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Cambridge IGCSE0606

Factors of polynomials

Mathematics Additional 0606 Chapter Notes

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Factors of polynomials
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In this chapter

  1. 1The remainder theorem
  2. 2The factor theorem
  3. 3Finding an unknown constant
  4. 4Finding the quadratic factor
  5. 5Choosing a factor to test
  6. 6Solving a cubic equation
  7. 7How many roots?
  8. 8A cubic in disguise

1. The remainder theorem

When a polynomial p(x) is divided by x - a, the remainder is p(a).

For a divisor ax + b, substitute x = -b/a. The divisor x + 2 gives x = -2.

remainder on dividing by x - a = p(a)

remainder on dividing by ax + b = p(-b/a)

Examiner insight

“A few candidates used long division but should be advised that use of the remainder theorem is an easier approach.”

Worked example 11 mark

The polynomial p is given by p(x) = 6x³ - x² - 5x + 2. Find the remainder when p(x) is divided by x + 2.

  1. 1

    divisor x + 2, so remainder = p(-2) = 6(-2)³ - (-2)² - 5(-2) + 2

  2. 2

    = -48 - 4 + 10 + 2 = -40 [1]

2. The factor theorem

If p(a) = 0, then x - a is a factor of p(x). If x - a is a factor, the remainder is 0.

To show that x + 2 is a factor, evaluate p(-2) and show the remainder to be 0. Then state the conclusion.

p(a) = 0 iff (x - a)is a factor of p(x)

Examiner insight

“A small number of candidates incorrectly used the factor theorem with x = 2.” “Some candidates used long or synthetic division and this was acceptable if done correctly and a remainder of 0 was indicated in some way.”

Worked example 11 mark

The polynomial p is given by p(x) = 4x³ - 5x² - 7x + 2. Show that x - 2 is a factor of p(x).

  1. 1

    p(2) = 4(2)³ - 5(2)² - 7(2) + 2

  2. 2

    = 32 - 20 - 14 + 2 = 0, so x - 2 is a factor [1]

3. Finding an unknown constant

A question may give a polynomial with an unknown constant and one fact about it: a factor, a remainder, or p(a) = 0. Substitute into p and form an equation.

With two unknowns, use two facts to form simultaneous equations. If the constants are stated to be integers, a non-integer answer signals an error.

Examiner insight

“Poor arithmetic manipulation let down many who had correct equations but did not solve the simultaneous equations.” “Obtaining non integer solutions should alert candidates to errors having been made and thus giving an opportunity to check their work.”

Worked example 14 marks

The polynomial p is such that p(x) = ax³ + 4x² - ax - 28, where a is a positive constant. It is given that p(a) = 0. Find the value of a.

  1. 1

    a × a³ + 4a² - a × a - 28 = 0, so a⁴ + 4a² - a² - 28 = 0 [1]

  2. 2

    (a² + 7)(a² - 4) = 0 [1]

  3. 3

    a² = 4 [1]

  4. 4

    a = 2 only, because a is positive [1]

4. Finding the quadratic factor

Once x - a is a factor, divide it into p(x) to get a quadratic factor. Use algebraic long division or observation.

Then factorise the quadratic. Write the answer as a product of factors, and check by multiplying out.

Examiner insight

“It was expected that candidates give their final answer as the product of two factors and not just state the quadratic factor.” “Candidates using synthetic division gave answers that were too great by a factor of 2.”

Worked example 13 marks

The polynomial p is given by p(x) = 4x³ - 5x² - 7x + 2, and x - 2 is a factor. Write p(x) as a product of linear factors.

  1. 1

    divide by x - 2: taking away 4x²(x - 2) leaves 3x² - 7x + 2; taking away 3x(x - 2) leaves -x + 2; taking away -1(x - 2) leaves 0 [1]

  2. 2

    p(x) = (x - 2)(4x² + 3x - 1) [1]

  3. 3

    4x² + 3x - 1 = (4x - 1)(x + 1)

  4. 4

    p(x) = (x - 2)(4x - 1)(x + 1) [1]

5. Choosing a factor to test

With no factor given, find a root first. Start with values of x that are factors of the constant term. For a constant term of 2, test x = 1, -1, 2 and -2.

Show each trial: evaluate p and state the result. Write each factor as ax + b with integer a and b, so 2x - 1, not x - 1/2. The answer is a product of linear factors.

Examiner insight

“It was not enough to make a statement such as ‘using hit and trial’ or ‘using trial and error method’ and then show that x – 4 or 2x – 1 was a factor using the factor theorem, or similar, as this did not justify the initial selection of these factors.” “Algebraic factors should be written in the form ax + b where a and b are integers.”

Worked example 14 marks

The polynomial p is given by p(x) = 6x³ - x² - 5x + 2. Write p(x) as a product of linear factors.

  1. 1

    test factors of 2: p(1) = 6 - 1 - 5 + 2 = 2, p(-1) = -6 - 1 + 5 + 2 = 0, so x + 1 is a factor [1]

  2. 2

    p(x) = (x + 1)(6x² - 7x + 2) [2]

  3. 3

    6x² - 7x + 2 = (3x - 2)(2x - 1)

  4. 4

    p(x) = (x + 1)(3x - 2)(2x - 1) [1]

6. Solving a cubic equation

To solve a cubic equation, make one side 0 and factorise. Set each factor equal to 0.

Solve the quadratic factor as well as the linear one, and show the working. Roots taken from a calculator, with no method shown, do not earn full marks. The answer is the values of x, not the factors.

Examiner insight

“A few candidates omitted to state x = –2 as a solution.” “A small number of candidates misinterpreted the question and gave (x + 2)(2x –1)(2x +1) = 0 as their answer.”

Worked example 13 marks

A polynomial p is given by p(x) = 4x³ + 8x² - x - 2. Solve the equation p(x) = 0.

  1. 1

    use the factor x + 2: (x + 2)(4x² - 1) = 0 [1]

  2. 2

    4x² - 1 = (2x + 1)(2x - 1), so x = ±1/2 [1]

  3. 3

    x = -2, 1/2, -1/2 [1]

7. How many roots?

After the linear factor is found, the quadratic factor decides whether the equation has more roots.

To show it has none, evaluate the discriminant b² - 4ac of the quadratic factor. State that it is negative, so there are no real roots.

b² - 4ac < 0 ⇒ no real roots

Examiner insight

“In both cases, candidates needed to state a conclusion such as ‘no solutions’, ‘no real roots’ or ‘the square root of a negative is not real’ or similar, based on their correct work.” “Some relied on a simple statement for the final mark but they did not show that the discriminant is less than zero.”

Worked example 12 marks

The polynomial p is such that p(x) = 2x³ + 4x² - 2x - 28 = (x - 2)(2x² + 8x + 14). Hence show that the equation p(x) = 0 has exactly one root.

  1. 1

    x - 2 = 0 gives the root x = 2

  2. 2

    discriminant of 2x² + 8x + 14 = 8² - 4 × 2 × 14 = 64 - 112 = -48 [1]

  3. 3

    discriminant is negative, so no real roots from the quadratic: x = 2 is the only root [1]

8. A cubic in disguise

A ‘Hence solve’ part may give an equation in u² with the same coefficients as p(x). Replace x with u² in the factors of p(x) and set each factor equal to 0.

A positive value of u² gives two values of u, one positive and one negative. A negative value of u² gives no real value of u. Leave a surd such as √2 exact.

u² = k, k > 0 ⇒ u = ±√k

Examiner insight

“A few candidates omitted x = –0.5 as they forgot to include the negative square root.”

Worked example 12 marks

The polynomial p is given by p(x) = 4x³ - 5x² - 7x + 2 = (x - 2)(4x - 1)(x + 1). Hence solve the equation 4u⁶ - 5u⁴ - 7u² + 2 = 0.

  1. 1

    the equation is p(u²) = 0, so (u² - 2)(4u² - 1)(u² + 1) = 0

  2. 2

    u² = 2 gives u = ±√2; u² = 1/4 gives u = ±1/2 [1]

  3. 3

    u² = -1 gives no real value, so u = ±√2, ±1/2 only [1]

Recap

  • Remainder theorem: dividing p(x) by x - a leaves p(a). For ax + b use x = -b/a.
  • Factor theorem: p(a) = 0 means x - a is a factor. Show the remainder is 0 and say so.
  • An unknown constant: substitute, form an equation, solve. Two unknowns give simultaneous equations.
  • Divide by the known factor to find the quadratic factor. Write the answer as a product, and multiply out to check.
  • Write factors as ax + b with integers. Show every trial that picks a factor.
  • Solve every factor set to 0. For no more roots, show the discriminant is negative.
  • An equation in u²: replace x with u². Each positive value gives ± a square root; a negative value gives nothing.

End-of-chapter exercise

Test yourself on the whole chapter. Work through these before moving on.

  1. Find the remainder when x³ - 2x + 5 is divided by x - 3.2 marks
  2. Do not use a calculator. Find the remainder when 4x³ - 3x² + x + 6 is divided by 2x + 1.3 marks
  3. Show that x + 3 is a factor of x³ + 2x² - 5x - 6.2 marks
  4. Given that x - 2 is a factor of x³ + kx² + x - 6, find the value of k.2 marks
  5. The polynomial p(x) = 2x³ + ax² + bx - 6 has a factor x - 1 and leaves a remainder of 12 when divided by x + 1. Find a and b.5 marks
  6. Given that x - 3 is a factor of p(x) = x³ - 2x² - 5x + 6, write p(x) as a product of a linear factor and a quadratic factor.3 marks
  7. Do not use a calculator. Given that 2x + 1 is a factor of p(x) = 6x³ + x² - 5x - 2, write p(x) as a product of three linear factors.4 marks
  8. Do not use a calculator. Solve 2x³ - 3x² - 8x - 3 = 0.5 marks
  9. Do not use a calculator. The polynomial p(x) = 4x³ + kx² - 23x + 6 leaves a remainder of -18 when divided by x - 1. (a) Find the value of k. (b) Write p(x) as a product of linear factors. Show the trials you use to find the first factor. (c) Hence solve the equation 4u⁶ + ku⁴ - 23u² + 6 = 0, giving exact answers.9 marks
  10. Do not use a calculator. A cuboid has width x cm, length (x + 3) cm and height (2x - 1) cm. Its volume is 55 cm³. (a) Show that 2x³ + 5x² - 3x - 55 = 0. (b) Show that 2x - 5 is a factor of 2x³ + 5x² - 3x - 55. (c) Show that the equation in part (a) has exactly one real root, and find the dimensions of the cuboid.9 marks

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