MathematicsCore18 min read

Division of Polynomials

Long division, and the two theorems that make most of it unnecessary

This topic appears in:

1st Year Mathematics · ch.9A Level Mathematics (9709) · ch.2A Level Mathematics (9709) · ch.3
01

Dividing polynomials is dividing numbers, with letters

When 17 is divided by 5 you get a quotient of 3 and a remainder of 2, and the whole statement is 17 = 5 × 3 + 2. Polynomials behave identically, with one extra condition: the remainder must be of lower degree than the divisor, which is the analogue of a remainder being smaller than the divisor.

So dividing by a linear divisor (x − a) always leaves a remainder that is a constant, because the only degree below 1 is 0. That single observation is what makes the remainder theorem possible.

f(x) = (divisor) × (quotient) + remainderdeg(remainder) < deg(divisor)the division algorithm — quote it whenever a question says "hence" or "write in the form"

Leave gaps for the missing powers

Dividing x³ − 7 by (x − 2), write the dividend as x³ + 0x² + 0x − 7. Long division lines up by degree, and a missing power that is not written as a zero throws every subsequent column out. This is by far the commonest source of a wrong quotient.

02

The remainder theorem

Long division to find a remainder is honest work, and almost always unnecessary. Write the division statement f(x) = (x − a)Q(x) + R and then substitute x = a. The whole quotient term vanishes because its bracket is zero, and you are left with f(a) = R.

So the remainder on dividing by (x − a) is simply the value of the polynomial at a — one substitution instead of a page of division.

f(x) ÷ (x − a) leaves remainder f(a)f(x) ÷ (ax − b) leaves remainder f(b/a)the second form catches out anyone who substitutes the wrong number: use the root of the divisor, not its coefficients

The dashed line shows f(a), and that number is the remainder. Drag a onto a point where the curve crosses the axis: the remainder becomes zero, and there the divisor is an exact factor.

03

The factor theorem

A remainder of zero means the division was exact — the divisor goes in a whole number of times, with nothing left over. So (x − a) is a factor of f(x) exactly when f(a) = 0. That is the factor theorem, and it is the remainder theorem in the special case R = 0.

It works in both directions, which is what makes it useful. Given a factor, you get a root for free. Given a root, you get a factor and can divide it out to reduce the degree of the problem.

Worked example

Factorise f(x) = x³ − 6x² + 11x − 6 completely.

  1. Try small factors of the constant term −6: the candidates are ±1, ±2, ±3, ±6.Any whole-number root must divide the constant term, so this short list is guaranteed to contain them.
  2. f(1) = 1 − 6 + 11 − 6 = 0, so (x − 1) is a factor.Start with x = 1; it is the fastest to evaluate and very often works in exam questions.
  3. Divide: x³ − 6x² + 11x − 6 ÷ (x − 1) = x² − 5x + 6.Long division or synthetic division — either is accepted, and the quotient is one degree lower.
  4. Factorise the quadratic: x² − 5x + 6 = (x − 2)(x − 3).Two numbers multiplying to 6 and adding to −5.
  5. f(x) = (x − 1)(x − 2)(x − 3).Check by expanding, or by confirming f(2) = 0 and f(3) = 0.

(x − 1)(x − 2)(x − 3)

04

Synthetic division

When the divisor is (x − a), the long division can be compressed into a single row of arithmetic. Write the coefficients of f(x) — including zeros for missing powers — put a to the left, bring down the first coefficient, then repeatedly multiply by a and add.

The numbers you generate are the coefficients of the quotient, and the last one is the remainder. It is faster than long division, less prone to slips, and perfectly acceptable in an exam provided the working is shown.

Worked example

Use synthetic division to divide 2x³ − 3x + 5 by (x + 2).

  1. Coefficients, with the missing x² written as zero: 2, 0, −3, 5.The gap must be filled or every column afterwards is wrong.
  2. The divisor is x + 2 = x − (−2), so a = −2.Use the root of the divisor, which is the sign opposite to the one written.
  3. Bring down 2. Then 2 × (−2) = −4, add to 0 → −4. Then −4 × (−2) = 8, add to −3 → 5. Then 5 × (−2) = −10, add to 5 → −5.Multiply by a, add to the next coefficient, repeat.
  4. Quotient 2x² − 4x + 5, remainder −5.Check with the remainder theorem: f(−2) = −16 + 6 + 5 = −5 ✓

Quotient 2x² − 4x + 5, remainder −5

Two conditions at once

A question giving two pieces of information — "when f(x) is divided by (x − 1) the remainder is 4, and (x + 2) is a factor" — is handing you two equations: f(1) = 4 and f(−2) = 0. Substitute both, and solve the pair simultaneously for the two unknown coefficients. That is the standard long question in this chapter.

Before you leave this chapter

  1. f(x) = divisor × quotient + remainder, with the remainder of lower degree than the divisor.
  2. Write zeros for missing powers before dividing.
  3. Remainder theorem: dividing by (x − a) leaves f(a). For (ax − b), use f(b/a).
  4. Factor theorem: (x − a) is a factor exactly when f(a) = 0.
  5. To factorise a cubic, find one root among the factors of the constant term, divide it out, then factorise the quadratic.

Practice questions

6 questions · 20 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 6 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State the remainder theorem.
Model answer

When a polynomial f(x) is divided by (x − a), the remainder is f(a) — the value obtained by substituting x = a into the polynomial.

Examiner tip. Note the sign: dividing by (x + 3) means a = −3, so the remainder is f(−3). Substituting +3 is the standard error.

SQ2[2 marks]
Find the remainder when x³ + 2x² − 5x + 1 is divided by (x − 2).
Model answer

f(2) = 8 + 8 − 10 + 1 = 7.

Examiner tip. One substitution replaces the whole long division. Only divide out when the question actually asks for the quotient.

SQ3[2 marks]
Show that (x + 1) is a factor of x³ + 3x² + 3x + 1.
Model answer

f(−1) = −1 + 3 − 3 + 1 = 0. Since the remainder is zero, by the factor theorem (x + 1) is a factor. (In fact the polynomial is (x + 1)³.)

Examiner tip. Finish with the sentence naming the theorem. "f(−1) = 0" alone leaves the examiner to supply the conclusion, and it is worth a mark.

Solved numericals

2 · 8 marks

Full working, one step per line, with the marks shown where they are awarded.

N1[4 marks]
When f(x) = 2x³ + ax² + bx − 6 is divided by (x − 1) the remainder is −6, and (x + 2) is a factor. Find a and b.
Full working
  1. f(1) = 2 + a + b − 6 = −6, so a + b = −2remainder theorem[1]
  2. f(−2) = −16 + 4a − 2b − 6 = 0, so 4a − 2b = 22 and 2a − b = 11factor theorem: the remainder is zero[1]
  3. Adding the two equations eliminates b: 3a = 9, so a = 3the b terms cancel because their coefficients are +1 and −1[1]
  4. b = −2 − 3 = −5check: f(1) = 2 + 3 − 5 − 6 = −6 ✓ and f(−2) = −16 + 12 + 10 − 6 = 0 ✓[1]

a = 3, b = −5

Examiner tip. Two conditions give two equations. Set both up completely before solving either — mixing the substitution and the elimination is where sign errors creep in.

N2[4 marks]
Divide x³ − 4x² + x + 6 by (x − 3) and hence factorise the polynomial completely.
Full working
  1. f(3) = 27 − 36 + 3 + 6 = 0, so (x − 3) is a factorconfirming the division will be exact[1]
  2. Division gives the quotient x² − x − 2by long or synthetic division[1]
  3. x² − x − 2 = (x − 2)(x + 1)[1]
  4. f(x) = (x − 3)(x − 2)(x + 1)[1]

(x − 3)(x − 2)(x + 1)

Examiner tip. Check the constant term of your factorisation against the original: (−3)(−2)(1) = 6 ✓. That takes two seconds and catches a wrong sign in the quotient.

Long questions

1 · 6 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
The polynomial p(x) = x³ + px² + qx + 12 has (x − 2) and (x + 3) as factors.
  1. Form two equations in p and q.
  2. Solve them.
  3. Find the third factor of p(x).
Mark scheme
  1. p(2) = 8 + 4p + 2q + 12 = 0, so 4p + 2q = −20 and 2p + q = −10factor theorem[1]
  2. p(−3) = −27 + 9p − 3q + 12 = 0, so 9p − 3q = 15 and 3p − q = 5[1]
  3. Adding: 5p = −5, so p = −1the q terms cancel[1]
  4. q = −10 − 2(−1) = −8so p(x) = x³ − x² − 8x + 12[1]
  5. The product of all three roots is −12 (from −constant/leading coefficient), and two roots are 2 and −3or divide out both known factors[1]
  6. Third root = −12 / (2 × −3) = 2, so the third factor is (x − 2) and p(x) = (x − 2)²(x + 3)check: p(x) = (x−2)²(x+3) expands to x³ − x² − 8x + 12 ✓[1]

(a) 2p + q = −10 and 3p − q = 5 (b) p = −1, q = −8 (c) (x − 2), giving p(x) = (x − 2)²(x + 3)

Examiner tip. Using the product of the roots is quicker than a second long division. For x³ + ax² + bx + c the product of the roots is −c, which here is −12.