Factors of polynomials for IGCSE Mathematics - Additional: Methods and Worked Examples

The central principle: evaluation equals remainder

Every polynomial can be divided by a linear expression, and the remainder that survives tells you something precise about the polynomial's value. This single idea underpins both the remainder theorem and the factor theorem, and it gives IGCSE Additional Mathematics candidates a powerful, systematic method for breaking polynomials apart.

Consider a polynomial f(x) of degree n divided by the linear divisor (x - a). The division produces a quotient Q(x) of degree n - 1 and a constant remainder R, so that f(x) = (x - a)Q(x) + R. Substituting x = a into both sides collapses the first term to zero, leaving f(a) = R. That equality is the remainder theorem: the remainder when f(x) is divided by (x - a) is simply f(a).

The factor theorem follows as a direct corollary. If f(a) = 0, then R = 0, and the division is exact: f(x) = (x - a)Q(x). In other words, (x - a) is a factor of f(x) precisely when a is a root. This logical chain, from division identity to evaluation to factorisation, is worth internalising because it converts the abstract question "does this factor work?" into a concrete arithmetic check.

The remainder theorem in practice

The remainder theorem applies to any polynomial divided by any linear expression. When the divisor is (x - a), the remainder is f(a). When the divisor takes the form (ax - b), the remainder is f(b/a), because ax - b = 0 when x = b/a.

Worked example 1: Finding a remainder

Find the remainder when f(x) = 2x³ - 5x² + 3x - 7 is divided by (x - 3).

1. Identify the value: the divisor is (x - 3), so evaluate f(3)
2. f(3) = 2(27) - 5(9) + 3(3) - 7 = 54 - 45 + 9 - 7 = 11
3. The remainder is 11

No long division was needed. The entire calculation reduces to substitution and arithmetic, which is both faster and less prone to sign errors than the full division algorithm.

Worked example 2: Remainder with a non-unit coefficient

Find the remainder when g(x) = 4x³ + 2x² - x + 5 is divided by (2x - 1).

1. Set 2x - 1 = 0, giving x = 1/2
2. g(1/2) = 4(1/8) + 2(1/4) - 1/2 + 5 = 1/2 + 1/2 - 1/2 + 5 = 5.5
3. The remainder is 5.5 (or 11/2)

Worked example 3: Using the remainder to find an unknown coefficient

The polynomial p(x) = x³ + kx² - 4x + 6 has remainder 18 when divided by (x - 2). Find the value of k.

1. By the remainder theorem, p(2) = 18
2. p(2) = 8 + 4k - 8 + 6 = 6 + 4k
3. Setting 6 + 4k = 18 gives 4k = 12, so k = 3

This type of question appears frequently on Cambridge papers. The logic flows in one direction: set up the equation from the theorem, substitute, solve for the unknown. Candidates who try to find k by performing full polynomial division with an unknown coefficient create unnecessary complexity and risk algebraic slips.

The factor theorem: from evaluation to factorisation

The factor theorem states that (x - a) is a factor of f(x) if and only if f(a) = 0. The "if and only if" is important: the result works in both directions. If you know (x - a) is a factor, you can conclude f(a) = 0 without calculating it. If you calculate f(a) = 0, you can conclude (x - a) is a factor without performing division.

Worked example 4: Verifying a factor

Show that (x + 2) is a factor of f(x) = x³ + 6x² + 11x + 6.

1. The factor (x + 2) corresponds to a = -2
2. f(-2) = (-8) + 6(4) + 11(-2) + 6 = -8 + 24 - 22 + 6 = 0
3. Since f(-2) = 0, the factor theorem confirms that (x + 2) is a factor

Notice the careful handling of signs. The expression (x + 2) means (x - (-2)), so the test value is -2, not +2. This sign reversal is the single most common source of error in factor theorem questions on the IGCSE Additional Mathematics exam.

Factorising cubics completely

The real power of the factor theorem emerges when it combines with polynomial division to break a cubic into three linear factors. The strategy has three stages: find one root by systematic trial, divide to obtain a quadratic quotient, then factorise that quadratic by inspection or by the quadratic formula.

Which values to try first

For a cubic f(x) = ax³ + bx² + cx + d with integer coefficients, any rational root p/q must satisfy: p divides d (the constant term) and q divides a (the leading coefficient). This narrows the candidates dramatically.

Start with the smallest values: +/-1, +/-2 are quick to evaluate and account for most roots in IGCSE-level cubics. If those fail, move outward to +/-3 and beyond.

Worked example 5: Full factorisation of a cubic

Factorise f(x) = 2x³ + 3x² - 11x - 6 completely.

Stage 1: Find a root.

• f(1) = 2 + 3 - 11 - 6 = -12 (not zero)
• f(-1) = -2 + 3 + 11 - 6 = 6 (not zero)
• f(2) = 16 + 12 - 22 - 6 = 0 (root found)

Since f(2) = 0, the factor theorem gives (x - 2) as a factor.

Stage 2: Divide to find the quadratic quotient.

Dividing 2x³ + 3x² - 11x - 6 by (x - 2) using long division or synthetic division:

1. 2x³ divided by x gives 2x². Multiply: 2x²(x - 2) = 2x³ - 4x²
2. Subtract: (2x³ + 3x²) - (2x³ - 4x²) = 7x². Bring down -11x to get 7x² - 11x
3. 7x² divided by x gives 7x. Multiply: 7x(x - 2) = 7x² - 14x
4. Subtract: (7x² - 11x) - (7x² - 14x) = 3x. Bring down -6 to get 3x - 6
5. 3x divided by x gives 3. Multiply: 3(x - 2) = 3x - 6
6. Subtract: (3x - 6) - (3x - 6) = 0. Division is exact, confirming the factor.

The quotient is 2x² + 7x + 3.

Stage 3: Factorise the quadratic.

2x² + 7x + 3 = (2x + 1)(x + 3), found by seeking two numbers that multiply to 2 x 3 = 6 and add to 7. Those numbers are 6 and 1, leading to the split 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3).

The complete factorisation is f(x) = (x - 2)(2x + 1)(x + 3).

Solving cubic equations

Once a cubic is fully factorised, setting each factor equal to zero yields all three roots. From the previous example, 2x³ + 3x² - 11x - 6 = 0 gives x = 2, x = -1/2, or x = -3.

Not every cubic has three distinct real roots. The quadratic factor obtained after the first division might have a repeated root (discriminant zero) or no real roots (discriminant negative). When the discriminant is negative, the cubic has one real root and two complex conjugate roots, and the IGCSE Additional Mathematics syllabus expects candidates to state that the quadratic factor "has no real roots" or "cannot be factorised further over the reals."

Worked example 6: A cubic with only one real root

Solve x³ - 4x² + 6x - 4 = 0.

1. Test x = 1: 1 - 4 + 6 - 4 = -1 (not a root)
2. Test x = 2: 8 - 16 + 12 - 4 = 0 (root found)
3. Divide by (x - 2): the quotient is x² - 2x + 2
4. Check the discriminant of x² - 2x + 2: b² - 4ac = 4 - 8 = -4
5. Since the discriminant is negative, x² - 2x + 2 has no real roots
6. The only real solution is x = 2

Polynomial long division: the full method

While the factor theorem reduces many problems to substitution, some exam questions explicitly require polynomial division, or the divisor is a quadratic rather than linear. The procedure mirrors numerical long division, adapted for algebraic terms.

The algorithm works as follows for dividing f(x) by d(x):

1. Divide the leading term of the current dividend by the leading term of d(x) to get the next term of the quotient
2. Multiply the entire divisor by that term
3. Subtract the result from the current dividend
4. Bring down the next term
5. Repeat until the degree of the remaining expression is less than the degree of d(x)

Worked example 7: Division by a quadratic

Divide f(x) = x⁴ + 2x³ - 7x² - 8x + 12 by (x² - 3).

1. x⁴ / x² = x². Multiply: x²(x² - 3) = x⁴ - 3x²
2. Subtract: (x⁴ + 2x³ - 7x²) - (x⁴ - 3x²) = 2x³ - 4x². Bring down -8x: 2x³ - 4x² - 8x
3. 2x³ / x² = 2x. Multiply: 2x(x² - 3) = 2x³ - 6x
4. Subtract: (2x³ - 4x² - 8x) - (2x³ - 6x) = -4x² - 2x. Bring down +12: -4x² - 2x + 12
5. -4x² / x² = -4. Multiply: -4(x² - 3) = -4x² + 12
6. Subtract: (-4x² - 2x + 12) - (-4x² + 12) = -2x

The quotient is x² + 2x - 4 with remainder -2x. This means x⁴ + 2x³ - 7x² - 8x + 12 = (x² - 3)(x² + 2x - 4) + (-2x).

Common errors and how to avoid them

Self-check questions

Work through each problem before consulting the answers.

1. Find the remainder when f(x) = 3x³ - 2x² + x - 5 is divided by (x - 1).
2. The polynomial p(x) = 2x³ + ax² - 7x - 6 has (x + 1) as a factor. Find the value of a.
3. Factorise x³ - 2x² - 5x + 6 completely.
4. Solve 2x³ - x² - 13x - 6 = 0.
5. Find the remainder when 4x³ - 3x + 1 is divided by (2x + 1).

Strategic approach to exam questions

Polynomial questions on the Cambridge IGCSE Additional Mathematics papers typically fall into three categories. The first asks for a remainder or unknown coefficient, requiring direct application of the remainder theorem. The second asks for complete factorisation of a cubic, combining the factor theorem with division and quadratic factorisation. The third gives simultaneous conditions (two different divisors producing specified remainders) and asks candidates to find two unknowns by setting up and solving a pair of simultaneous equations.

For the third type, suppose f(x) = x³ + ax² + bx + c has remainder 5 when divided by (x - 1) and remainder -1 when divided by (x + 1). The remainder theorem gives f(1) = 5 and f(-1) = -1, producing two equations in a and b (assuming c is known). Solving these simultaneously yields both unknowns. The technique is algebraically straightforward, but candidates who don't recognise the pattern waste time attempting alternative approaches.

Across all three question types, the factor and remainder theorems convert division problems into substitution problems. The candidate who understands this conversion, and who executes the arithmetic carefully, holds a significant advantage. Polynomial factorisation in 0606 rewards method and precision in equal measure, and both are trainable skills.