Factorization • Topic 1 of 3

Factor Theorem

What is the Factor Theorem? The Factor Theorem is a powerful tool that helps us find factors of a polynomial without performing long division. It states:

If p(x) is a polynomial and p(a) = 0, then (x − a) is a factor of p(x).

Conversely, if (x − a) is a factor of p(x), then p(a) = 0.

Why is this important?

  • It connects the zeros of a polynomial with its factors
  • To check if (x − a) is a factor, simply evaluate p(a)
  • If p(a) = 0, then (x − a) divides p(x) exactly
  • Saves time compared to polynomial division

Real-life analogy: Think of a polynomial as a lock and a number 'a' as a key. If p(a) = 0, the key "fits" perfectly and (x − a) is the "key shape" that opens the lock (divides the polynomial). If p(a) ≠ 0, the key doesn't fit.

Key relationships: | If this is a factor | Then this is a zero | And p( ) = 0 | |---------------------|---------------------|---------------| | (x − 2) | x = 2 | p(2) = 0 | | (x + 3) | x = −3 | p(−3) = 0 | | (2x − 1) | x = 1/2 | p(1/2) = 0 |

┌─────────────────────────────────────────────────────────────┐
│              FACTOR THEOREM - VISUAL REPRESENTATION          │
└─────────────────────────────────────────────────────────────┘

FACTOR THEOREM FLOWCHART:

    Start: Given polynomial p(x) and value 'a'
                  │
                  ▼
           Calculate p(a)
                  │
        ┌─────────┴─────────┐
        ▼                   ▼
    p(a) = 0            p(a) ≠ 0
        │                   │
        ▼                   ▼
    (x - a) IS a        (x - a) is NOT
    factor of p(x)      a factor of p(x)


VERIFICATION WITH EXAMPLE:

    p(x) = x² - 5x + 6
    
    Check if (x - 2) is a factor:
    
    p(2) = (2)² - 5(2) + 6
         = 4 - 10 + 6
         = 0  ✓
    
    Therefore, (x - 2) is a factor!
    
    Check if (x - 3) is a factor:
    
    p(3) = (3)² - 5(3) + 6
         = 9 - 15 + 6
         = 0  ✓
    
    Therefore, (x - 3) is also a factor!
    
    p(x) = (x - 2)(x - 3)


FACTOR THEOREM FOR DIFFERENT FORMS OF FACTORS:

    Factor Form          Zero to Test        Calculation
    ─────────────────────────────────────────────────────
    (x - a)              x = a               p(a)
    (x + a)              x = -a              p(-a)
    (ax - b)             x = b/a             p(b/a)
    (ax + b)             x = -b/a            p(-b/a)


TESTING POSSIBLE FACTORS:

    For p(x) = x³ - 6x² + 11x - 6
    
    Try x = 1: p(1) = 1 - 6 + 11 - 6 = 0 → (x-1) is a factor
    Try x = 2: p(2) = 8 - 24 + 22 - 6 = 0 → (x-2) is a factor
    Try x = 3: p(3) = 27 - 54 + 33 - 6 = 0 → (x-3) is a factor
    
    p(x) = (x-1)(x-2)(x-3)
Solved Examples
1
Worked Example
Use the Factor Theorem to determine whether (x − 3) is a factor of p(x) = \(x^{3} - 4x^{2} + x + 6\).
Solution
  1. Step 1: To check (x − 3), evaluate p(3)
  2. Step 2: p(3) = (3)\(^{3} - 4\)(3)\(^{2} + 3 + 6\) = 27 − 36 + 3 + 6
  3. Step 3: = 27 − 36 = −9, −9 + 3 = −6, −6 + 6 = 0
  4. Step 4: Since p(3) = 0, (x − 3) is a factor

Answer: Yes, (x − 3) is a factor

2
Worked Example
Find the value of k if (x − 2) is a factor of p(x) = \(x^{3} - 3x^{2} + kx - 4\).
Solution
  1. Step 1: If (x − 2) is a factor, then p(2) = 0
  2. Step 2: p(2) = (2)\(^{3} - 3\)(2)\(^{2} + k\)(2) − 4 = 8 − 12 + 2k − 4
  3. Step 3: Simplify: 8 − 12 = −4, −4 − 4 = −8, so p(2) = 2k − 8
  4. Step 4: Set p(2) = 0: 2k − 8 = 0 → 2k = 8 → k = 4

Answer: k = 4

3
Worked Example
Without actual division, prove that (2x + 1) is a factor of p(x) = \(4x^{3} + 4x^{2} - x - 1\).
Solution
  1. Step 1: For (2x + 1), set 2x + 1 = 0 → x = −1/2
  2. Step 2: Evaluate p(−1/2): p(−1/2) = 4(−1/2)\(^{3} + 4\)(−1/2)\(^{2}\) − (−1/2) − 1
  3. Step 3: (−1/2)\(^{3}\) = −1/8 → 4 × (−1/8) = −4/8 = −1/2
  4. Step 4: (−1/2)\(^{2}\) = 1/4 → 4 × 1/4 = 1
  5. Step 5: −(−1/2) = +1/2
  6. Step 6: Sum: −1/2 + 1 + 1/2 − 1 = (−1/2 + 1/2) + (1 − 1) = 0 + 0 = 0

Answer: Since p(−1/2) = 0, (2x + 1) is a factor

Key Points

  • Factor Theorem: p(a) = 0 ⇔ (x − a) is a factor of p(x)
  • To test (x − a), evaluate p(a)
  • To test (ax − b), evaluate p(b/a)
  • Factor Theorem saves time compared to polynomial division
  • It helps find factors of cubic and higher-degree polynomials
  • Works for any polynomial with real coefficients
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.$(x-a)$ is a factor of $p(x)$ if and only if:
Explanation: Factor theorem.
Q2.If $p(2)=0$, then a factor of $p(x)$ is:
Explanation: $(x-2)$ is a factor.
Q3.The factor theorem is a special case of the:
Explanation: Remainder $0\Rightarrow$ factor.
Q4.If $(x+1)$ is a factor of $p(x)$, then:
Explanation: $(x+1)=0\Rightarrow x=-1$.
Practice more MCQs → 📝 Assignment · 35 marks