Factor Theorem

Understand the Factor Theorem, how it connects zeros and factors of a polynomial, and learn to check and find factors using simple steps and examples.

1. Introduction

The Factor Theorem is a direct extension of the Remainder Theorem. It gives a powerful connection between the zeros of a polynomial and the factors of that polynomial. This theorem helps us check whether a given expression divides a polynomial and also assists in factorising polynomials.

It is one of the most important tools for polynomial factorisation.

2. Statement of the Factor Theorem

The Factor Theorem states:

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

In simple words, a number is a zero of the polynomial if and only if \(x - a\) is a factor.

So, zero ↔ factor.

3. How to Use the Factor Theorem

To check whether \(x - a\) is a factor of a polynomial:

  1. Find \(a\) from the divisor \(x - a\).
  2. Substitute \(x = a\) into \(p(x)\).
  3. If \(p(a) = 0\), then \(x - a\) is a factor.
  4. If \(p(a) \neq 0\), then it is not a factor.

3.1. Example of Steps

Check if \(x - 3\) is a factor of \(p(x) = x^2 - 5x + 6\).

Step 1: Here, \(a = 3\).

Step 2: Evaluate \(p(3)\):

\(p(3) = 3^2 - 5(3) + 6 = 9 - 15 + 6 = 0\)

Since \(p(3) = 0\), \(x - 3\) is a factor.

4. Finding Factors Using Factor Theorem

If we know a zero of the polynomial, we can immediately write one factor and then divide to find remaining factors.

4.1. Example

Given \(p(x) = x^2 - x - 6\), check if \(x - 3\) is a factor and then factorise the polynomial.

Step 1: Check \(p(3)\):

\(p(3) = 9 - 3 - 6 = 0\)

So, \(x - 3\) is a factor.

Step 2: Divide the polynomial by \(x - 3\):

The quotient is \(x + 2\).

Factorised form: \((x - 3)(x + 2)\)

5. Worked Examples

Here are some examples that use the Factor Theorem:

5.1. Example 1

Check if \(x + 2\) is a factor of \(p(x) = x^3 + 8\).

Divisor: \(x + 2 = x - (-2)\), so \(a = -2\).

\(p(-2) = (-2)^3 + 8 = -8 + 8 = 0\)

Yes, \(x + 2\) is a factor.

5.2. Example 2

Check if \(x - 1\) is a factor of \(p(x) = 2x^3 - 3x + 5\).

\(p(1) = 2(1)^3 - 3(1) + 5 = 2 - 3 + 5 = 4\)

No, \(x - 1\) is not a factor.

5.3. Example 3

Factorise \(p(x) = x^3 - 7x + 6\) using the Factor Theorem.

Try possible zeros: 1, 2, 3, -1, -2, -3.

\(p(1) = 1 - 7 + 6 = 0 → (x - 1)\) is a factor.

Divide to get quotient: \(x^2 + x - 6\).

Factor further: \((x + 3)(x - 2)\).

Complete factorisation: \((x - 1)(x + 3)(x - 2)\)

6. Common Mistakes

  • Incorrectly identifying the value of \(a\) from the divisor. For \(x + 3\), \(a = -3\).
  • Substitution errors while evaluating \(p(a)\).
  • Thinking any polynomial must have a factor of the form \(x - a\) (not always true).
  • Skipping division after finding one factor and assuming factorisation is complete.

7. Quick Practice

Use the Factor Theorem to answer the following:

  1. Check if \(x - 4\) is a factor of \(p(x) = x^3 - 4x^2 + 2x + 8\).
  2. Check if \(x + 1\) is a factor of \(p(x) = x^2 - 1\).
  3. Factorise \(p(x) = x^2 + x - 6\) using a suitable factor.
  4. Find a zero of \(p(x) = x^3 - 3x^2 - 4x + 12\).

8. Summary

  • Factor Theorem states that if \(p(a) = 0\), then \(x - a\) is a factor of \(p(x)\).
  • This theorem links zeros and factors of a polynomial.
  • It helps in checking factors and finding complete factorisations.
  • Always identify \(a\) correctly and substitute carefully.