HSA.APR.D.6: Rewriting Rational Expressions as Quotient Plus Remainder
In plain English: HSA.APR.D.6 is the Common Core algebra standard that asks students to rewrite simple rational expressions in different forms, especially to write a(x)/b(x) as q(x) + r(x)/b(x), where the remainder r(x) has lower degree than b(x). Students use inspection, polynomial long division or, for complicated cases, a computer algebra system. It is usually taught in Algebra II.
Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x), where a(x), b(x), q(x), and r(x) are polynomials with the degree of r(x) less than the degree of b(x), using inspection, long division, or, for the more complicated examples, a computer algebra system.
Common Core State Standards for Mathematics · Domain: Arithmetic with Polynomials and Rational Expressions (APR) · Cluster: Rewrite rational expressions Also written as HSA-APR.D.6 or A-APR.6 · Official standard
Students rewrite a rational expression a(x)/b(x) in the form q(x) + r(x)/b(x), the algebraic version of writing 47/6 as the mixed number 7 + 5/6. The key fact is the division statement a(x) = b(x) · q(x) + r(x), where the degree of the remainder r(x) is less than the degree of b(x). Dividing both sides by b(x) gives the new form, which is equal to the original expression wherever b(x) ≠ 0.
Students use three methods and learn when each one fits: inspection when the numerator is visibly close to a multiple of the denominator, polynomial long division for general cases, and a computer algebra system (CAS) for complicated examples. They also move in the other direction, combining q(x) + r(x)/b(x) back into a single fraction, and see what the new form tells them, such as the value an average cost approaches.
Learning Objectives
By the end of this lesson, students will be able to:
Explain the division statement a(x) = b(x) · q(x) + r(x) and why the degree of r(x) must be less than the degree of b(x)
Rewrite a(x)/b(x) as q(x) + r(x)/b(x) by inspection when the numerator is close to a multiple of the denominator
Use polynomial long division, including divisors of degree 2 and dividends with missing terms
Use a computer algebra system for complicated divisions and verify its output by multiplying back
Rewrite an expression of the form q(x) + r(x)/b(x) as a single fraction and interpret each form in context
Prior Knowledge Required
Students should already be comfortable with:
Adding, subtracting and multiplying polynomials HSA.APR.A.1
The Remainder Theorem for division by x - a HSA.APR.B.2
Rewriting improper fractions as mixed numbers and long division of whole numbers
The degree of a polynomial and writing polynomials in standard form
Start with numbers. Students work alone for three minutes, then compare.
Warm-Up Prompt
"Write 47/6 and 123/8 as a whole number plus a fraction that is less than 1. Then look at (x + 5)/(x + 2). Can you write it as a polynomial plus a fraction whose numerator is 'smaller' than its denominator?"
Collect 47/6 = 7 + 5/6 and 123/8 = 15 + 3/8, and ask how students know they are done (the remainder is smaller than the divisor). For (x + 5)/(x + 2), some students will split it as x/(x + 2) + 5/(x + 2), which is a correct but different form. Guide them to see x + 5 = (x + 2) + 3, so (x + 5)/(x + 2) = 1 + 3/(x + 2). Tell students that "smaller" for polynomials means lower degree, and that 123/8 will come back later in the lesson.
Direct Instruction20-25 minutes
The division statement. For polynomials a(x) and b(x) with b(x) not zero, there are polynomials q(x) and r(x) with a(x) = b(x) · q(x) + r(x) and the degree of r(x) less than the degree of b(x). Dividing by b(x) gives a(x)/b(x) = q(x) + r(x)/b(x). Show the first two examples by inspection: rewrite the numerator as a multiple of the denominator plus a leftover. Then model long division with Diagram 1, using these steps:
Write both polynomials in descending powers, with 0 as the coefficient of any missing power, such as x³ + 0x² + 0x - 8.
Divide the leading term of what is left by the leading term of b(x), and write the result in the quotient.
Multiply b(x) by that term and subtract the product, changing every sign.
Repeat until the degree of what is left is less than the degree of b(x). What is left is r(x).
Write and check: a(x)/b(x) = q(x) + r(x)/b(x), and expand b(x) · q(x) + r(x) to get a(x) back.
Inspection
Rewrite (x + 5)/(x + 2). The numerator is the denominator plus 3: x + 5 = (x + 2) + 3.
Equation: (x + 5)/(x + 2) = 1 + 3/(x + 2)
Inspection with a coefficient
Rewrite (2x - 1)/(x - 3). Twice the denominator is 2x - 6, and 2x - 1 = 2(x - 3) + 5.
Equation: (2x - 1)/(x - 3) = 2 + 5/(x - 3)
Long division, linear divisor
Divide x² + 3x - 7 by x - 2 (Diagram 1). The quotient is x + 5 and the remainder is 3.
Divide x⁴ - 3x³ + 2x - 5 by x² - x + 2 with a CAS. In GeoGebra's CAS view, Division(x^4 - 3x^3 + 2x - 5, x^2 - x + 2) returns the quotient and the remainder.
Equation: Quotient x² - 2x - 4, remainder 2x + 3, so the expression equals x² - 2x - 4 + (2x + 3)/(x² - x + 2)
Connect the third example to the warm-up: substituting x = 10 turns x² + 3x - 7 into 123 and x - 2 into 8, and the result x + 5 + 3/(x - 2) becomes 15 + 3/8. For the CAS example, have students multiply (x² - x + 2)(x² - 2x - 4) + 2x + 3 to confirm the output. Use Diagram 2 to show one reason the new form is useful: in 1 + 3/(x + 2), the fraction part gets close to 0 when x is far from 0, so the graph approaches the line y = 1. Stress that both forms are equal for every x except x = -2, where neither is defined.
Guided Practice15 minutes
Work through three problems with the class, asking which method fits before starting each one. (1) By inspection, (3x + 4)/(x + 1) = 3 + 1/(x + 1), because 3x + 4 = 3(x + 1) + 1. (2) By inspection, (x - 7)/(x - 4) = 1 - 3/(x - 4); point out that a remainder can be negative. (3) By long division, (x³ - 2x² + 5)/(x + 1) = x² - 3x + 3 + 2/(x + 1); students must write the missing term as 0x. Check each answer by multiplying back. Listen for these errors: forgetting to change all signs when subtracting, leaving out the 0x placeholder, and writing the remainder over the quotient instead of over the divisor.
Independent Practice15 minutes
Students work alone on four problems: (1) (4x + 9)/(x + 2) = 4 + 1/(x + 2) by inspection; (2) (x² - 5x + 1)/(x - 3) = x - 2 - 5/(x - 3) by long division; (3) (3x³ + 2x - 1)/(x² - 2) = 3x + (8x - 1)/(x² - 2), with a remainder of degree 1; (4) the other direction: write 2 + 3/(x - 1) as the single fraction (2x + 1)/(x - 1). Early finishers check problem 3 with a CAS.
Closure5-10 minutes
Exit ticket: (1) Rewrite (x + 8)/(x + 3) by inspection. (1 + 5/(x + 3).) (2) Is x + 1 + (x² + 4)/(x² + 3) in the form q(x) + r(x)/b(x) required by the standard? Explain. (No: the numerator x² + 4 has the same degree as x² + 3, so the division is not finished.) (3) Name one division from today that you would give to a CAS, and say why.
Differentiation Strategies
For Struggling Students
Start every problem with a numerical twin, such as 29/4 = 7 + 1/4, and keep it next to the algebraic problem
Provide a long division template with columns for each power of x and the placeholder 0 already printed
Have students check each answer by multiplying b(x) · q(x) and adding r(x), so they can find their own errors
For Advanced Students
Explain why the quotient and remainder are unique: if two pairs worked, what would their difference have to satisfy?
Find all values of k for which x - 2 leaves remainder 0 when dividing x³ - 3x² + kx + 6, and compare long division with the Remainder Theorem
Use the form q(x) + r(x)/b(x) to predict the slant asymptote of y = (x² + 3x - 7)/(x - 2), then check with a graphing tool (a preview of HSF.IF.C.7d)
Assessment Guidance
What to Look For
A finished answer has the form q(x) + r(x)/b(x) with the degree of r(x) less than the degree of b(x), and the fraction part keeps the original denominator. Ask students how they know they are done: "the degree of what is left is less than the degree of the divisor" is the answer to listen for. Check that students verify results by computing b(x) · q(x) + r(x). With a CAS, students should be able to read the output, write it in the required form and confirm it, not only copy the screen.
02
Classroom Activities
3 Activities
1
Mixed Numbers, Mixed Expressions
15 minPairs
Pairs match 8 cards: 4 rational expressions and 4 expressions in the form q(x) + r(x)/b(x). They find each match by inspection, then substitute x = 10 to see the matching mixed numbers.
The 8 Cards (4 Pairs)
(x + 6)/(x + 1) matches 1 + 5/(x + 1); at x = 10: 16/11 = 1 + 5/11
(5x - 2)/(x - 1) matches 5 + 3/(x - 1); at x = 10: 48/9 = 5 + 3/9
(x² + 1)/(x + 1) matches x - 1 + 2/(x + 1); at x = 10: 101/11 = 9 + 2/11
(x² + 4x + 3)/(x + 4) matches x + 3/(x + 4); at x = 10: 143/14 = 10 + 3/14
Procedure
Shuffle the cards and match them by rewriting each numerator as a multiple of its denominator plus a leftover
Check each match by multiplying back
Substitute x = 10 into both cards of each pair and confirm that the numbers agree
Discussion Questions
Why does x = 10 turn these expressions into ordinary mixed numbers?
At x = 1, the second pair gives 3/0. What does that tell you about where the two forms are equal?
Why is x² + 4x = x(x + 4) useful for the fourth pair?
2
Long Division Stations
25 minGroups of 3-4
Groups rotate through 4 stations, spending about 6 minutes at each. Every station has one division; the group writes the result as q(x) + r(x)/b(x) and checks it by multiplying back.
Stations
Station 1: (x³ + 4x² - x + 2)/(x + 3) = x² + x - 4 + 14/(x + 3)
Station 2: (6x² - x - 4)/(2x + 1) = 3x - 2 - 2/(2x + 1)
Station 3: (x⁴ - 1)/(x - 1) = x³ + x² + x + 1, with remainder 0 (placeholders needed)
Station 4: (x³ + 2x - 9)/(x² + x - 1) = x - 1 + (4x - 10)/(x² + x - 1)
Procedure
One student divides, one checks by multiplying back, and one records the answer on the station poster; rotate roles at each station
If the group's answer differs from an earlier group's answer on the poster, the group finds the error before moving on
Challenge Variation
At Station 3, ask groups to predict (x⁵ - 1)/(x - 1) and (x⁶ - 1)/(x - 1) without dividing, then explain the pattern.
3
By Hand or by CAS?
20 minPairs
For the more complicated examples, the standard allows a computer algebra system. Partners race on two divisions: one partner divides by hand, the other uses a CAS such as GeoGebra, and both verify the result.
The Two Divisions
(x⁵ - 2x⁴ + 3x² - x + 6)/(x³ + x - 2) = x² - 2x - 1 + (7x² - 4x + 4)/(x³ + x - 2)
In GeoGebra's CAS view, enter Division(numerator, denominator) to get the quotient and remainder
Write the CAS result in the form q(x) + r(x)/b(x), then check it by expanding b(x) · q(x) + r(x) (in GeoGebra, with Expand)
Partners compare their answers, then switch roles for the second division
Discussion Questions
Where did the hand method slow down or go wrong?
A CAS may display its answer in a different but equivalent form. How can you tell that two forms are equal?
Which divisions from today would you still do by hand, and why?
Modification for Distance Learning
Students use the online GeoGebra CAS calculator and paste a screenshot of their Division and Expand commands into a shared document.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Long Division of x² + 3x - 7 by x - 2
Each step divides the leading term, multiplies the divisor and subtracts. The division stops when the degree of what is left (3, degree 0) is less than the degree of x - 2 (degree 1).
Diagram 2: What the Form q(x) + r(x)/b(x) Shows
Graph of y = (x + 5)/(x + 2) = 1 + 3/(x + 2), drawn to scale (30 px per unit). The quotient 1 is the horizontal line the graph approaches, and the remainder part 3/(x + 2) shrinks toward 0 as x moves away from -2. Labeled points: (-5, 0), (0, 2.5) and (1, 2).
04
Homework Assignment
~30 min
HSA.APR.D.6 Homework: Writing a(x)/b(x) as q(x) + r(x)/b(x)
Directions: Write every answer in the form q(x) + r(x)/b(x) with the degree of r(x) less than the degree of b(x), unless the problem asks for a single fraction. Name the method you used (inspection, long division or CAS), and check each answer by expanding b(x) · q(x) + r(x).
Part 1: Inspection and Long Division (Problems 1-3)
Use long division to rewrite (x² + 7x + 4)/(x + 3).
Use long division to rewrite (2x³ - 3x² + x + 6)/(x - 2). Then use the Remainder Theorem to confirm your remainder.
Part 2: Harder Divisors, CAS and Other Forms (Problems 4-6)
Use long division to rewrite (x⁴ + 2x² - x + 3)/(x² + 3). Explain why your remainder is allowed to contain an x term.
Use a CAS to rewrite (3x⁵ - x³ + 4x² - 2)/(x³ - 2x + 1). Record the command you used, write the result in the required form, and verify it by expanding b(x) · q(x) + r(x).
(a) Write 4 - 7/(x + 2) as a single fraction. (b) A club orders custom T-shirts: the printer charges a $120 setup fee plus $6.50 per shirt, so the average cost per shirt for n shirts is (6.5n + 120)/n dollars. Rewrite this in the form q + r/n, explain what each part means, find the average cost for 40 shirts, and find the smallest order that brings the average cost to $8.00 or less.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Required Form
Every answer is q(x) + r(x)/b(x) with degree of r less than degree of b
Form correct on most problems
Form missing or division unfinished
Method
Method named and carried out correctly, placeholders used
Correct method with sign or placeholder errors
No clear method
Verification
b(x) · q(x) + r(x) expanded and matches a(x) for every problem
Some answers checked
No checks
Interpretation
Context parts explained correctly with units
Numbers correct, meaning unclear
Missing or incorrect
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Choose an answer for each multiple-choice question and read the explanation that appears. Work the short-answer questions on paper before opening their answers. Reset quiz clears the page for another try.
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
Rewrite (x + 7)/(x + 3) by inspection.
Answer: B
x + 7 = (x + 3) + 4, so (x + 7)/(x + 3) = 1 + 4/(x + 3). Choice A cancels the x terms, which is not allowed because x is a term, not a factor. Choice C has the wrong sign: 1 - 4/(x + 3) = (x - 1)/(x + 3). Choice D drops the denominator.
Question 2 of 20 · Multiple Choice
Rewrite (2x + 5)/(x + 1) in the form q(x) + r(x)/b(x).
Answer: C
2x + 5 = 2(x + 1) + 3, so the expression equals 2 + 3/(x + 1). Choice A keeps the original constant 5 as the remainder, forgetting that 2(x + 1) already contains 2. Choice D subtracts only 1 from 5.
Question 3 of 20 · Multiple Choice
Which is (x² + 2x - 5)/(x - 1) written as q(x) + r(x)/b(x)?
Answer: A
Long division gives quotient x + 3: (x - 1)(x + 3) = x² + 2x - 3, and (x² + 2x - 5) - (x² + 2x - 3) = -2. So the result is x + 3 - 2/(x - 1). Choice C has the wrong sign on the remainder. Choice D ignores the remainder. Choice B does not multiply back to the numerator: (x - 1)(x + 1) - 4 = x² - 5.
Question 4 of 20 · Multiple Choice
What is the remainder when x³ + x - 4 is divided by x - 2?
Answer: C
Long division gives x³ + x - 4 = (x - 2)(x² + 2x + 5) + 6. The Remainder Theorem agrees: 2³ + 2 - 4 = 6. Choice A has the wrong sign. Choice B comes from a sign error on the x term: 8 - 2 - 4 = 2. Choice D forgets the -4: 8 + 2 = 10.
Question 5 of 20 · Multiple Choice
Which expression is written in the form q(x) + r(x)/b(x) with the degree of r(x) less than the degree of b(x)?
Answer: D
In choice D the remainder 3x + 2 has degree 1, less than degree 2 of x² + 1. In choice A the remainder x² + 2 has the same degree as the divisor, so one more division step is needed. Choice B has not been rewritten, and in choice C the numerator x³ has a higher degree than the divisor.
Question 6 of 20 · Multiple Choice
Use long division to rewrite (x² - 4x + 9)/(x - 3).
Answer: A
(x - 3)(x - 1) = x² - 4x + 3, and 9 - 3 = 6, so the result is x - 1 + 6/(x - 3). Choice B adds 3 instead of subtracting it in the last step. Choice C comes from dividing by x + 3 by mistake. Choice D has the wrong sign in the quotient.
Question 7 of 20 · Multiple Choice
Rewrite (6x² + 5x - 1)/(2x + 1) in the form q(x) + r(x)/b(x).
Answer: D
(2x + 1)(3x + 1) = 6x² + 5x + 1, and (6x² + 5x - 1) - (6x² + 5x + 1) = -2. So the result is 3x + 1 - 2/(2x + 1). Choice A drops the remainder. Choice B has the wrong sign on the remainder. Choice C divides 6x² by 2x correctly but then subtracts incorrectly.
Question 8 of 20 · Multiple Choice
What is (x³ - 8)/(x - 2) written as a polynomial?
Answer: B
Write the dividend as x³ + 0x² + 0x - 8. Long division gives x² + 2x + 4 with remainder 0, which matches the identity x³ - 8 = (x - 2)(x² + 2x + 4). Choice A has a sign error in the middle term. Choice C comes from skipping the 0x² placeholder. Choice D invents a remainder: the Remainder Theorem gives 2³ - 8 = 0.
Question 9 of 20 · Multiple Choice
Write 3 + 2/(x - 4) as a single fraction.
Answer: A
3 = 3(x - 4)/(x - 4), so the sum is (3x - 12 + 2)/(x - 4) = (3x - 10)/(x - 4). Choice B adds numerators without a common denominator. Choice C multiplies 3 by x but not by -4. Choice D subtracts 2 instead of adding it.
Question 10 of 20 · Multiple Choice
Use long division to rewrite (x³ + 2x² + 3)/(x² + 1).
Answer: C
(x² + 1)(x + 2) = x³ + 2x² + x + 2, and (x³ + 2x² + 0x + 3) - (x³ + 2x² + x + 2) = -x + 1. So the result is x + 2 + (1 - x)/(x² + 1). Choice A loses the -x term. Choice B does not change the sign of x when subtracting. Choice D adds the 2 instead of subtracting it.
Question 11 of 20 · Multiple Choice
The standard says to use a computer algebra system "for the more complicated examples." Which division is the best candidate for a CAS?
Answer: D
Choice D has a degree 6 dividend and a degree 3 divisor, so long division takes many steps with many chances for sign errors: a CAS is the sensible tool, followed by a check. Choices A and C can be done by inspection, and choice B is the difference of squares (x + 3)(x - 3) divided by x - 3.
Question 12 of 20 · Multiple Choice
A student writes (x² + 5)/(x + 2) = x + 5/2. What is the correct form?
Answer: B
The student divided term by term, which only works when the denominator is a single term. Long division: (x + 2)(x - 2) = x² - 4, and 5 - (-4) = 9, so the result is x - 2 + 9/(x + 2). Choice A leaves out the quotient term -2. Choice D subtracts 4 instead of adding it.
Question 13 of 20 · Multiple Choice
Given that (x² + 3x + 4)/(x + 1) = x + 2 + 2/(x + 1), what does substituting x = 10 tell you about 134/11?
Answer: C
At x = 10 the numerator is 100 + 30 + 4 = 134, the denominator is 11, the quotient x + 2 is 12 and the remainder part is 2/11. So 134/11 = 12 + 2/11, and indeed 11 · 12 + 2 = 134. Choice A uses x instead of x + 2 as the quotient. Choice B changes the denominator. Choice D is not equal to 134/11: 13 + 1/11 = 144/11.
Question 14 of 20 · Multiple Choice
A student group prints posters for a fundraiser. The cost is a $150 design fee plus $8 per poster, so the average cost per poster is (8n + 150)/n = 8 + 150/n dollars. What does 150/n represent?
Answer: A
Rewriting shows the average cost as the $8 printing cost plus the design fee divided among n posters. As n grows, 150/n gets smaller, so the average cost gets closer to $8. For 50 posters, 150/50 = $3 and the average is $11. Choice B is the quotient 8, not the remainder part. Choice C is 8n + 150.
Question 15 of 20 · Short Answer
Use long division to rewrite (x² - 6x + 2)/(x - 4).
Divide x² by x to get x, multiply x(x - 4) = x² - 4x and subtract: -2x + 2. Divide -2x by x to get -2, multiply -2(x - 4) = -2x + 8 and subtract: -6. So the result is x - 2 - 6/(x - 4). Check: (x - 4)(x - 2) - 6 = x² - 6x + 8 - 6 = x² - 6x + 2.
Question 16 of 20 · Short Answer
Rewrite (4x + 7)/(2x + 1) by inspection.
Twice the denominator is 4x + 2, and 4x + 7 = 2(2x + 1) + 5. So (4x + 7)/(2x + 1) = 2 + 5/(2x + 1).
Question 17 of 20 · Short Answer
Rewrite (2x³ + 5x² - x + 1)/(x + 3) in the form q(x) + r(x)/b(x).
Long division (or synthetic division with -3) gives quotient 2x² - x + 2 and remainder -5. So the result is 2x² - x + 2 - 5/(x + 3). Check: (x + 3)(2x² - x + 2) - 5 = 2x³ + 5x² - x + 6 - 5 = 2x³ + 5x² - x + 1. The Remainder Theorem agrees: 2(-27) + 5(9) + 3 + 1 = -5.
Question 18 of 20 · Short Answer
Rewrite (x⁴ + 3x - 2)/(x² - x) in the form q(x) + r(x)/b(x).
Write the dividend as x⁴ + 0x³ + 0x² + 3x - 2. The quotient is x² + x + 1, because (x² - x)(x² + x + 1) = x⁴ - x. Subtracting leaves (x⁴ + 3x - 2) - (x⁴ - x) = 4x - 2, which has degree 1 < 2. So the result is x² + x + 1 + (4x - 2)/(x² - x).
Question 19 of 20 · Short Answer
A CAS reports that dividing a(x) by b(x) = x² + 2x - 1 gives quotient x² - 3 and remainder 5x + 2. Find a(x) in standard form and write a(x)/b(x) in the form q(x) + r(x)/b(x).
Rewrite (x² + 10)/(x² + 4) in the form q(x) + r(x)/b(x). Which method is fastest?
Inspection: x² + 10 = (x² + 4) + 6, so (x² + 10)/(x² + 4) = 1 + 6/(x² + 4). The remainder 6 has degree 0, less than degree 2 of the divisor, so the form is finished.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSA.APR.D.6 mean?
HSA.APR.D.6 means students can rewrite a rational expression a(x)/b(x) as a polynomial plus a proper fraction: q(x) + r(x)/b(x), with the degree of r(x) less than the degree of b(x). It is the polynomial version of writing an improper fraction as a mixed number. Students use inspection, long division or a computer algebra system.
Is HSA.APR.D.6 Algebra 1 or Algebra 2?
It is usually taught in Algebra II, in the unit on polynomial and rational expressions, after the Remainder Theorem (HSA.APR.B.2). It prepares students for graphing rational functions and for work with rational expressions in Precalculus.
Why rewrite a(x)/b(x) as q(x) + r(x)/b(x)?
The new form shows the structure of the expression. The quotient q(x) describes the behavior when x is far from 0, because the fraction r(x)/b(x) gets close to 0 there. In context, an average cost such as (5n + 200)/n = 5 + 200/n shows the per-item cost and the shared fixed cost separately. Later, in calculus, this form is often the first step in integrating a rational function.
When can students use inspection instead of long division?
Use inspection when you can see how to write the numerator as a multiple of the denominator plus a leftover. For example, 3x + 1 = 3(x - 2) + 7, so (3x + 1)/(x - 2) = 3 + 7/(x - 2). This works well when the numerator and denominator have the same degree, or when the numerator is a small change from a product you recognize.
Can students use synthetic division for HSA.APR.D.6?
Yes, when the divisor has the form x - a. Synthetic division is a shortcut for long division by x - a, and it gives the same quotient and remainder. It does not work directly for divisors such as x² + 1 or 2x + 1 without adjusting, so students still need long division.
Why must the degree of r(x) be less than the degree of b(x)?
That condition is what makes the division finished and the answer unique. If the remainder had the same degree as b(x), you could divide once more and move another term into the quotient. It plays the same role as requiring the remainder in whole-number division to be less than the divisor: 47/6 = 7 + 5/6, not 6 + 11/6.
What are common mistakes in polynomial long division?
Frequent errors are forgetting the 0 placeholder for a missing power (such as 0x² in x³ - 8), subtracting only the first term of the product instead of changing every sign, and writing the remainder over the quotient instead of over the divisor. Checking the answer by expanding b(x) · q(x) + r(x) catches each of these.
Which computer algebra system can students use?
Any CAS that divides polynomials works. In the free GeoGebra CAS, Division(a, b) returns the quotient and the remainder, and Expand can check the result. The standard reserves a CAS for the more complicated examples, and students should still write the output in the form q(x) + r(x)/b(x) and verify it.
Are a(x)/b(x) and q(x) + r(x)/b(x) always equal?
They are equal for every x where b(x) ≠ 0. At a zero of b(x), both forms are undefined. For example, (x + 5)/(x + 2) and 1 + 3/(x + 2) agree everywhere except x = -2, where neither has a value.
How can parents help with rewriting rational expressions?
Start with numbers: ask your student to turn 29/4 into a mixed number and explain each step, then do the same with a polynomial division. A strong habit to encourage is the check: multiply the divisor by the quotient and add the remainder, and see whether you get the original numerator back.
07
Related Standards
6 standards
These standards connect to HSA.APR.D.6: 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
