When a polynomial is divided by a quadratic, the remainder may be a linear expression Ax + B—not just a number. The rule is that the remainder must have lower degree than the divisor. To recover A and B, either perform polynomial division or, when the quadratic has two distinct known roots, evaluate the original polynomial at those two roots and solve two simple equations.
For example, dividing P(x) = x³ + 2x² + 4x + 5 by x² − 1 leaves remainder 5x + 7. The two evaluation conditions P(1) = 12 and P(−1) = 2 determine its coefficients. This guide explains why that works, how repeated or non-real divisor roots change the method, and how to check the result using the polynomial-division identity.
A quadratic divisor allows a linear remainder
Polynomial division has a governing rule: when P(x) is divided by a nonzero polynomial D(x), the result is P(x) = D(x)Q(x) + R(x), with the remainder R either zero or of degree strictly less than D.
If the divisor has degree two, the remainder can be written Ax + B. It may happen that A = 0, leaving a constant, or both A and B are zero, leaving no remainder. But the general remainder form must allow a linear term.
This differs from division by x − c, where the remainder has degree less than one and is therefore constant. A learner who automatically writes R for every divisor is applying a linear-divisor rule to a quadratic-divisor problem.
The first step is always to compare polynomial degrees and identify the possible remainder form before trying to calculate its coefficients.
Why evaluating the divisor’s roots helps
Suppose P(x) = D(x)Q(x) + Ax + B and D(r) = 0. At x = r, the product D(r)Q(r) vanishes, so P(r) = Ar + B.
If D has two distinct roots r and s, then P(r) = Ar + B and P(s) = As + B. Those are two simultaneous linear equations in A and B.
This method avoids calculating the quotient Q when the question asks only for the remainder. It works because the divisor becomes zero at each chosen input, leaving precisely the remainder.
The roots must be distinct to give independent equations. When a quadratic has a repeated root, a second evaluation at that same root does not supply new information, and another technique is needed.
Worked example: division by x² − 1
Let P(x) = x³ + 2x² + 4x + 5, and let the divisor be x² − 1 = (x − 1)(x + 1). Write the unknown remainder as Ax + B.
At x = 1, P(1) = 1 + 2 + 4 + 5 = 12. Therefore A + B = 12.
At x = −1, P(−1) = −1 + 2 − 4 + 5 = 2. Therefore −A + B = 2.
Subtract the second equation from the first: 2A = 10, giving A = 5. Then B = 7. The remainder is 5x + 7.
Notice that we did not find the quotient. The roots of the quadratic divisor provided just enough information to recover the two unknown remainder coefficients.
Check the example through polynomial division
Divide x³ + 2x² + 4x + 5 by x² − 1. The leading quotient term is x, because x³ divided by x² is x. Subtract x(x² − 1) = x³ − x, leaving 2x² + 5x + 5.
The next quotient term is +2. Subtract 2(x² − 1) = 2x² − 2, leaving 5x + 7.
Thus P(x) = (x² − 1)(x + 2) + 5x + 7. The remainder from long division agrees with the root-evaluation method.
This identity is a complete algebraic verification. Expanding the right side yields x³ + 2x² − x − 2 + 5x + 7 = x³ + 2x² + 4x + 5, exactly the original polynomial.
Learning to reconcile both approaches makes the remainder concept more reliable than memorising an evaluation shortcut alone.
A different divisor produces different evaluation points
Suppose P(x) = 2x³ − 3x² + 4x − 5 is divided by x² − 4. The divisor factors as (x − 2)(x + 2), so the relevant inputs are 2 and −2.
Let the remainder be Ax + B. Evaluate P(2) = 16 − 12 + 8 − 5 = 7, giving 2A + B = 7.
Evaluate P(−2) = −16 − 12 − 8 − 5 = −41, giving −2A + B = −41. Subtracting gives 4A = 48, so A = 12. Then B = −17.
Therefore the remainder is 12x − 17. A check by division gives P(x) = (x² − 4)(2x − 3) + 12x − 17. The degree of the remainder is one, as required.
Use the general formula for a divisor with two roots
Suppose the quadratic divisor factors as (x − r)(x − s), with r ≠ s, and P has unknown remainder Ax + B. The conditions are Ar + B = P(r) and As + B = P(s).
Subtract to get A(r − s) = P(r) − P(s). Therefore A = [P(r) − P(s)]/(r − s). Once A is known, B = P(r) − Ar.
This gives a useful interpretation: the linear remainder is the unique straight-line expression whose values agree with P at r and s. It is a two-point interpolation problem.
Students need not memorise the formula in every school task. Writing the two evaluation equations is usually transparent enough. The formula explains why distinct roots determine a unique remainder and why repeated roots require a different idea.
Why a remainder can be constant even under a quadratic divisor
A quadratic divisor allows a remainder of degree at most one, but it does not require an x term. Sometimes A = 0.
Consider P(x) = x⁴ − 2x² + 3 and D(x) = x² − 1. Observe that P(x) = (x² − 1)² + 2.
Therefore the quotient is x² − 1 and the remainder is the constant 2. It is a valid quadratic-divisor remainder because degree zero is less than degree two.
Evaluating at the roots confirms P(1) = 2 and P(−1) = 2. An unknown remainder Ax + B satisfying A + B = 2 and −A + B = 2 gives A = 0 and B = 2.
The correct general form provides enough freedom for the answer, even when one coefficient turns out to vanish.
The zero remainder means exact divisibility
If both remainder coefficients vanish, the polynomial is exactly divisible by the quadratic. For example P(x) = x³ − x can be factored as x(x² − 1).
Dividing P by x² − 1 gives quotient x and remainder zero. Root evaluation agrees: P(1) = 0 and P(−1) = 0, so the linear remainder must satisfy A + B = 0 and −A + B = 0, yielding A = B = 0.
This connects directly with the factor theorem. Both distinct roots of the quadratic divisor are roots of P, so their linear factors divide P. In this example the product of the factors is the entire quadratic divisor.
Do not confuse a zero remainder with an undefined quotient. A zero remainder means division finishes exactly, with no leftover polynomial term.
A repeated quadratic root requires more than one value
Consider division by (x − 1)². Its only distinct root is x = 1, repeated. If the unknown remainder is Ax + B, evaluating P at 1 gives only P(1) = A + B.
One equation cannot generally determine two unknowns. Writing P(1) twice does not create a second independent condition.
A reliable route is ordinary polynomial division. Another useful extension uses differentiation: if P(x) = (x − 1)²Q(x) + Ax + B, then P′(1) = A, because the derivative of the squared-factor product is zero at x = 1.
Thus the derivative supplies the second condition when that method is available and appropriate to the learner’s syllabus.
Worked repeated-factor example
Let P(x) = x³ + 2x² + 4x + 5 and divide by (x − 1)² = x² − 2x + 1. Write P(x) = (x − 1)²Q(x) + Ax + B.
First evaluate P(1) = 12, giving A + B = 12. Differentiate P′(x) = 3x² + 4x + 4, so P′(1) = 11. Hence A = 11 and B = 1.
The remainder is 11x + 1. Check with division: (x² − 2x + 1)(x + 4) = x³ + 2x² − 7x + 4. Subtracting from P leaves 11x + 1.
This worked example illustrates both why repeated roots are different and how the derivative can provide missing independent information. It is an optional extension where polynomial derivatives are part of the student’s taught material.
When the quadratic has no real roots
A divisor such as x² + 1 has no real zeros. The method of evaluating at two real roots is therefore unavailable. That does not mean polynomial division is impossible.
Use long division or polynomial identities instead. For P(x) = x³ + 2x² + 4x + 5, divide by x² + 1. The quotient is x + 2, because (x² + 1)(x + 2) = x³ + 2x² + x + 2.
Subtract to obtain the remainder 3x + 3. Thus P(x) = (x² + 1)(x + 2) + 3x + 3.
This route stays entirely within real polynomial algebra. It avoids unnecessarily introducing complex roots when the school task asks only for a real polynomial quotient and remainder.
A shortcut using equivalence modulo the divisor
When dividing by x² + 1, the relation x² + 1 has zero remainder. In computations of the remainder, we may replace x² by −1, because their difference is divisible by x² + 1.
For P(x) = x³ + 2x² + 4x + 5, this means x³ = x·x² behaves like −x, while 2x² behaves like −2 in the remainder calculation.
The expression reduces to −x − 2 + 4x + 5 = 3x + 3, exactly the remainder found by division.
This notation is a polynomial-remainder equivalence, not a claim that x² equals −1 for ordinary real x. The distinction is important. We are replacing polynomial expressions by others with the same remainder under a fixed divisor, not solving x² = −1 as a real equation.
For x² − c, even and odd powers simplify differently
If the divisor is x² − c, where c is a constant, then x² and c have the same remainder. Therefore x³ and cx have the same remainder, and x⁴ and c² have the same remainder.
For example, under division by x² − 4, we can replace x² by 4, x³ by 4x and x⁴ by 16 when calculating remainders.
This often turns a higher-degree polynomial into a linear expression without completing a long division. It is especially convenient for expressions containing only a few powers.
The shortcut should still be verified through the polynomial identity or direct division if the result is uncertain. It is an efficient representation of the same algebra, not a separate theorem that overrides the degree rule.
A fourth-degree example without full division
Find the remainder when P(x) = x⁴ + 3x³ − 2x² + 5x − 1 is divided by x² − 4.
For remainder purposes, x² is equivalent to 4, x³ to 4x and x⁴ to 16. Therefore P is equivalent to 16 + 3(4x) − 2(4) + 5x − 1.
Collect like terms: the remainder is 17x + 7. To check, evaluate at x = 2: P(2) = 16 + 24 − 8 + 10 − 1 = 41, and the remainder gives 17(2) + 7 = 41.
At x = −2, P(−2) = 16 − 24 − 8 − 10 − 1 = −27, and the remainder gives 17(−2) + 7 = −27. Both divisor roots agree, confirming the linear remainder.
The quotient is optional only when the question allows it
If a task says “find the remainder”, evaluating at roots or using polynomial equivalences can avoid finding the quotient. But if it asks for the complete quotient and remainder, those shortcuts alone do not supply all the requested information.
For the opening example, knowing the remainder is 5x + 7 does not tell us the quotient until we divide or reason further. The full identity is P(x) = (x² − 1)(x + 2) + 5x + 7.
The required output determines the most appropriate method. A short remainder calculation is efficient for a remainder-only question; polynomial division is necessary when the quotient must also be displayed.
Always return to the task wording before stopping. An accurate intermediate result is not a complete solution if a later requested quantity remains unreported.
A quotient–remainder identity is the strongest check
To verify a proposed quotient Q(x) and remainder R(x), multiply D(x)Q(x), then add R(x). The result should match the original polynomial coefficient by coefficient.
For P(x) = x³ + 2x² + 4x + 5, D(x) = x² − 1, Q(x) = x + 2 and R(x) = 5x + 7, the calculation is (x² − 1)(x + 2) + 5x + 7.
Expanding gives x³ + 2x² − x − 2 + 5x + 7 = x³ + 2x² + 4x + 5. Every term agrees.
This is stronger than a single numerical check, which can agree by coincidence. The identity proves the polynomial relationship for all real inputs and confirms that the remainder degree is within the allowed range.
Keep the sign correct when the divisor is x² − 4
The zeros of x² − 4 are x = 2 and x = −2. Both roots must be used to determine a general linear remainder.
Students sometimes use P(2) twice, or confuse the negative evaluation with substituting −2 into x² and obtaining −4. In fact (−2)² = 4, so both are genuine roots of the divisor.
Protect negative substitutions with brackets. For a cubic term, (−2)³ = −8; for a square term, (−2)² = +4. When calculating P(−2), alternating signs can change several terms.
A small two-row table with inputs 2 and −2, their polynomial values and equations 2A + B and −2A + B is enough to make this reasoning transparent.
What if the divisor includes a leading coefficient?
Suppose the divisor is 2x² − 2 rather than x² − 1. These divisors have the same roots, x = ±1, but different leading coefficients.
The remainder is unchanged when dividing a fixed polynomial by a non-zero constant multiple of a divisor: the quotient scales reciprocally, while the remainder with degree less than two remains the same.
For P(x) = x³ + 2x² + 4x + 5, division by 2x² − 2 still leaves 5x + 7. The quotient becomes (x + 2)/2, giving P(x) = (2x² − 2)[(x + 2)/2] + 5x + 7.
This demonstrates that the remainder is tied to the factors and roots of the divisor, while the quotient compensates for the divisor’s scaling.
The limit of the simple root method
A quadratic divisor does not always factor into two distinct real linear terms. It may have a repeated real root or no real root. Using P(r) at two different real zeros is only available in the distinct-real-root case.
For repeated roots, polynomial division works universally, and a derivative condition can be useful enrichment. For irreducible quadratics such as x² + 1 over the reals, direct division or reducing powers modulo the divisor is natural.
A student should therefore inspect the divisor first, identify the roots or factorisation if possible and choose a method appropriate to its structure.
This habit prevents a generalisation error: the theorem for a linear divisor is valid, but applying its constant-remainder form blindly to every higher-degree divisor is not.
A seven-step exam routine
First identify the degree of the divisor. Second write the most general remainder with lower degree. For a quadratic, use Ax + B.
Third examine whether the divisor has two distinct convenient roots. If so, evaluate P at each root. If not, choose polynomial division or another justified identity method. Fourth solve the resulting equations for A and B.
Fifth confirm the remainder has degree less than the divisor. Sixth check its values at the divisor roots when available. Seventh, if the quotient was also requested, provide it and verify P = DQ + R.
This is a short, reusable structure for unfamiliar polynomial questions. It replaces guesswork with a sequence of decisions grounded in the polynomial-division definition.
Five questions for independent practice
Question one: divide x³ + 2x² + 4x + 5 by x² − 1. Remainder: 5x + 7.
Question two: divide 2x³ − 3x² + 4x − 5 by x² − 4. Remainder: 12x − 17.
Question three: divide x⁴ − 2x² + 3 by x² − 1. Remainder: 2.
Question four: divide x³ + 2x² + 4x + 5 by x² + 1. Remainder: 3x + 3.
Question five: divide x³ + 2x² + 4x + 5 by (x − 1)². Remainder: 11x + 1.
Before reading the answers, classify each divisor as two distinct real roots, repeated root or no real roots. Then decide whether root evaluation, division or power reduction is the clearest route.
Syllabus scope and related learning
The polynomial-division identity, factor theorem and remainder theorem connect to algebraic factorisation, unknown coefficient problems and partial-fraction preparation. The specific level of remainder work expected differs by school and examination pathway.
Check the relevant SEAB references for 2026 O-Level Additional Mathematics 4049, 2027 SEC G3 K341 or 2027 SEC G2 K232. Some repeated-root and polynomial-equivalence techniques are useful extensions rather than universal assessment requirements.
For connected reading, see How to Recover Polynomial Coefficients from Two Remainders, Why Polynomial Division Comes Before Improper Partial Fractions and How to Disprove a False Algebraic Rule with One Counterexample.
Further study guides are in the eduKateSG Additional Mathematics archive. Families may enquire about suitable eduKate small-group Mathematics tutorials near Sixth Avenue MRT. The aim is to understand which information about a divisor is enough to reconstruct its exact remainder. Properly taught kids shine a bright light into the future.