Lesson 4: Polynomial Division and the Remainder Theorem

Dividing polynomials looks like bookkeeping, but it powers three big ideas: it factors polynomials, it evaluates them quickly, and — via the Remainder and Factor Theorems — it finds their roots. This lesson practises synthetic division until it is fast, then uses the theorems to dismantle cubics completely, including cubics that model real objects.

Long division and synthetic division

Dividing p(x) by (x − c) produces a quotient q(x) and a constant remainder R with p(x) = (x − c)q(x) + R. Synthetic division compresses the arithmetic into the coefficients alone. Two warnings before the mechanics: include a 0 coefficient for every missing power, and the divisor must be in the form (x − c) — dividing by (x + 3) means c = −3.

Worked Example 1 — synthetic division. Divide p(x) = x^3 − 2x^2 − 5x + 6 by (x − 1).

Coefficients:  1   −2   −5    6        c = 1
Bring down 1; multiply by 1 → 1; add: −2 + 1 = −1
Multiply by 1 → −1; add: −5 + (−1) = −6
Multiply by 1 → −6; add: 6 + (−6) = 0
Quotient: x^2 − x − 6, remainder 0

Remainder 0 means (x − 1) is a factor, and the quotient factors further: x^2 − x − 6 = (x − 3)(x + 2). So p(x) = (x − 1)(x − 3)(x + 2), with roots 1, 3 and −2. Verify numerically: p(3) = 27 − 18 − 15 + 6 = 0 ✓ and p(−2) = −8 − 8 + 10 + 6 = 0 ✓.

The Remainder Theorem

Remainder Theorem: the remainder when p(x) is divided by (x − c) equals p(c). Evaluation and division are the same computation wearing two coats.

Worked Example 2 — evaluating by synthetic division. Find p(3) for p(x) = 2x^3 − x^2 + 4x − 5. Synthetic division with c = 3 on coefficients 2, −1, 4, −5: bring down 2; 2·3 = 6, −1 + 6 = 5; 5·3 = 15, 4 + 15 = 19; 19·3 = 57, −5 + 57 = 52. The remainder is 52, so p(3) = 52. Direct check: 2(27) − 9 + 12 − 5 = 54 − 9 + 12 − 5 = 52 ✓. Synthetic division is the faster and less error-prone route, especially for large c.

The Factor Theorem and complete factorisation

Factor Theorem: (x − c) is a factor of p(x) if and only if p(c) = 0. To factor a cubic, hunt for a root among the factors of the constant term divided by the factors of the leading coefficient (the Rational Root Theorem), test candidates with the Remainder Theorem, divide, then factor the quadratic.

Worked Example 3 — dismantling a cubic. Factor p(x) = x^3 + 2x^2 − 5x − 6 completely. Candidates: ±1, ±2, ±3, ±6. Test x = −1: p(−1) = −1 + 2 + 5 − 6 = 0 ✓, so (x + 1) is a factor. Synthetic division with c = −1 on 1, 2, −5, −6: bring 1; −1·1 = −1, 2 − 1 = 1; −1·1 = −1, −5 − 1 = −6; −1·(−6) = 6, −6 + 6 = 0. Quotient x^2 + x − 6 = (x + 3)(x − 2). Final answer: p(x) = (x + 1)(x + 3)(x − 2), roots −1, −3, 2.

Graphically these roots are the x-intercepts; the factorised form also hands you the sign chart and the end behaviour (degree 3, positive leading coefficient: falls left, rises right) for free.

Modelling: the dimensions of a box

A packaging company folds a box whose volume, in cubic centimetres, is modelled by V(x) = x^3 + 3x^2 − 6x − 8, where x is a design parameter in centimetres. Factoring finds the dimensions. Test x = −1: V(−1) = −1 + 3 + 6 − 8 = 0, so (x + 1) is a factor. Synthetic division (c = −1; coefficients 1, 3, −6, −8) gives quotient x^2 + 2x − 8 = (x + 4)(x − 2), so

V(x) = (x + 1)(x + 4)(x − 2)

The three factors are the box's length, width and height. Physical dimensions must be positive, so the model only makes sense for x > 2. At x = 5 the dimensions are 6 cm, 9 cm and 3 cm, and the volume check is 6 · 9 · 3 = 162 cm^3 — matching the direct evaluation V(5) = 125 + 75 − 30 − 8 = 162 ✓. Factorisation turned an opaque formula into a design the company can actually build from.

Common mistakes

• Forgetting placeholder zeros. Dividing x^3 − 5x + 6 requires coefficients 1, 0, −5, 6. Omitting the 0 shifts every column and ruins the quotient.
• Using the wrong sign for c. Dividing by (x + 2) means c = −2. Most synthetic-division errors are this one sign.
• Confusing remainder with root. A remainder of 0 means c is a root and (x − c) is a factor. A remainder of 5 means p(c) = 5 — the point (c, 5) is on the graph, nothing more.
• Stopping after one factor. A cubic with one linear factor leaves a quadratic, which may factor again — or have complex roots (Lesson 1). Always examine the quotient.
• Dividing by (2x − 1) synthetically with c = 1. The synthetic shortcut needs the divisor in the form (x − c); for 2x − 1, c = 1/2.

Exam Experience

1. When p(x) = 3x^3 − 2x^2 + 5 is divided by (x − 2), the remainder is

• (A) 5
• (B) 21
• (C) 13
• (D) 29

Answer: (B). By the Remainder Theorem, the remainder is p(2) = 3(8) − 2(4) + 5 = 24 − 8 + 5 = 21. Note the missing x-term needs a 0 coefficient in synthetic division; forgetting it gives (C)-style errors.

2. Which of the following is a factor of x^3 − 4x^2 + x + 6?

• (A) (x − 1)
• (B) (x + 1)
• (C) (x + 2)
• (D) (x − 4)

Answer: (B). p(−1) = −1 − 4 − 1 + 6 = 0, so (x + 1) is a factor. Check the others: p(1) = 4, p(−2) = −16, p(4) = 10 — none zero.

3. Dividing x^3 + 2x^2 − 5x − 6 by (x − 2) gives quotient

• (A) x^2 + 4x + 3 with remainder 0
• (B) x^2 + 4x + 3 with remainder −6
• (C) x^2 − 3 with remainder 0
• (D) x^2 + 4x − 3 with remainder 0

Answer: (A). Synthetic with c = 2 on 1, 2, −5, −6: bring 1; 2, 2 + 2 = 4; 8, −5 + 8 = 3; 6, −6 + 6 = 0. So 2 is a root (consistent with Worked Example 3) and the quotient is x^2 + 4x + 3 = (x + 1)(x + 3).

Constructed response. (a) Find the value of k such that (x − 2) is a factor of p(x) = x^3 − 3x^2 + kx − 8. (b) With this value of k, factor p(x) completely over the complex numbers. (c) State all roots of p(x) = 0.

Scoring guide. (a) 2 points: 1 for applying the Factor Theorem, p(2) = 0; 1 for solving 8 − 12 + 2k − 8 = 0 to get 2k = 12, k = 6. (b) 3 points: 1 for synthetic division by (x − 2) on 1, −3, 6, −8 giving quotient x^2 − x + 4; 1 for applying the quadratic formula to x^2 − x + 4 = 0: discriminant 1 − 16 = −15; 1 for the complete factorisation p(x) = (x − 2)(x − (1 + i√15)/2)(x − (1 − i√15)/2), or equivalently (x − 2)(x^2 − x + 4) with the quadratic's roots stated. (c) 1 point: roots 2 and (1 ± i√15)/2 — a conjugate pair, as Lesson 1 promised.

Going deeper: long division with missing terms, and counting roots

Synthetic division requires the divisor (x − c); long division handles everything else and shows its working more transparently. Divide 2x^3 + 3x − 5 by (x − 2):

2x^3 + 0x^2 + 3x − 5 ÷ (x − 2):
First term: 2x^3/x = 2x^2. Multiply back: 2x^2(x − 2) = 2x^3 − 4x^2. Subtract: 4x^2 + 3x.
Next: 4x^2/x = 4x. Multiply: 4x(x − 2) = 4x^2 − 8x. Subtract: 11x − 5.
Next: 11x/x = 11. Multiply: 11(x − 2) = 11x − 22. Subtract: remainder 17.
Quotient 2x^2 + 4x + 11, remainder 17.

Three audits: the placeholder 0x^2 was written in before starting; the remainder 17 equals p(2) = 2(8) + 3(2) − 5 = 17 ✓ (Remainder Theorem, as it must); and synthetic division with c = 2 on coefficients 2, 0, 3, −5 gives the same 2, 4, 11, 17 in a single row. Choose synthetic for speed, long division when the divisor is not (x − c) — for example (2x − 1), where c = 1/2 synthetically is legal but error-prone.

Counting roots. The Fundamental Theorem of Algebra guarantees a degree-n polynomial exactly n roots in the complex numbers, counting multiplicity. The cubic of Worked Example 1 in this lesson has roots 1, 3, −2 — three distinct. The quadratic x^2 − x + 4 from the Exam Experience has roots (1 ± i√15)/2 — two, both non-real. A repeated factor like (x − 2)^2 contributes the root 2 twice to the count. This is the bookkeeping that makes "solve" a well-posed instruction: you know when you are finished.

One more division: a quartic with two missing terms

Divide x^4 − 3x^2 + 2 by (x − 1). Coefficients with placeholders: 1, 0, −3, 0, 2; c = 1. Synthetic row: bring 1; add 1 → 1; add 1 → −2; add −2 → −2; add −2 → 0. Quotient x^3 + x^2 − 2x − 2, remainder 0 — so x = 1 is a root, and indeed 1 − 3 + 2 = 0 ✓. The quotient still has structure: x^3 + x^2 − 2x − 2 = x^2(x + 1) − 2(x + 1) = (x + 1)(x^2 − 2), so the quartic's roots are 1, −1, ±√2 — four roots for degree 4, exactly as the Fundamental Theorem promises, found by alternating division with grouping.