Practice Maths

L10 — Dividing Polynomials & the Remainder Theorem

Key Terms

  • Division algorithm — P(x) = D(x) × Q(x) + R(x); dividend = divisor × quotient + remainder
  • Quotient Q(x) — the polynomial result of dividing P(x) by D(x)
  • Remainder R(x) — what is left after division; the degree of R must be less than the degree of D
  • Remainder Theorem — when P(x) is divided by (x − a), the remainder equals P(a)
  • Factor Theorem — (x − a) is a factor of P(x) if and only if P(a) = 0
  • Zero of a polynomial — a value x = a where P(a) = 0; corresponds to an x-intercept of the graph

The Division Algorithm

When polynomial P(x) is divided by divisor D(x), the result is:

P(x) = D(x) × Q(x) + R(x)

where Q(x) is the quotient and R(x) is the remainder. The degree of R must be less than the degree of D.

Polynomial Long Division

Example: Divide P(x) = x³ − 3x² + 5x − 4 by (x − 1).

Step 1: Divide the leading term: x³ ÷ x = x².
      Multiply: x²(x − 1) = x³ − x².
      Subtract: (x³ − 3x²) − (x³ − x²) = −2x². Bring down: −2x² + 5x.

Step 2: Divide: −2x² ÷ x = −2x.
      Multiply: −2x(x − 1) = −2x² + 2x.
      Subtract: (−2x² + 5x) − (−2x² + 2x) = 3x. Bring down: 3x − 4.

Step 3: Divide: 3x ÷ x = 3.
      Multiply: 3(x − 1) = 3x − 3.
      Subtract: (−4) − (−3) = −1 (remainder).

Result: x³ − 3x² + 5x − 4 = (x − 1)(x² − 2x + 3) + (−1)

Remainder Theorem

When P(x) is divided by (x − a), the remainder equals P(a).

Check from the example above: P(1) = 1 − 3 + 5 − 4 = −1. ✓

This means you can find the remainder without performing long division — just evaluate the polynomial at x = a.

Factor Theorem

(x − a) is a factor of P(x)  ⇔  P(a) = 0.

That is: if substituting x = a gives zero, then (x − a) divides P(x) exactly (zero remainder).
Hot Tip: To find a factor of a cubic P(x), test integer factors of the constant term: try x = ±1, ±2, ±3, … until you find a value where P(a) = 0. Once you have one factor, use long division to find the quadratic quotient and factorise it further.

Strategy: Fully Factorise a Cubic

  1. Find a zero by testing simple values: x = ±1, ±2, ±3, … (factors of the constant term).
  2. Use long division (or inspection) to find the quotient Q(x).
  3. Factorise Q(x) further if possible (usually a quadratic).
Example: Factorise P(x) = x³ − 6x² + 11x − 6.
Test x = 1: P(1) = 1 − 6 + 11 − 6 = 0 ✓. So (x−1) is a factor.
Divide: P(x) = (x−1)(x²−5x+6) = (x−1)(x−2)(x−3).

Division of Polynomials

Dividing one polynomial by another follows the same algorithm as long division in arithmetic. You divide, multiply, subtract, bring down — and repeat until the remainder has a smaller degree than the divisor.

When dividing by a linear factor (x − a), the remainder is always a constant. If that constant is zero, then (x − a) is an exact factor.

Worked Example 1: Polynomial long division

Divide P(x) = x³ + 2x² − 5x − 6 by (x + 3).

Solution:
Step 1: x³ ÷ x = x².   x²(x+3) = x³ + 3x².
        Subtract: x³ + 2x² − (x³ + 3x²) = −x². Bring down: −x² − 5x.
Step 2: −x² ÷ x = −x.   −x(x+3) = −x² − 3x.
        Subtract: −x² − 5x − (−x² − 3x) = −2x. Bring down: −2x − 6.
Step 3: −2x ÷ x = −2.   −2(x+3) = −2x − 6.
        Subtract: −6 − (−6) = 0 (no remainder).

Result: x³ + 2x² − 5x − 6 = (x + 3)(x² − x − 2) = (x+3)(x−2)(x+1)

The Remainder Theorem

When you divide P(x) by (x − a), the remainder is simply P(a). You can verify this from the division algorithm: P(x) = (x − a)Q(x) + R. Setting x = a gives P(a) = 0 × Q(a) + R = R.

This means you don't need to do long division just to find the remainder — substitute x = a directly.

Worked Example 2: Remainder theorem shortcut

Find the remainder when P(x) = x4 − 2x³ + 3x − 1 is divided by (x − 2).

Solution:
By the remainder theorem, remainder = P(2).
P(2) = 24 − 2(2)³ + 3(2) − 1 = 16 − 16 + 6 − 1 = 5

The remainder is 5 (no long division needed).

The Factor Theorem

The factor theorem is the key to finding and verifying factors: (x − a) is a factor of P(x) if and only if P(a) = 0. This is just the remainder theorem with R = 0.

Worked Example 3: Verify a factor and fully factorise

Show that (x − 2) is a factor of P(x) = x³ − 7x + 6, then factorise P(x) completely.

Solution:
P(2) = 8 − 14 + 6 = 0 ✓  ⇒  (x − 2) is a factor.

Divide P(x) by (x − 2):
x³ − 7x + 6 = (x − 2)(x² + 2x − 3)
Factorise the quadratic: x² + 2x − 3 = (x + 3)(x − 1)

Complete factorisation: P(x) = (x − 2)(x + 3)(x − 1)

  1. Long division. Fluency

    Perform polynomial long division. Express your answer in the form P(x) = D(x) × Q(x) + R.

    • (a) (x² − 5x + 6) ÷ (x − 2)
    • (b) (x³ − 2x² + x − 3) ÷ (x − 1)
    • (c) (x³ + x² − 4x − 4) ÷ (x + 2)
    • (d) (2x³ − x² + 3x − 1) ÷ (x − 1)

  2. Remainder theorem. Fluency

    Use the remainder theorem to find the remainder when each polynomial is divided by the given divisor. Do not perform long division.

    • (a) P(x) = x³ − 2x² + x − 5, divisor (x − 3)
    • (b) P(x) = 2x³ + x² − x + 4, divisor (x + 1)
    • (c) P(x) = x4 − 3x² + 2, divisor (x − 2)
    • (d) P(x) = x³ − 4x + 1, divisor (x + 2)

  3. Factor theorem. Fluency

    Use the factor theorem to determine whether the given linear expression is a factor of P(x).

    • (a) P(x) = x³ − x² − 4x + 4, test (x − 2)
    • (b) P(x) = x³ + 2x² − x − 2, test (x + 2)
    • (c) P(x) = x³ − 3x² + x − 3, test (x − 3)
    • (d) P(x) = x³ − 2x² + x − 4, test (x − 1)

  4. Find the unknown coefficient. Fluency

    Each polynomial has the given factor. Use the factor theorem to find the unknown value k.

    • (a) P(x) = x² + kx − 6 has factor (x − 2)
    • (b) P(x) = x³ − kx² + x − 4 has factor (x + 1)
    • (c) P(x) = 2x³ + x² − kx + 3 has factor (x − 3)
    • (d) P(x) = x³ + kx − 2 has factor (x − 1)

  5. Fully factorise. Understanding

    Use the factor theorem to find one linear factor, then factorise completely.

    • (a) P(x) = x³ − 6x² + 11x − 6
    • (b) P(x) = x³ + 2x² − 5x − 6
    • (c) P(x) = x³ − 4x² + x + 6
    • (d) P(x) = 2x³ − x² − 2x + 1

  6. Build the polynomial from its zeros. Understanding

    A cubic polynomial P(x) = x³ + ax² + bx + c has zeros at x = −1, x = 2, and x = 3.

    • (a) Write P(x) in factored form.
    • (b) Expand to standard form and state the values of a, b, and c.
    • (c) Verify the y-intercept P(0) matches the constant term c.
    • (d) Find P(4).

  7. Find k using the remainder theorem. Understanding

    P(x) = x³ − 3x² + kx + 4 leaves a remainder of 12 when divided by (x − 2).

    • (a) Use the remainder theorem to write an equation involving k.
    • (b) Solve for k.
    • (c) Write the full polynomial and verify the remainder by evaluating P(2).

  8. Possible remainders. Understanding

    When a cubic polynomial is divided by a quadratic divisor (degree 2), the remainder must have degree less than 2. Which of the following expressions can be the remainder? Give a reason for each answer.

    • (a) 5
    • (b) x² − 2
    • (c) 3x + 1
    • (d) x² + 3x

  9. Find a and b from two factor conditions. Problem Solving

    The quadratic x² + ax + b has both (x − 3) and (x + 2) as factors.

    • (a) Use the factor theorem to write two equations in a and b.
    • (b) Solve the simultaneous equations to find a and b.
    • (c) Write the fully factorised form of x² + ax + b and verify.

  10. Two conditions — find a, b, and fully factorise. Problem Solving

    P(x) = x³ + ax² + bx − 12. It is known that (x − 2) is a factor, and that P(1) = −6.

    • (a) Use the factor condition and the remainder condition to write two equations in a and b.
    • (b) Solve simultaneously to find a and b.
    • (c) Write the full polynomial and factorise it completely over the reals.
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