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Algebra and Functions

Polynomials and the Factor Theorem

Pearson Edexcel A-level Mathematics

The factor theorem

  • If f(ba) = 0, then (ax − b) is a factor of f(x).
  • Conversely, a factor (x − p) gives f(p) = 0, and a factor (x + p) gives f(−p) = 0.
Method: find an unknown constant from a given factor
  1. Set the factor equal to zero to find the root.
  2. Substitute the root into f(x) and write the result = 0.
  3. Solve the equation for the constant.
  • In a show that, write = 0 on a line before the given value appears.
  • Write f(−p) = 0 and solve for the constant. Substitute the root of the factor, not the given value of the constant.
Worked example: find a constant from a given factor

f(x) = 2x3 + kx2 − 13x + 6, where k is a constant. Given that (x + 3) is a factor of f(x), find the value of k.

f(−3) = 0

2(−3)3 + k(−3)2 − 13(−3) + 6 = 0

−54 + 9k + 39 + 6 = 0, so 9k − 9 = 0

k = 1

Finding the quadratic factor

Method: factorise a cubic with a known factor
  1. Write f(x) ≡ (x − p)(ax2 + bx + c).
  2. Read a from the x3 term and c from the constant term.
  3. Find b by comparing the x2 or x coefficients, or use algebraic long division.
  4. Factorise the quadratic factor if it has real roots.
  5. Write f(x) as the product of all its factors.

Factorise completely: a quadratic factor with no real roots stays as a quadratic in the final product.

Worked example: show that a constant takes a value, then factorise completely

h(x) = 2x3 + bx2 − 11x − 6, where b is a constant. Given that (2x + 1) is a factor of h(x),
(a) show that b = 3
(b) hence factorise h(x) completely.

(a) h(−12) = 0

2(−12)3 + b(−12)2 − 11(−12) − 6 = 0

−14 + b4 + 112 − 6 = 0, so b4 − 34 = 0

b = 3

(b) 2x3 + 3x2 − 11x − 6 ≡ (2x + 1)(x2 + cx − 6)

Comparing x2 terms: 2c + 1 = 3, so c = 1

h(x) = (2x + 1)(x2 + x − 6)

h(x) = (2x + 1)(x − 2)(x + 3)

Solving a cubic

Method: solve a cubic equation
  1. Rearrange so that one side is zero.
  2. Take out the known factor, or one found by the factor theorem.
  3. Solve the quadratic factor by factorising, completing the square or the quadratic formula.
  4. List every root: the one from the linear factor and those from the quadratic.
Worked example: hence solve a cubic equation

f(x) = x3 + 2x2 − 7x + 4. Given that (x − 2) is a factor of f(x) − 6, hence solve the equation f(x) = 6, giving your answers in exact form.

x3 + 2x2 − 7x − 2 = 0

(x − 2)(x2 + 4x + 1) = 0

x2 + 4x + 1 = 0, so (x + 2)2 = 3

x = 2, x = −2 ± √3

Algebraic division

Method: divide a polynomial by a linear term
  1. Write f(x) ≡ (ax + b) × quotient + remainder, with a quotient one degree lower than f(x) and a constant remainder.
  2. Compare coefficients to find the quotient and the remainder, or use algebraic long division.
  3. For an improper fraction, write f(x)ax + b = quotient + remainderax + b.

Rational expressions

Method: simplify a rational expression
  1. Factorise the numerator and the denominator fully, using the factor theorem for a cubic.
  2. Cancel factors common to both.
  3. Where the numerator still has a degree at least that of a linear denominator, divide.
  4. Write the result as a single fraction or as a quotient plus a fraction.

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