + ___. (2) Let p(x) = x² + 2x + 5. Find p(3). (3) Multiply (x - 3)(x + 5) and add 20. What do you notice?"
Answers: 247 = 12 × 20 + 7, p(3) = 20, and (x - 3)(x + 5) + 20 = x² + 2x + 5. Write the last result as x² + 2x + 5 = (x - 3)(x + 5) + 20 directly under 247 = 12 × 20 + 7. Ask: "Which number plays the role of the remainder? Where else did 20 show up?" Leave the question open. The lesson answers it.
Direct Instruction20 minutes
Model one synthetic division (Diagram 1), then build the argument in four short steps:
Write the division statement. Dividing p(x) by x - a gives p(x) = (x - a)q(x) + r. Because the divisor has degree 1, the remainder r has degree 0: it is a number.
Substitute x = a. p(a) = (a - a)q(a) + r = 0 · q(a) + r = r. So the remainder is p(a). This is the Remainder Theorem.
Read off the special case. If p(a) = 0, then r = 0 and p(x) = (x - a)q(x), so x - a is a factor. If x - a is a factor, then p(x) = (x - a)q(x) and p(a) = 0. Both directions together are the "if and only if" (often called the Factor Theorem).
Watch the sign of a. Dividing by x + 2 means x - (-2), so a = -2 and the remainder is p(-2).
Work these five examples on the board. For the first one, do both routes (divide, then evaluate) so students see the same number come out twice:
Remainder two ways
Divide p(x) = x³ - 4x² + 2x + 5 by x - 3. Then compute p(3).
Find the remainder when p(x) = x⁴ - 3x² + 7 is divided by x + 1. For the synthetic division, use the row 1, 0, -3, 0, 7.
Equation: p(-1) = 1 - 3 + 7 = 5, so the remainder is 5 (quotient x³ - x² - 2x + 2)
Unknown coefficient
When p(x) = x³ + kx² - 5x + 2 is divided by x - 2, the remainder is 12. Find k.
Equation: p(2) = 8 + 4k - 10 + 2 = 4k = 12, so k = 3
Guided Practice15 minutes
Pairs work through these four problems. One partner divides, the other evaluates, and they switch roles after each problem. Stop after every two problems to compare answers as a class.
Remainder when x³ + x² - 10 is divided by x - 2 (answer: 2)
Remainder when 3x³ - x + 4 is divided by x + 1 (answer: 2)
Is x + 3 a factor of x³ + 2x² - 5x - 6? (p(-3) = 0, so yes)
Find k so that x - 1 is a factor of x³ + kx² - 5x + 2 (k = 2)
Listen for students who substitute +2 for a divisor of x + 2, and for synthetic division rows that skip a missing x term.
Independent Practice10-15 minutes
Students complete four problems alone: two remainders found by evaluation (one with a divisor of the form x + a), one factor test that ends with writing p(x) = (x - a)q(x), and one unknown-coefficient problem. Require one sentence per problem naming the theorem used, for example "Because p(4) = 0, x - 4 is a factor."
Closure5-10 minutes
Exit ticket: "When p(x) is divided by x - 5, the quotient is x² + 1 and the remainder is -3. (a) What is p(5)? (b) Is x - 5 a factor of p(x)? Explain in one sentence." Answers: p(5) = -3, and x - 5 is not a factor because p(5) is not 0.
Differentiation Strategies
For Struggling Students
Give a synthetic division template with the box for a and one slot per power of x already drawn, including slots for missing terms
Have students rewrite every divisor as x - (number) before they start, for example x + 4 = x - (-4)
Start with quadratics, where students can check the division by multiplying back
For Advanced Students
Find the remainder when x¹⁰⁰ - 3x⁵⁰ + 2 is divided by x + 1, and explain why evaluating is the only practical route (answer: 0)
Find a and b so that x - 1 and x + 2 are both factors of x³ + ax² + bx - 6 (a = 4, b = 1)
Prove that x - 1 is a factor of xⁿ - 1 for every positive integer n
Assessment Guidance
What to Look For
Check that students can say why the remainder equals p(a), not only use it: ask them to point to the term that becomes 0. Check the sign of a on every divisor of the form x + a, and check that students state a conclusion about factors in both directions (p(a) = 0 means x - a is a factor; p(a) not 0 means it is not).
02
Classroom Activities
3 Activities
1
Two Routes, One Remainder
15 minPairs
Each pair gets six division cards. Partner A finds the remainder by synthetic division while Partner B finds it by evaluating p(a). They compare, then switch roles for the next card. The point is for students to see the match happen again and again before they are asked to explain it.
Card Set
(x² + 6x + 1) ÷ (x - 1): remainder 8
(x³ - 2x² + 3) ÷ (x - 2): remainder 3
(2x³ + x² - 5) ÷ (x + 1): remainder -6
(x³ - 8) ÷ (x - 2): remainder 0
(x⁴ - x + 4) ÷ (x + 2): remainder 22
(3x³ - 4x² + x - 7) ÷ (x - 3): remainder 41
Debrief Questions
Which route was faster on each card? When would you still want to divide? (When you need the quotient.)
On the card with remainder 0, what can you say about x - 2 and x³ - 8?
Write the division statement p(x) = (x - a)q(x) + r for one card and substitute x = a. What happens to the first term?
Modification for Distance Learning
Post the cards on a shared slide. In breakout rooms, one student shares a screen with the division while the other types p(a) into a graphing app and reads the value off the graph.
2
Factor or Not? Card Sort
20 minGroups of 3-4
Groups receive polynomial cards and candidate-factor cards. For each pairing they compute p(a), place the card in a "Factor" or "Not a Factor" column, and write the remainder on every card in the second column.
Pairings
p(x) = x³ - 7x + 6 with x - 1 (p(1) = 0: factor), x - 3 (p(3) = 12: not a factor) and x + 3 (p(-3) = 0: factor)
q(x) = x³ + 2x² - x - 2 with x + 2 (q(-2) = 0: factor), x - 2 (q(2) = 12: not a factor) and x + 1 (q(-1) = 0: factor)
r(x) = 2x³ - 3x² - 11x + 6 with x - 3 (r(3) = 0: factor), x + 1 (r(-1) = 12: not a factor) and x + 2 (r(-2) = 0: factor)
Procedure
Each group member takes one polynomial and tests its three candidates, then the group checks each other's arithmetic
For each polynomial with two factors found, the group writes p(x) as (first factor)(quotient) using synthetic division, and checks by multiplying
Groups present one "Not a Factor" card and explain what the nonzero remainder means
Variation: Speed Round
Read out a polynomial and a divisor. Students hold up a card saying "Factor" or "Not" within 30 seconds, then one student justifies the answer with p(a).
3
Reverse Engineering: Find the Missing Coefficient
20 minIndividual then share
Students solve puzzles where the remainder or a factor is known but one coefficient is not. Then each student writes a puzzle of their own for a partner. Writing a puzzle forces students to use the theorem in the forward direction to build the answer key.
Starter Puzzles
Find k so that x + 3 is a factor of 2x³ + kx² - 7x + 6 (k = 3)
Find k so that the remainder is 3 when 2x³ + kx - 5 is divided by x + 2 (k = -12)
Find k so that x - 4 is a factor of x³ - kx² + 16 (k = 5)
Write Your Own
Choose a polynomial with one unknown coefficient and a divisor x - a
Decide the remainder you want (0 or not), then solve for the coefficient that makes it happen, so you have the answer key
Trade with a partner; the partner solves it and checks by synthetic division
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Synthetic Division and p(a) Give the Same Remainder
Dividing x³ - 4x² + 2x + 5 by x - 3 leaves remainder 2, and evaluating p(3) also gives 2. The division statement on the right explains why: the factor x - 3 is 0 when x = 3.
Diagram 2: What the Remainder Means on a Graph
The remainder on division by x - a is the height of the graph at x = a. At x = 1 the height is 6, so x - 1 is not a factor. At x = 2 the height is 0, so x - 2 is a factor. The other zeros, -2 and 3, give the factors x + 2 and x - 3.
04
Homework Assignment
~30 min
HSA.APR.B.2 Homework: Remainders and Factors
Directions: Show your work for every problem. When you use the Remainder Theorem, write the value of a you substituted. When you decide whether something is a factor, finish with a sentence that names the value of p(a).
Part 1: Finding Remainders (Problems 1-3)
Use synthetic division to divide p(x) = 2x³ - 3x² + x - 4 by x - 2. State the quotient and the remainder, then check the remainder by computing p(2).
Without dividing, find the remainder when p(x) = x⁴ + 2x³ - 5x + 1 is divided by x + 3.
Find the remainder when p(x) = x⁵⁰ + 3x - 2 is divided by x - 1. Explain why the Remainder Theorem is much faster than dividing here.
Part 2: Factors and Unknown Coefficients (Problems 4-6)
Show that x - 4 is a factor of p(x) = x³ - 6x² + 5x + 12. Then write p(x) as (x - 4) times a quadratic.
Which of x - 1, x + 1 and x - 3 are factors of p(x) = x³ + x² - 9x - 9? For each one that is not a factor, give the remainder.
Find the value of k so that x + 2 is a factor of p(x) = x³ + kx² + x + 6. Then explain, using the words "if and only if", how you know your answer is right.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Choice of a
Correct a for every divisor, including x + a forms
Sign error on one divisor
Sign errors on several divisors, or a not identified
Division and Evaluation
Synthetic division and p(a) both correct, missing terms shown as 0
Correct method, one arithmetic error
Method incorrect or no work shown
Factor Conclusions
Each conclusion stated and justified with the value of p(a)
Conclusion correct but not justified
Conclusion missing or wrong
Reasoning
Problems 3 and 6 explained clearly using the theorem
Explanation present but vague
No explanation
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
According to the Remainder Theorem, what is the remainder when a polynomial p(x) is divided by x - a?
Answer: B
Write p(x) = (x - a)q(x) + r and substitute x = a: the first term becomes 0, so r = p(a). Choice A comes from using the opposite sign, which is right only for a divisor of the form x + a. Choice C is the constant term of p(x), which is the remainder only when a = 0.
Question 2 of 20 · Multiple Choice
What is the remainder when x³ - 2x² + 4 is divided by x - 3?
Answer: A
p(3) = 27 - 18 + 4 = 13. Choice B is p(-3) = -27 - 18 + 4 = -41, which comes from using the wrong sign for a. Choice C is p(0), the constant term.
Question 3 of 20 · Multiple Choice
What is the remainder when x² + 5x - 2 is divided by x + 4?
Answer: C
The divisor x + 4 is x - (-4), so a = -4 and p(-4) = 16 - 20 - 2 = -6. Choice A is p(4) = 34, the result of substituting +4. Choice D drops the negative sign of the answer.
Question 4 of 20 · Multiple Choice
Which of the following is a factor of p(x) = x³ - 13x + 12?
Answer: D
p(3) = 27 - 39 + 12 = 0, so x - 3 is a factor. The others leave nonzero remainders: p(-1) = 24, p(-3) = 24 and p(2) = -6. A common error is to test x + 3 by substituting 3 instead of -3; that gives 0 and makes choice B look correct.
Question 5 of 20 · Multiple Choice
A polynomial p(x) has p(5) = 0. Which statement must be true?
Answer: B
By the Remainder Theorem, the remainder on division by x - 5 is p(5) = 0, so x - 5 divides p(x) evenly and is a factor. Choice A has the sign reversed: x + 5 is a factor exactly when p(-5) = 0. Choice D confuses p(5) with p(0), which is the remainder on division by x.
Question 6 of 20 · Multiple Choice
When p(x) is divided by x - 2, the quotient is x² + 3x - 1 and the remainder is 4. What is p(2)?
Answer: D
p(x) = (x - 2)(x² + 3x - 1) + 4, so p(2) = 0 · 9 + 4 = 4. Choice A evaluates the quotient at 2 instead of p. Choice B adds the quotient value and the remainder, forgetting that the quotient is multiplied by x - 2, which is 0 at x = 2.
Question 7 of 20 · Multiple Choice
Which setup is correct for synthetic division of 3x³ - 5x + 2 by x + 1?
Answer: A
The divisor x + 1 means a = -1, and the missing x² term needs a placeholder 0, so the row is 3, 0, -5, 2. Choice B makes both errors: wrong sign and no placeholder. Choice C has the right sign but skips the 0, so every later column is misaligned.
Question 8 of 20 · Multiple Choice
For what value of k is x - 3 a factor of x³ - 2x² + kx - 6?
Answer: D
x - 3 is a factor if and only if p(3) = 0: 27 - 18 + 3k - 6 = 3 + 3k = 0, so k = -1. Choice C comes from substituting -3: -27 - 18 - 3k - 6 = 0 gives k = -17. Choice A is a sign slip when solving 3k = -3.
Question 9 of 20 · Multiple Choice
What is the remainder when x⁴ - x³ + 2 is divided by x + 1?
Answer: B
a = -1: (-1)⁴ - (-1)³ + 2 = 1 + 1 + 2 = 4. Choice A is p(1) = 1 - 1 + 2 = 2, the result of using the wrong sign for a. Choice A also comes out if a student evaluates -(-1)³ as -1, so watch the signs of odd powers.
Question 10 of 20 · Multiple Choice
Which statement is equivalent to "x - a is a factor of p(x)"?
Answer: C
The standard says p(a) = 0 if and only if x - a is a factor of p(x), so the two statements are equivalent. Choice A describes the factor x + a. Choice B mixes up the input and output of p.
Question 11 of 20 · Multiple Choice
Given that p(2) = 0 for p(x) = 2x³ + x² - 13x + 6, which of these equals p(x)?
Answer: A
Since p(2) = 0, x - 2 is a factor. Synthetic division with 2 on the row 2, 1, -13, 6 gives 2, 5, -3 with remainder 0, so the quotient is 2x² + 5x - 3. Choice C copies the first three coefficients of p instead of dividing. Choice D has a sign error in the last step; multiplying it out gives 2x³ + x² - 7x - 6, not p(x).
Question 12 of 20 · Multiple Choice
What is the remainder when 4x³ - 2x + 7 is divided by x?
Answer: D
Dividing by x is dividing by x - 0, so the remainder is p(0) = 7, the constant term. Choice A is p(1) = 9. Choice B assumes x divides every polynomial evenly, which is true only when the constant term is 0.
Question 13 of 20 · Multiple Choice
A student divides p(x) by x - 4 and gets remainder 0. Which must be true?
Answer: C
The remainder on division by x - 4 is p(4), so p(4) = 0 and 4 is a zero of p. Choices B and D both use the opposite sign; they would follow from a zero remainder on division by x + 4.
Question 14 of 20 · Multiple Choice
If p(-1) = 6, what is the remainder when p(x) is divided by x + 1?
Answer: B
x + 1 = x - (-1), so the remainder is p(-1) = 6. Choice D is a common instinct, but the Remainder Theorem says the one value p(-1) is all you need. Choice A flips the sign of the answer along with the sign of the divisor.
Question 15 of 20 · Short Answer
Use synthetic division to divide p(x) = x³ - x² - 10x - 8 by x - 4. State the quotient and the remainder, then check the remainder with the Remainder Theorem. What does the result tell you about x - 4?
Using 4 on the row 1, -1, -10, -8 gives 1, 3, 2 with remainder 0, so the quotient is x² + 3x + 2 and the remainder is 0. Check: p(4) = 64 - 16 - 40 - 8 = 0. Because the remainder is 0, x - 4 is a factor and p(x) = (x - 4)(x² + 3x + 2).
Question 16 of 20 · Short Answer
Explain why the remainder on division by x - a is always a number (not an expression in x), and then explain why that number equals p(a).
The remainder must have lower degree than the divisor. The divisor x - a has degree 1, so the remainder has degree 0: it is a constant r. That gives p(x) = (x - a)q(x) + r for every x. Substituting x = a gives p(a) = (a - a)q(a) + r = 0 + r, so r = p(a).
Question 17 of 20 · Short Answer
Let p(x) = x³ - 4x² + x + 6. Find the remainder when p(x) is divided by x - 3 and when it is divided by x - 1. What does each result tell you?
p(3) = 27 - 36 + 3 + 6 = 0, so x - 3 is a factor of p(x). p(1) = 1 - 4 + 1 + 6 = 4, so x - 1 is not a factor; dividing by x - 1 leaves remainder 4.
Question 18 of 20 · Short Answer
When p(x) = 3x³ + kx + 4 is divided by x + 2, the remainder is -6. Find k.
The remainder is p(-2) = 3(-8) + k(-2) + 4 = -20 - 2k. Set -20 - 2k = -6, so -2k = 14 and k = -7. Check: 3(-8) - 7(-2) + 4 = -24 + 14 + 4 = -6.
Question 19 of 20 · Short Answer
A student says, "x - 3 is a factor of x³ + 27, because 27 = 3³." Is the student right? Use the Factor Theorem to decide, and find the linear factor of this form that does work.
The student is not right. For x - 3 to be a factor, p(3) must be 0, but p(3) = 27 + 27 = 54, so dividing by x - 3 leaves remainder 54. Try x + 3 instead: p(-3) = -27 + 27 = 0, so x + 3 is a factor, and synthetic division gives x³ + 27 = (x + 3)(x² - 3x + 9).
Question 20 of 20 · Short Answer
The volume of a box, in cubic inches, is V(x) = x³ + 6x² + 11x + 6, and one edge has length x + 1 inches. Use the Factor Theorem to show that x + 1 is a factor of V(x), then find the expression for the area of the face formed by the other two edges.
V(-1) = -1 + 6 - 11 + 6 = 0, so by the Factor Theorem x + 1 is a factor. Synthetic division with -1 on the row 1, 6, 11, 6 gives 1, 5, 6 with remainder 0, so V(x) = (x + 1)(x² + 5x + 6). The other face has area x² + 5x + 6 square inches, which factors further as (x + 2)(x + 3). Note that x = -1 makes the edge length 0, so it is not a real box size; the theorem uses x = -1 only as an algebraic test.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What is the difference between the Remainder Theorem and the Factor Theorem?
The Remainder Theorem says the remainder on division by x - a is p(a). The Factor Theorem is the special case where that remainder is 0: p(a) = 0 if and only if x - a is a factor of p(x). HSA.APR.B.2 states both in one sentence, so students should learn them together.
Why does dividing by x + 3 mean I substitute -3?
The theorem is stated for divisors of the form x - a. Rewrite x + 3 as x - (-3), so a = -3. Another way to see it: x + 3 is 0 when x = -3, and the value that makes the divisor 0 is the value that wipes out the (x - a)q(x) term.
Should students use synthetic division, long division, or just substitute?
If only the remainder is needed, substitute. If the quotient is needed too, for example to finish factoring, use synthetic or long division. Synthetic division is shorter but works only for divisors of the form x - a; long division always works and shows where the method comes from.
Why is the remainder a number and not an expression?
In polynomial division the remainder always has a lower degree than the divisor. The divisor x - a has degree 1, so the remainder has degree 0, which means it is a constant. That is why p(x) = (x - a)q(x) + r has a plain number r at the end.
What does "if and only if" mean in this standard?
It means the statement works in both directions. If p(a) = 0, then x - a is a factor. And if x - a is a factor, then p(a) = 0. Students should be able to use either direction: the first to find factors, the second to find zeros from a factored form.
Can I use the same idea to divide by 2x - 1 or by x² + 1?
The standard covers divisors x - a only. As an extension: for 2x - 1 the remainder is p(1/2), because 2x - 1 is 0 at x = 1/2, but synthetic division needs an adjustment. For a quadratic divisor such as x² + 1 the remainder can be a linear expression, so plain substitution does not give it. Division by general polynomials belongs to HSA.APR.D.6.
What mistakes do students often make on HSA.APR.B.2?
Using the wrong sign for a, especially with divisors like x + 2
Leaving out placeholder zeros for missing terms in synthetic division
Confusing the last number in the synthetic division row (the remainder) with the quotient
Sign errors with negative inputs, such as writing (-2)³ as 8
Saying "x - a is a factor" when p(a) is small but not 0
How does this standard connect to graphs of polynomials?
p(a) is the height of the graph at x = a, so the remainder on division by x - a is that height. When p(a) = 0 the graph meets the x-axis at x = a, and x - a is a factor. The next standard, HSA.APR.B.3, uses factors to find zeros and sketch graphs.
Is HSA.APR.B.2 (the Remainder Theorem) tested on the SAT?
The theorem is rarely named on the SAT, but the idea behind it appears in the Advanced Math domain: a question may give a value such as p(3) = 0 and ask which expression must be a factor, or give a factor and ask for a zero. Students who know the "if and only if" can answer these without dividing.
How can a parent or student check this work at home?
Two quick checks: multiply the quotient by the divisor and add the remainder, which should give back the original polynomial; or type p(x) into a free graphing app and read the height of the graph at x = a, which should equal the remainder.
07
Related Standards
6 standards
These standards connect to HSA.APR.B.2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSA.APR.A.1Prerequisite
Add, subtract and multiply polynomials; polynomials are closed under these operations
