Learning Objectives
Write a polynomial division as dividend = divisor × quotient + remainder.
Divide by linear polynomials, including non-monic divisors.
Divide by quadratic polynomials and identify the permitted remainder degree.
Verify a division by multiplying back.
Prerequisites: Section 1B expansion, coefficient comparison and rational-expression restrictions. All practice questions in this lesson are original, not official TMUA questions.
What This Lesson Is About
Polynomial division rewrites a polynomial as a multiple of a chosen divisor plus a smaller remainder. This lesson explains what the quotient and remainder mean, how to calculate them, and how to check that division is complete.
We begin with degree, which determines the allowed remainder. We then practise the division algorithm with linear and quadratic divisors, including divisors whose leading coefficient is not 1.
1. Understanding the Division Identity
Before carrying out a division, identify the size of the divisor and the form the answer must take.
1.1 Degree and leading coefficient
The degree of a nonzero polynomial is the largest power with a nonzero coefficient. For example, 3x³ − x + 2 has degree 3. A polynomial with leading coefficient 1 is called monic; 2x − 1 is non-monic.
1.2 Dividend, divisor, quotient and remainder
For a nonzero polynomial divisor D(x), division gives P(x) = D(x)Q(x) + R(x). Here P is the dividend, Q is the quotient and R is the remainder. Either R is zero or its degree is smaller than the degree of D. The notation P(x) means the value of the polynomial at x.
1.3 How degree determines when to stop
A linear divisor leaves a constant remainder. A quadratic divisor leaves a remainder of the form ax + b, including a constant or zero as special cases. A remainder of equal or higher degree means division is not finished.
1.4 An identity and a quotient have different domains
The polynomial identity holds for every real x. The fraction form P(x)/D(x) = Q(x) + R(x)/D(x) requires D(x) ≠ 0. Exact division does not remove that restriction from an original quotient.
Worked Example: Check the Form First
Question: A proposed division by x² + 1 has remainder 3x² − 2. Can it be finished?
Method: Compare the degree of the proposed remainder with the divisor. If they are equal, transfer another multiple of the divisor into the quotient.
No. The remainder still has degree 2, the same as the divisor. Subtract 3(x² + 1) from this remainder to leave −5, adding 3 to the quotient.
The new remainder is constant and has degree below 2.
2. Carrying Out Polynomial Long Division
The division identity gives the target form. Long division constructs the quotient one term at a time by removing the largest remaining power.
2.1 Setting out missing powers and choosing the next term
Arrange powers in descending order and insert zero coefficients for any missing powers. Divide the leading term of the current remainder by the leading term of the divisor. Add that term to the quotient, multiply the entire divisor by it, and subtract the whole product.
2.2 Subtracting a whole product and repeating
Each subtraction removes the highest remaining power. Repeat until the remainder has smaller degree than the divisor. Brackets around the product being subtracted prevent sign errors.
Worked Example: A Monic Linear Divisor
Question: Divide x³ − 2x² − 5x + 6 by x − 1.
Method: Divide the leading terms, multiply the whole divisor, and subtract. Repeat until no power of x remains in the remainder.
First x³/x = x². Subtract x²(x − 1) = x³ − x² to leave −x² − 5x + 6.
Next −x²/x = −x. Subtract −x(x − 1) = −x² + x to leave −6x + 6.
Next −6x/x = −6. Subtract −6(x − 1) = −6x + 6 to leave 0. The quotient is x² − x − 6 and the remainder is 0.
Check: (x − 1)(x² − x − 6) = x³ − 2x² − 5x + 6.
3. Dividing by a Non-Monic Linear Polynomial
The same procedure applies when the divisor begins with a coefficient other than 1. That coefficient must be included whenever you choose a quotient term.
3.1 Using the actual leading term
With a non-monic divisor, divide by its actual leading term. Dividing 6x³ by 2x gives 3x², not 6x². Fractional coefficients in a quotient or remainder are valid; do not force integers where the calculation does not give them.
Worked Example: A Non-Monic Linear Divisor
Question: Divide 6x³ + x² − 4x + 5 by 2x − 1.
Method: At each stage divide by 2x, the leading term of the divisor. Check the result by multiplying the quotient by 2x − 1 and adding the remainder.
The first quotient term is 3x². Subtract 6x³ − 3x² to leave 4x² − 4x + 5.
The next term is 2x. Subtract 4x² − 2x to leave −2x + 5. The next term is −1. Subtract −2x + 1 to leave 4.
Thus Q(x) = 3x² + 2x − 1 and R(x) = 4. Multiplying (2x − 1)(3x² + 2x − 1) and adding 4 recovers every original coefficient.
4. Dividing by a Quadratic Polynomial
A quadratic divisor changes the stopping point: a linear remainder is already small enough. Use the same leading-term calculation, then check the degree of what remains.
4.1 Keeping missing powers and accepting a linear remainder
The same algorithm works for a quadratic divisor. Stop when the remaining expression has degree at most 1; there is no need to force the remainder to be constant. Missing powers in the dividend should keep their place throughout the subtraction.
Worked Example: A Missing Cubic Term
Question: Divide x⁴ + 2x² − x + 3 by x² + 1.
Method: Write the missing cubic coefficient as zero. Stop as soon as the remainder has degree below 2.
Write the dividend as x⁴ + 0x³ + 2x² − x + 3. The first quotient term is x². Subtract x⁴ + x² to leave x² − x + 3.
The next quotient term is 1. Subtract x² + 1 to leave −x + 2. Stop because this has degree 1.
The identity is x⁴ + 2x² − x + 3 = (x² + 1)(x² + 1) − x + 2. Expanding the square verifies it.
Worked Example: A Non-Monic Quadratic Divisor
Question: Divide 2x³ + 3x² + 4x + 5 by 2x² + 1.
Method: Divide by the leading term 2x² and allow fractional coefficients. A linear remainder is a completed answer.
The first term is x. Subtract x(2x² + 1) to leave 3x² + 3x + 5. The next term is 3/2. Subtract (3/2)(2x² + 1) to leave 3x + 7/2.
Therefore Q(x) = x + 3/2 and R(x) = 3x + 7/2. Multiplying the divisor and quotient gives 2x³ + 3x² + x + 3/2; adding the remainder recovers 2x³ + 3x² + 4x + 5.
Common Mistakes
Use zero placeholders rather than letting a missing power shift the remaining coefficients.
Subtract every term of the product, including negative terms.
Divide by the whole leading term of a non-monic divisor.
Check both the reconstruction identity and the degree of the remainder.
5. Practice
For each division, state both the quotient and remainder, then check by reconstruction. The final questions test when division is complete and why simplifying a fraction preserves its original restriction.
1. Divide x² + 5x + 7 by x + 2.
2. Divide 2x² + 3x − 2 by 2x − 1.
3. Divide x³ + 1 by x + 1.
4. Divide x³ + 2x² + 3x + 4 by x² + 1.
5. Divide 2x³ − x² + 3x + 2 by 2x² + 1.
6. A learner writes x³ = (x² + 1)x − x. Is this a completed division by x² + 1? Is the remainder allowed to be negative?
7. Simplify (x³ − 1)/(x − 1) by division and state the original domain.
Worked Solutions
1. A linear quotient with a constant remainder
Q = x + 3, R = 1. Check: (x + 2)(x + 3) + 1 = x² + 5x + 7. A constant remainder is permitted for a linear divisor.
2. Exact division by a non-monic divisor
Q = x + 2, R = 0. Check: (2x − 1)(x + 2) = 2x² + 3x − 2.
3. Keeping the missing powers in a cubic
Keep the zero coefficients of x² and x. Division gives Q = x² − x + 1, R = 0. Check: (x + 1)(x² − x + 1) = x³ + 1.
4. A quadratic divisor with a linear remainder
The first term x leaves 2x² + 2x + 4; the next term 2 leaves 2x + 2. Thus Q = x + 2 and R = 2x + 2. Reconstruction gives x³ + 2x² + 3x + 4.
5. Fractional coefficients in the quotient
The first term x leaves −x² + 2x + 2; the next term −1/2 leaves 2x + 5/2. Thus Q = x − 1/2 and R = 2x + 5/2. Check: (2x² + 1)(x − 1/2) + 2x + 5/2 = 2x³ − x² + 3x + 2.
6. Recognising a completed division
Yes. The identity expands to x³ + x − x = x³, and the remainder −x has degree 1, below the divisor degree 2. Polynomial remainders need not be positive.
7. Simplifying a quotient while keeping its restriction
Division gives quotient x² + x + 1 and remainder 0, so the fraction equals x² + x + 1 for x ≠ 1. Multiplication by x − 1 reconstructs x³ − 1; the original quotient is still undefined at 1.
6. Summary
- Write P = DQ + R and keep the degree of R below that of D.
- Cancel one leading term at a time, keeping missing powers and subtraction signs organised.
- Non-monic divisors can give fractional coefficients.
- Multiply back to verify the result and retain restrictions in a quotient expression.