Additional Mathematics | How to Find a Remainder When the Divisor Is a Repeated Linear Factor

The ordinary Remainder Theorem says that the remainder when a polynomial P(x) is divided by x−a is P(a). But what if the divisor is (x−a)²? The remainder can now be linear rather than constant, and one function value is not enough. The missing information comes from differentiation.

For division by (x−a)², the exact remainder is R(x)=P(a)+P′(a)(x−a). For example, take P(x)=x⁴−3x³+2x²+5x+1 and divisor (x−2)². We obtain P(2)=11 and P′(2)=9, so R(x)=11+9(x−2)=9x−7. Indeed P(x)=(x−2)²(x²+x+2)+9x−7.

Did you know that this formula is closely related to the tangent line of a polynomial graph at x=a? This guide derives the result without long division, verifies it algebraically, explores repeated roots and parameter questions, and explains why a zero remainder for a squared divisor requires both P(a)=0 and P′(a)=0.

Recall what polynomial division guarantees

When a polynomial P(x) is divided by a nonzero polynomial D(x), it can be written uniquely as P(x)=D(x)Q(x)+R(x), where R is zero or has degree strictly less than D.

For a linear divisor x−a, the remainder must be a constant. For a quadratic divisor (x−a)², the remainder may be a linear polynomial.

This difference in allowed remainder degree is the starting point. A student who assumes that every remainder must be a single number has applied the ordinary linear Remainder Theorem outside its setting.

We want to find the coefficients of the linear remainder directly from a polynomial value and its derivative, avoiding unnecessary long division when the divisor has a repeated linear factor.

Write the correct remainder form

Let D(x)=(x−a)². Its degree is two, so the remainder has degree less than two.

Therefore write R(x)=mx+n, where m and n are constants to be determined. An equivalent and often more convenient form is R(x)=A+B(x−a).

The latter form records the remainder’s value and gradient at x=a directly: R(a)=A and R′(a)=B.

A good method chooses the form that makes the supplied divisor structure useful. Since the divisor vanishes to second order at x=a, centring the remainder at x=a makes the conditions especially simple.

Substitute x equal to the repeated factor root

Write P(x)=(x−a)²Q(x)+R(x). Substitute x=a into this division identity.

The product (a−a)²Q(a) equals zero. Therefore P(a)=R(a).

In the centred linear form R(x)=A+B(x−a), substituting x=a gives R(a)=A.

Thus A=P(a). This first condition is the familiar Remainder Theorem idea, but it identifies only the constant value of a linear remainder. One more independent condition is still required to determine B.

Differentiate the division identity

Differentiate P(x)=(x−a)²Q(x)+R(x) using the product rule.

We obtain P′(x)=2(x−a)Q(x)+(x−a)²Q′(x)+R′(x).

Now substitute x=a. Both terms containing x−a vanish, leaving P′(a)=R′(a).

Since R′(x)=B in the centred form, B=P′(a). Differentiating has supplied the missing remainder coefficient.

Combining both results gives R(x)=P(a)+P′(a)(x−a). This is an exact polynomial-division formula for the repeated linear divisor, not a numerical approximation.

State the derivative remainder formula clearly

For any polynomial P and any real a, the remainder on division by (x−a)² is

R(x)=P(a)+P′(a)(x−a).

Its degree is at most one, so it satisfies the remainder-degree requirement. It also matches both the original polynomial’s value and its first derivative at x=a.

The formula works whether or not a is a root of P. When P(a)=0, the remainder may still contain a nonzero linear term.

The key is to evaluate P(a) and P′(a) accurately. A sign error in a or in the derivative can change both coefficients, so substitute carefully and verify the result whenever a direct factorisation is possible.

Introduce the worked fourth-degree polynomial

Let P(x)=x⁴−3x³+2x²+5x+1. We want the remainder when P is divided by (x−2)².

The repeated factor root is a=2. The remainder will have form R(x)=P(2)+P′(2)(x−2).

There is no need to carry out fourth-degree polynomial long division initially. We need only one function evaluation and one derivative evaluation.

Keeping the divisor as (x−2)² rather than expanding it immediately makes the relevant point a=2 clear. The method is particularly efficient for high-degree polynomials where long division would create many opportunities for arithmetic errors.

Calculate P of two exactly

Substitute x=2 into P(x)=x⁴−3x³+2x²+5x+1.

The terms are 2⁴=16, −3(2³)=−24, 2(2²)=8, 5(2)=10 and the constant 1.

Add them: 16−24+8+10+1=11. Therefore P(2)=11.

This is the value the remainder must take at x=2. It does not yet determine the slope of the remainder line, but it fixes one of the two pieces of information needed for a linear remainder.

Differentiate and evaluate at two

Differentiate the polynomial term by term to obtain P′(x)=4x³−9x²+4x+5.

At x=2, this becomes 4(8)−9(4)+4(2)+5.

Calculate 32−36+8+5=9. Thus P′(2)=9.

This tells us that the remainder has gradient nine. The derivative is not the remainder itself; it is the coefficient of x−2 in the centred remainder form. The next step combines the value and slope into one exact linear expression.

Construct the remainder in centred form

Use R(x)=P(2)+P′(2)(x−2). Substitute the evaluated values.

Then R(x)=11+9(x−2). Expand to obtain R(x)=11+9x−18=9x−7.

Either form is correct. The centred form makes the value and gradient at x=2 clear; the expanded form may be preferred if the question asks for mx+n.

Check immediately: R(2)=18−7=11 and R′(x)=9. Both agree with P(2) and P′(2). This establishes the defining conditions for the polynomial-division remainder.

Verify the answer through exact factorisation

A stronger verification is to check that P(x)−R(x) really has (x−2)² as a factor.

Subtract 9x−7 from the original polynomial. The result is x⁴−3x³+2x²−4x+8.

This factors as (x−2)²(x²+x+2). Expand the product to confirm: (x²−4x+4)(x²+x+2) reproduces x⁴−3x³+2x²−4x+8.

Therefore P(x)=(x−2)²(x²+x+2)+9x−7. The quotient and remainder identity is exact, proving the result without relying only on matching two sample values.

Why the remainder is not just eleven

If the divisor were x−2, the remainder would indeed be the constant P(2)=11.

But the divisor (x−2)² has degree two, so a linear remainder is permitted. Here its slope is P′(2)=9, not zero.

Using R=11 would leave P(x)−11 with derivative P′(2)=9 at x=2. Therefore that difference cannot be divisible by (x−2)².

This is a precise diagnosis of the shortcut error: a constant remainder matches P at a=2 but not its derivative. A repeated divisor requires matching one additional local condition.

A second example with a cubic polynomial

Let P(x)=x³−2x²+4x+1 and divide by (x−1)².

Calculate P(1)=1−2+4+1=4. Differentiate: P′(x)=3x²−4x+4, so P′(1)=3−4+4=3.

The remainder is R(x)=4+3(x−1)=3x+1.

Check by subtraction: P(x)−(3x+1)=x³−2x²+x=x(x−1)². Thus P(x)=(x−1)²x+3x+1.

The quotient is simply x, and the linear remainder matches the derivative formula. This is an accessible example for students practising the principle before applying it to a quartic.

A root of P may still give a nonzero remainder

Consider P(x)=x²−5x+6=(x−2)(x−3), and divide by (x−2)².

Here P(2)=0 because x=2 is a root. But P′(x)=2x−5, so P′(2)=−1.

The remainder is R(x)=0−(x−2)=−x+2, not zero.

Indeed P(x)=(x−2)²−(x−2). The repeated divisor does not divide P exactly, because x−2 occurs only once as a factor. Knowing P(a)=0 proves one factor, not a double factor.

A zero remainder requires two conditions

The divisor (x−a)² divides P(x) exactly if and only if the remainder R(x)=P(a)+P′(a)(x−a) is identically zero.

That happens exactly when P(a)=0 and P′(a)=0.

Thus a is a repeated root of multiplicity at least two precisely when both the polynomial value and its first derivative vanish there.

This explains a familiar calculus connection: repeated polynomial roots are also stationary zeros of the polynomial graph, although whether the graph touches or crosses the axis depends on the exact multiplicity. The remainder formula makes the algebraic reason explicit.

Distinguish simple and repeated polynomial roots

For P(x)=(x−2)(x−3), the root at x=2 is simple. We saw P(2)=0 but P′(2)=−1, so the squared divisor fails to divide.

For Q(x)=(x−2)²(x+1), both Q(2)=0 and Q′(2)=0. Therefore its remainder on division by (x−2)² is zero.

A student who checks only Q(2)=0 has established a factor x−2 but not its multiplicity.

The derivative condition is an efficient test when the polynomial is given in expanded form and repeated factorisation would be cumbersome. It also connects algebraic multiplicity to the local slope of a graph.

The formula resembles a tangent line

The tangent to y=P(x) at x=a has equation y=P(a)+P′(a)(x−a).

That is exactly the same expression as the remainder on division by (x−a)².

This is not accidental. Subtracting the tangent line from the polynomial leaves a function whose value and first derivative both vanish at x=a. For a polynomial, that difference is divisible by (x−a)².

Thus the repeated-factor remainder is the first-degree polynomial that matches P’s local value and slope at the repeated root of the divisor. It is a beautiful connection between polynomial algebra and differential calculus.

The tangent is exact as a remainder, not a whole-curve approximation

A tangent line generally approximates a curved graph only near its contact point. But in this polynomial division problem, the tangent expression is the exact remainder in a global identity.

For the quartic example, P(x)=(x−2)²(x²+x+2)+9x−7 for every real x.

The line 9x−7 is the tangent at x=2, while the factorised remainder difference accounts for all other values.

This distinction matters. The tangent line is not equal to the original polynomial everywhere; rather, the original polynomial equals that tangent plus a term divisible by the squared factor. The division identity itself is exact.

Use the method to recover an unknown coefficient

Suppose P(x)=x³+kx²+3x+2, and we are told its remainder on division by (x−1)² is 7+8(x−1).

The remainder formula requires P(1)=7 and P′(1)=8.

Calculate P(1)=1+k+3+2=k+6. Thus k+6=7, giving k=1.

Differentiate P′(x)=3x²+2kx+3. At x=1, P′(1)=6+2k, which equals 8 when k=1.

Both conditions agree, so k=1 is a valid coefficient. The remainder information provides two local constraints, even though only one unknown k is present.

A conflicting remainder can prove no parameter exists

Take the same P(x)=x³+kx²+3x+2, but suppose someone claims the remainder is 7+5(x−1).

The value requirement P(1)=7 again forces k=1. But the derivative requirement P′(1)=5 gives 6+2k=5, or k=−1/2.

No single k can equal both 1 and −1/2. Therefore no parameter makes the polynomial have that claimed remainder.

A student should not use only the first remainder condition and accept k=1 without checking the second. Inconsistent data are a valid mathematical outcome, not a request to invent an average of conflicting answers.

Recover two unknown coefficients from one linear remainder

Let P(x)=x³+ax²+bx+2. Suppose its remainder on division by (x−1)² is 7+8(x−1).

The value condition P(1)=7 gives 1+a+b+2=7, so a+b=4.

The derivative is P′(x)=3x²+2ax+b. At x=1, the derivative condition P′(1)=8 gives 3+2a+b=8, so 2a+b=5.

Subtract the first equation from the second to get a=1. Then b=3. Thus P(x)=x³+x²+3x+2. Two independent remainder conditions have recovered two unknown coefficients exactly.

Check the two-coefficient reconstruction

For P(x)=x³+x²+3x+2, evaluate P(1)=1+1+3+2=7.

Differentiate to get P′(x)=3x²+2x+3, so P′(1)=3+2+3=8.

Therefore the remainder is 7+8(x−1)=8x−1.

A long division could verify the quotient, but these two equalities are enough because a linear polynomial is uniquely determined by its value and gradient at x=1.

This illustrates why the repeated divisor encodes two pieces of independent local information. It is not simply a more complicated way of asking for one function value.

The general coefficient form mx plus n

Some questions request the remainder as R(x)=mx+n rather than the centred form A+B(x−a).

The formula R(x)=P(a)+P′(a)(x−a) expands to R(x)=P′(a)x+[P(a)−aP′(a)].

Therefore m=P′(a) and n=P(a)−aP′(a).

For the quartic example, m=9 and n=11−2(9)=−7, giving R(x)=9x−7.

Both forms are algebraically identical. The centred form is usually simpler for deriving the result, while the expanded form may match the answer style expected on a worksheet.

A constant polynomial and a linear polynomial are special cases

If P(x) is constant, then P′(a)=0. The remainder on division by a squared linear factor is simply that constant, which has degree less than two.

If P(x)=mx+n is already linear, then P′(a)=m and P(a)=ma+n. The formula gives R(x)=ma+n+m(x−a)=mx+n, exactly the original polynomial.

That is correct: a polynomial of degree less than two is already its own remainder when divided by a degree-two divisor, with quotient zero.

These edge cases verify that the formula behaves properly without requiring the dividend degree to exceed the divisor degree.

A repeated factor of degree three needs more information

Suppose the divisor is (x−a)³. The remainder may now have degree up to two.

To determine it, match P(a), P′(a) and P″(a). The resulting formula is R(x)=P(a)+P′(a)(x−a)+[P″(a)/2](x−a)².

This is the next term of the polynomial’s Taylor expansion around a. The identity is exact as a remainder modulo the cubic repeated factor.

The squared-divisor formula is therefore the first nontrivial case of a wider pattern. Students should understand the degree requirement and derivative matching rather than memorising unrelated formulas for every power.

A worked cubic-divisor extension

Let P(x)=x³+2x²+x+1 and divide by (x−1)³.

Calculate P(1)=1+2+1+1=5. Differentiate: P′(x)=3x²+4x+1, so P′(1)=8.

Differentiate again: P″(x)=6x+4, so P″(1)=10.

The quadratic remainder is R(x)=5+8(x−1)+5(x−1)².

Expand if desired: 5+8x−8+5x²−10x+5=5x²−2x+2. Since P(x)−R(x)=(x−1)³, the quotient is one and the result is verified exactly.

Why a zero double-factor remainder is not enough for multiplicity three

If P(a)=0 and P′(a)=0, then (x−a)² divides P. But (x−a)³ may still fail to divide.

For example, P(x)=(x−2)² has P(2)=0 and P′(2)=0, yet its second derivative is P″(x)=2, so P″(2)≠0.

Thus the repeated factor has multiplicity exactly two, not three.

A third-order factor requires P(a)=P′(a)=P″(a)=0. The derivative conditions increase with the requested multiplicity, matching the number of coefficients of the allowed remainder polynomial.

Long division is still a valid verification

The derivative formula is efficient, but ordinary polynomial long division remains mathematically valid.

For the quartic example, dividing by x²−4x+4 gives quotient x²+x+2 and remainder 9x−7.

The quotient–remainder identity can be multiplied out to confirm that the result is correct.

In a timed examination, the derivative method usually requires less arithmetic when the divisor is exactly (x−a)². But if the divisor is a general irreducible quadratic not written as a repeated real linear factor, the same one-point derivative shortcut does not directly apply. Choose the method suited to the divisor’s structure.

Do not use the derivative shortcut for two distinct linear factors

A divisor such as (x−1)(x−3) has roots at two different inputs, not one repeated input.

Its remainder is linear, but the appropriate conditions are R(1)=P(1) and R(3)=P(3), obtained by substituting the two distinct roots.

There is no general requirement that R′(1)=P′(1), because the divisor has only a single factor x−1.

This is the key distinction. Repeated factors give value and derivative matching at one location; distinct factors give value matching at multiple locations. Confusing them can produce an incorrect remainder even when both divisors are quadratic.

A useful geometric interpretation of the difference

For the quartic example, the difference P(x)−R(x)=(x−2)²(x²+x+2).

The factor (x−2)² makes the difference vanish at x=2 and prevents it from changing sign due to that factor alone.

The second factor x²+x+2 is always positive because its discriminant 1−8 is negative and its leading coefficient is positive.

Therefore P(x)≥R(x) for all real x, with equality only at x=2. Here the remainder line is also a global supporting tangent beneath the quartic.

That inequality is a special property of this example, not a general rule about the remainder of every polynomial divided by a squared factor.

The remainder formula gives an easy computer-free check

Given a proposed remainder mx+n for divisor (x−a)², two numerical conditions establish whether it is correct.

Check ma+n=P(a), then check m=P′(a). A linear remainder satisfying both must equal the true remainder because its two coefficients are uniquely determined.

For the candidate 9x−7 with a=2, the value check is 18−7=11 and the gradient check is 9.

These match P(2) and P′(2). This small verification is often quicker than performing a full second polynomial division and catches copied coefficient errors immediately.

A concise examination routine

First identify the repeated root a from divisor (x−a)² and write a linear remainder R=P(a)+P′(a)(x−a).

Second evaluate the polynomial and its first derivative at a using exact arithmetic. Third substitute both values into the formula and simplify to the requested linear form.

If parameters are unknown, equate both the value and derivative conditions to the stated remainder and solve the resulting equations.

Finally check either the two matching conditions or the exact factorisation P−R=(x−a)²Q. Remember that a zero remainder requires both P(a)=0 and P′(a)=0.

Practice questions with checked answers

Question one: find the remainder of x⁴−3x³+2x²+5x+1 divided by (x−2)². Answer: 9x−7.

Question two: verify the quotient in Question one. Answer: x²+x+2.

Question three: find the remainder of x³−2x²+4x+1 divided by (x−1)². Answer: 3x+1.

Question four: find the remainder of x²−5x+6 divided by (x−2)². Answer: −x+2.

Question five: state when (x−a)² divides P exactly. Answer: P(a)=0 and P′(a)=0.

Question six: for P=x³+kx²+3x+2, find k if the remainder modulo (x−1)² is 7+8(x−1). Answer: k=1.

Question seven: find a,b for P=x³+ax²+bx+2 when the remainder modulo (x−1)² is 7+8(x−1). Answer: a=1,b=3.

For each, show both the function-value and derivative checks.

The formula can be derived from a local polynomial expansion

Expand P(x) around x=a in powers of x−a. Every polynomial can be written exactly as P(x)=c₀+c₁(x−a)+c₂(x−a)²+… .

Substituting x=a gives c₀=P(a). Differentiating and then evaluating at a gives c₁=P′(a), since all higher-power derivative terms still contain a factor x−a and vanish there.

When dividing by (x−a)², every term beginning with c₂(x−a)² belongs to the divisible quotient part. Only c₀+c₁(x−a) remains.

Thus the remainder is P(a)+P′(a)(x−a). This derivation reaches the same formula through polynomial expansion and makes its connection to Taylor coefficients explicit.

A simple repeated-root counterexample using a cubic

Consider Q(x)=(x−1)²(x+3). The root at x=1 has multiplicity exactly two.

Because Q already contains (x−1)², its remainder on division by that squared factor is zero. Both Q(1) and Q′(1) vanish.

But the graph does not vanish identically; away from x=1, the quotient x+3 is generally nonzero. This reminds students that zero remainder means exact divisibility, not that the entire polynomial must be zero.

For division by (x−1)³, the remainder is not zero, because the repeated factor has insufficient multiplicity. A zero remainder depends on the specific divisor’s power.

A derivative-based remainder check can reveal an inconsistent question

Suppose a worksheet claims that P(x)=x³−2x²+4x+1 leaves remainder 2x+2 when divided by (x−1)².

The proposed remainder at x=1 is four, agreeing with P(1)=4. But its derivative is two, whereas P′(1)=3.

Therefore the proposed remainder cannot be correct. The actual one is 3x+1, which matches both the value and slope.

This two-condition check is especially useful when assessing a partially completed solution. Matching one value is necessary but insufficient for a repeated divisor; the derivative comparison pinpoints the missing requirement.

Repeated-factor remainders connect the polynomial Remainder Theorem, factor multiplicity, differentiation and tangent-line approximation. Some higher-power divisor extensions are enrichment, so they should be matched to the learner’s assessment level.

For Singapore, refer to the applicable SEAB syllabus lists for 2026 O-Level Additional Mathematics or 2027 SEC G3, as relevant.

Families can enquire about suitable eduKate Singapore three-student Mathematics tutorials near Sixth Avenue MRT. The enduring insight is that a double factor requires matching both the function’s value and its slope, so a derivative can replace a much longer polynomial division. Properly taught kids shine a bright light into the future.