Learning Objectives
Find the remainder for a linear divisor by evaluating a polynomial.
Use a zero remainder to find a factor and solve a factorable polynomial equation.
Determine unknown coefficients from factor or remainder conditions.
Find a linear remainder for a quadratic with two distinct known roots.
Prerequisites: Section 1B polynomial division, simultaneous equations and the zero-product rule. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
The remainder and factor theorems connect polynomial division to substitution. They let you find a remainder or test a factor without carrying out a full division.
This lesson derives that connection, uses it to find unknown coefficients, and then extends the idea to a quadratic divisor with two distinct roots.
1. Finding a Remainder by Substitution
Start from the division identity. Choosing an input that makes the divisor zero removes the quotient term and exposes the remainder.
1.1 Deriving the remainder theorem
Division by x − a gives P(x) = (x − a)Q(x) + r, where the remainder r is a constant. Put x = a into this polynomial identity: the quotient term vanishes, leaving P(a) = r. This is the remainder theorem.
1.2 Finding the zero of a non-monic divisor
The substitution is into the polynomial identity, not into a quotient with zero denominator. P(a) is defined even though P(x)/(x − a) is not defined at a. For a divisor ux + v with u ≠ 0, the zero is −v/u, so the remainder is P(−v/u).
Worked Example: Use the Zero of the Divisor
Question: Find the remainder when 2x³ − 3x + 1 is divided by 2x − 1.
Method: Set the divisor equal to zero and substitute that input into the dividend polynomial.
The divisor vanishes at x = 1/2. Evaluate P(1/2) = 2(1/8) − 3/2 + 1 = −1/4.
Check by division: P(x) = (2x − 1)(x² + x/2 − 5/4) − 1/4. Expanding recovers 2x³ − 3x + 1. Substituting x = 1 would use the wrong divisor zero.
2. Recognising Factors and Finding Roots
A remainder of zero means the division is exact. This turns the remainder theorem into a test for a linear factor.
2.1 The factor theorem works in both directions
The factor theorem says that x − a is a factor of P(x) exactly when P(a) = 0. One direction follows from substitution into P(x) = (x − a)Q(x). The other follows because the remainder theorem then gives a zero remainder.
2.2 Continuing from one factor to all roots
After finding one factor, divide to obtain the quotient and factor that if possible. To solve P(x) = 0, set each factor equal to zero. This gives all roots once the factorisation is complete.
2.3 Understanding the limits of trial substitutions
Testing small integers can help locate a factor, but unsuccessful trials do not prove that no real roots exist. Use information given by the problem; general quadratic methods will be developed next section.
Worked Example: From One Root to a Complete Factorisation
Question: Factorise P(x) = x³ − 3x² − 4x + 12 and solve P(x) = 0.
Method: Find one zero, divide out its linear factor, then factor the remaining quadratic to obtain all roots.
P(3) = 27 − 27 − 12 + 12 = 0, so x − 3 is a factor. Division gives x² − 4, which is (x − 2)(x + 2).
Thus P(x) = (x − 3)(x − 2)(x + 2), giving roots 3, 2 and −2. Multiplying (x − 3)(x² − 4) reconstructs the original, checking the full factorisation.
3. Finding Unknown Coefficients
A stated factor or remainder provides an equation involving the coefficients. Several conditions can therefore determine several unknowns.
3.1 Translating each condition into an equation
An unknown coefficient does not stop you using the theorem. A factor condition gives an equation with right side zero; a stated remainder gives that remainder as the right side. Two independent conditions can determine two coefficients using simultaneous equations.
3.2 Choosing the input with the correct sign
Keep the signs in the divisor straight. The factor x + 1 means evaluate at −1, because x + 1 = x − (−1).
Worked Example: One Factor and One Remainder
Question: Let P(x) = x³ + ax + b. It has factor x − 1 and remainder 4 on division by x + 1. Find a and b.
Method: Write one equation for the factor condition and another for the stated remainder, then solve them together.
P(1) = 0 gives 1 + a + b = 0. P(−1) = 4 gives −1 − a + b = 4. Add the equations: 2b = 4, so b = 2; then a = −3.
Check both conditions: P(1) = 1 − 3 + 2 = 0 and P(−1) = −1 + 3 + 2 = 4.
4. Finding Remainders for Quadratic Divisors
A quadratic divisor can leave a linear remainder, so one value is usually insufficient. Two distinct zeros of the divisor provide two equations for that remainder.
4.1 Determining the two coefficients of a linear remainder
For division by (x − a)(x − b), with a ≠ b, the remainder has form mx + c. Evaluate the division identity at a and b to get P(a) = ma + c and P(b) = mb + c. Solving these determines the remainder without finding the full quotient.
4.2 Why a repeated zero supplies only one substitution condition
The two zeros must be distinct for this two-equation method. For a repeated factor such as (x − a)², evaluating at a twice gives only one condition. Divide directly, or use other information in the question.
Worked Example: A Quadratic Remainder
Question: P(1) = 3 and P(2) = 5. Find the remainder when P is divided by (x − 1)(x − 2).
Method: Write the remainder as mx + c and use the two given values to form simultaneous equations.
Write the remainder as mx + c. The conditions are m + c = 3 and 2m + c = 5, so m = 2 and c = 1.
The remainder is 2x + 1, which has degree below 2 and takes the required values 3 and 5. Its uniqueness follows because two distinct linear equations determine m and c.
Worked Example: A Root Need Not Be Repeated
Question: Does P(1) = 0 imply that (x − 1)² is a factor?
Method: Divide out the known factor and test whether the quotient still has the same root.
No. For P(x) = x² − 1, P(1) = 0, but division by x − 1 gives x + 1. This quotient is 2 at x = 1, so it is not divisible by x − 1 again.
A single zero test proves one factor only. If the quotient also vanishes there, a second use of the factor theorem proves a repeated factor.
Common Mistakes
Use the root of the divisor, not the constant written in it.
A nonzero remainder is valid information; it means the divisor is not a factor.
After finding one factor, inspect the quotient before claiming all roots are known.
Repeated factors need an extra check; evaluating twice at the same input gives no new information.
5. Practice
Choose the input that makes each divisor zero before calculating. For quadratic remainders, state the general linear form first; for claims about roots, explain what the given substitutions do and do not establish.
1. Find the remainder when x³ − 2x + 5 is divided by x + 2.
2. Find the remainder when 2x³ − x + 4 is divided by 2x − 1.
3. Find k if x − 2 is a factor of x³ + kx² − 4.
4. Factorise x³ + x² − 4x − 4 completely and list its real roots.
5. Find a and b if x − 2 and x + 1 are factors of x³ + ax + b.
6. P has remainder 4 on division by x − 1 and remainder 1 on division by x + 2. Find its remainder on division by (x − 1)(x + 2).
7. A learner argues that P(0) = P(1) = 0 proves P is the zero polynomial. Disprove the claim and state what it does imply.
Worked Solutions
1. Remainder for a divisor with a negative zero
Evaluate at −2: −8 + 4 + 5 = 1. As a check, P(x) = (x + 2)(x² − 2x + 2) + 1.
2. Remainder for a non-monic divisor
Evaluate at 1/2: 1/4 − 1/2 + 4 = 15/4. Check: P(x) = (2x − 1)(x² + x/2 − 1/4) + 15/4.
3. Finding a coefficient from a factor condition
P(2) = 8 + 4k − 4 = 0, so k = −1. Check: x³ − x² − 4 = (x − 2)(x² + x + 2).
4. Completing a cubic factorisation
Group x²(x + 1) − 4(x + 1) = (x + 1)(x² − 4) = (x + 1)(x − 2)(x + 2). The roots are −1, 2 and −2. Expansion recovers x³ + x² − 4x − 4.
5. Using two factor conditions together
P(2) = 0 gives 8 + 2a + b = 0; P(−1) = 0 gives −1 − a + b = 0. Subtract to obtain 9 + 3a = 0, so a = −3 and b = −2. Check: x³ − 3x − 2 = (x − 2)(x + 1)², containing both required factors.
6. Determining a linear remainder
Write R(x) = mx + c. Then m + c = 4 and −2m + c = 1. Subtraction gives 3m = 3, so m = 1 and c = 3. R(x) = x + 3 has the correct values 4 and 1 and degree below 2.
7. Distinguishing two roots from an identity
P(x) = x(x − 1) is zero at 0 and 1 but has P(2) = 2, so it is not the zero polynomial. The conditions do imply factors x and x − 1, and hence their product divides P. To see the product, write P = xQ; then P(1) = Q(1) = 0, so Q has factor x − 1.
6. Summary
- A linear divisor’s zero gives the constant remainder by substitution.
- A zero remainder proves a factor, and division reveals what remains.
- Turn stated factors and remainders into equations for unknown coefficients.
- Use two distinct divisor zeros for a linear remainder; check repeated factors separately.