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O-Level Additional Mathematics · Secondary 3 and 4

Polynomials and Partial Fractions

A polynomial is an expression made of whole-number powers of x, such as 2x³ - 5x² + x + 6. This topic covers the remainder and factor theorems, solving cubic equations, the sum and difference of two cubes, and splitting a fraction into partial fractions. Factorising cubics is needed when you sketch graphs and find stationary points, and partial fractions turn a complicated fraction into simpler pieces, which becomes important for integration in H2 Mathematics.

What the syllabus expects

  • Multiply and divide polynomials, including long division to find a quotient and a remainder
  • Use the remainder theorem: the remainder when f(x) is divided by (ax - b) is f(b/a)
  • Use the factor theorem to find factors and to solve cubic equations
  • Factorise using the sum and difference of two cubes, a³ + b³ and a³ - b³
  • Express proper fractions in partial fractions where the denominator has distinct linear factors, a repeated linear factor, or a linear factor and a quadratic factor of the form x² + c²
  • Divide first when a fraction is improper, then express the proper part in partial fractions

Remainder and factor theorems

When a polynomial f(x) is divided by (x - a), we can write f(x) = (x - a)Q(x) + R, where Q(x) is the quotient and R is the remainder. Putting x = a makes the first term zero, so R = f(a). This is the remainder theorem. More generally, the remainder on dividing by (ax - b) is f(b/a). If the remainder is zero, then (x - a) is a factor, which is the factor theorem. For example, if f(x) = x³ - 2x² - 5x + 6, then f(1) = 1 - 2 - 5 + 6 = 0, so (x - 1) is a factor. Questions often give two pieces of information, such as a remainder and a factor, so that you can set up two simultaneous equations for two unknown coefficients.

Solving cubic equations

To solve a cubic equation f(x) = 0, find one root by trial using the factor theorem. Try factors of the constant term, such as ±1, ±2 and ±3. Once you have a factor (x - a), divide f(x) by it, either by long division or by comparing coefficients, to get a quadratic. Then factorise the quadratic or use the quadratic formula. Using the example above, x³ - 2x² - 5x + 6 = (x - 1)(x² - x - 6) = (x - 1)(x - 3)(x + 2), so the roots are 1, 3 and -2. To compare coefficients, write x³ - 2x² - 5x + 6 = (x - 1)(x² + px + q), expand the right side, and match the x² terms and the constants. If the quadratic factor has a negative discriminant, the cubic has only one real root, and you should say so.

Sum and difference of two cubes

Two identities are worth knowing by heart: a³ + b³ = (a + b)(a² - ab + b²) and a³ - b³ = (a - b)(a² + ab + b²). The quadratic factor in each cannot be factorised further using real numbers. For example, 8x³ - 27 = (2x)³ - 3³ = (2x - 3)(4x² + 6x + 9), and x³ + 64 = (x + 4)(x² - 4x + 16). Look for these patterns whenever you see two cube terms and nothing else. A useful check is to expand: (2x - 3)(4x² + 6x + 9) = 8x³ + 12x² + 18x - 12x² - 18x - 27 = 8x³ - 27. Notice how the middle terms cancel, which is why the identities work.

Partial fractions

A proper fraction, where the degree of the numerator is less than the degree of the denominator, can be split into simpler fractions. The form depends on the denominator. Distinct linear factors: (px + q)/((ax + b)(cx + d)) = A/(ax + b) + B/(cx + d). A repeated linear factor: (px² + qx + r)/((ax + b)(cx + d)²) = A/(ax + b) + B/(cx + d) + C/(cx + d)². A quadratic factor: (px² + qx + r)/((ax + b)(x² + c²)) = A/(ax + b) + (Bx + C)/(x² + c²). Multiply both sides by the denominator, then substitute convenient values of x and compare coefficients to find the constants. For example, (5x + 1)/((x - 1)(x + 2)) = A/(x - 1) + B/(x + 2) gives 5x + 1 = A(x + 2) + B(x - 1). Putting x = 1 gives 6 = 3A, so A = 2. Putting x = -2 gives -9 = -3B, so B = 3. If the fraction is improper, divide first to get a polynomial plus a proper fraction.

Worked example

Express (3x² - 5x + 8)/((x - 2)(x² + 1)) in partial fractions.

  1. The denominator has a linear factor and the quadratic factor x² + 1, so write the fraction as A/(x - 2) + (Bx + C)/(x² + 1).
  2. Multiply both sides by (x - 2)(x² + 1): 3x² - 5x + 8 = A(x² + 1) + (Bx + C)(x - 2).
  3. Put x = 2: 12 - 10 + 8 = 5A, so 10 = 5A and A = 2.
  4. Compare the coefficients of x²: 3 = A + B, so B = 1.
  5. Compare the constants: 8 = A - 2C, so 8 = 2 - 2C and C = -3.
  6. Check with the x terms: the coefficient of x on the right is -2B + C = -2 - 3 = -5, which matches the left side.

Answer: 2/(x - 2) + (x - 3)/(x² + 1)

Common mistakes

  • Using f(a) instead of f(-a) for the remainder when dividing by (x + a). The divisor x + a is zero when x = -a, so the remainder is f(-a).
  • Substituting x = 2 instead of x = 1/2 for a divisor such as (2x - 1). Set the divisor equal to zero and solve for x.
  • Putting a single constant over a quadratic factor such as x² + 4. A quadratic factor needs a linear numerator Bx + C.
  • Starting partial fractions on an improper fraction. If the degree of the numerator is equal to or greater than that of the denominator, long division comes first.
  • Factorising a³ - b³ as (a - b)(a² - ab + b²). The middle sign in the quadratic factor is opposite to the sign between the cubes, so a³ - b³ = (a - b)(a² + ab + b²).
  • Stopping at one root of a cubic equation. After finding one factor, divide to get the quadratic factor, then find the remaining roots or show that there are none.

Exam tips

  • Substituting the root of each linear factor finds most constants in one line each. Use comparing coefficients for the rest, then use one unused coefficient as a check.
  • In factor theorem questions with two unknowns, label your equations (1) and (2), solve them simultaneously, then write out the polynomial in full before doing the next part.
  • When the quadratic factor of a cubic has a negative discriminant, write 'b² - 4ac < 0, so there are no further real roots' rather than leaving the quadratic unsolved.

Practice questions

1. Find the remainder when 2x³ + x² - 4x + 7 is divided by (x + 2).

Show answer

The remainder is 3. By the remainder theorem it is f(-2) = 2(-8) + 4 + 8 + 7 = -16 + 4 + 8 + 7 = 3.

2. The polynomial f(x) = x³ + ax² + bx - 6 has a factor (x - 1) and leaves a remainder of -8 when divided by (x + 1). Find a and b, and show that f(x) = 0 has only one real root.

Show answer

a = 2 and b = 3. From f(1) = 0: 1 + a + b - 6 = 0, so a + b = 5. From f(-1) = -8: -1 + a - b - 6 = -8, so a - b = -1. Solving gives a = 2 and b = 3. Then f(x) = (x - 1)(x² + 3x + 6). The quadratic factor has discriminant 9 - 24 = -15 < 0, so x = 1 is the only real root.

3. Express (2x² + x + 3)/((x + 1)(x - 1)²) in partial fractions.

Show answer

1/(x + 1) + 1/(x - 1) + 3/(x - 1)². Write 2x² + x + 3 = A(x - 1)² + B(x + 1)(x - 1) + C(x + 1). Put x = 1: 6 = 2C, so C = 3. Put x = -1: 4 = 4A, so A = 1. Compare x²: 2 = A + B, so B = 1. Check the constants: A - B + C = 1 - 1 + 3 = 3, which matches.

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