Factor Theorem

The factor theorem is a fundamental concept in algebra that helps determine if a linear expression is a factor of a polynomial, making factorisation of polynomials easier. It establishes a connection between the factors of the polynomial and the roots of the polynomial. The Factor Theorem is essential for factorising cubic and higher-degree polynomials that are difficult to factorise by grouping or splitting the middle term. In this guide, we will learn the factor theorem, its statement, and related examples to build a strong understanding of the concept.

Table of Contents

• What is Factor Theorem
• Solved Examples on Factor Theorem

What is Factor Theorem

Statement: Let p(x) be a polynomial of degree n ≥ 1 and ‘a’ be a real number. (x - a) is a factor of p(x) if and only if p(a) = 0.

Proof:

Let p(x) be a polynomial with degree n ≥ 1 and ‘a’ be a real number. Let (x - a) divide p(x).
To prove that if p(a) = 0, then (x - a) is a factor.
By the division algorithm, p(x) = q(x)(x-a) + r, where q(x) is the quotient and r is the remainder.
By the remainder theorem, we know that p(a) = r.
Here we assume that p(a) = 0 ⇒ r = 0.
Hence, p(x) = q(x)(x - a)
So, (x - a) is a factor of p(x).

Conversely,
To prove that if (x-a) is a factor, then p(a) = 0.
If (x-a) is a factor of p(x), we can write p(x) as p(x) = (x-a)q(x).
Substituting x = a, we get p(a) = (a-a)q(x) = 0.
Hence proved.

NOTE: 

• (ax − b) is a factor ⇔ p(b/a) = 0

• (ax + b) is a factor ⇔ p(−b/a) = 0

• If p(x) = (x − r₁)(x − r₂)...(x − rₙ), then r₁, r₂, ..., rₙ are the zeroes.
  
  

Solved Examples on Factor Theorem

Example 1: Is (x + 2) a factor of p(x) = 2x³ + 5x² − x − 6?
Solution: By the factor theorem, (x - a) is a factor of p(x) if and only if p(a) = 0.
Substituting x = -2 in 2x³ + 5x² − x − 6, we get,
p(-2) = 2(-2)³ + 5(-2)² − (-2) − 6 =  0
Since p(-2) = 0, (x + 2) is a factor of 2x³ + 5x² − x − 6.

Example 2: Factorise p(x) = x³ − 6x² + 11x − 6 completely.
Solution: By the factor theorem, (x - a) is a factor of p(x) if and only if p(a) = 0.
Find a factor by trial and error.
Let x = 1, p(1) = 1³ − 6(1²) + 11 − 6 = 0
∴(x - 1) is a factor of x³ − 6x² + 11x − 6. Divide x³ − 6x² + 11x − 6 by (x - 1) using the long division method. We get x² − 5x + 6 as the quotient.
Now factorise x² − 5x + 6 by middle term splitting: x² − 5x + 6 = (x − 2)(x − 3).
∴ p(x) =  x³ − 6x² + 11x − 6 = (x - 1)(x - 2)(x - 3)

Example 3: If both (x − 1) and (x + 2) are factors of p(x) = x³ + ax² + bx + 6, find a and b.
Solution: By the factor theorem, (x - a) is a factor of p(x) if and only if p(a) = 0.
Substituting x = 1 in x³ + ax² + bx + 6, we get p(1) = 1 + a + b + 6 = a + b + 7
Substituting x = -2 in x³ + ax² + bx + 6, we get p(-2) = -8 + 4a - 2b + 6 = 4a - 2b - 2.
Since both (x − 1) and (x + 2) are factors of p(x), p(1) = 0 and p(-2) = 0
⇒ a+b = -7 and 4a - 2b = 2. Solving these equations, we get a = -2 and b = -5.