Chapter 2: Polynomials

Factor Theorem

The Connection to Zeroes

Factor Theorem: (xa)(x - a) is a factor of p(x)p(x) if and only if p(a)=0p(a) = 0.

In other words: Zeroes give us factors!


Using Factor Theorem

Example 1: Is (x+2)(x + 2) a factor of x3+3x2+5x+6x^3 + 3x^2 + 5x + 6?

Step 1: Find the zero of (x+2)(x + 2) x+2=0x=2x + 2 = 0 \Rightarrow x = -2

Step 2: Calculate p(2)p(-2) p(2)=(2)3+3(2)2+5(2)+6p(-2) = (-2)^3 + 3(-2)^2 + 5(-2) + 6 =8+1210+6= -8 + 12 - 10 + 6 =0= 0

Step 3: Conclude Since p(2)=0p(-2) = 0, yes (x+2)(x + 2) IS a factor! ✓


Example 2: Find kk if (x1)(x - 1) is factor of x2+x+kx^2 + x + k

Step 1: If (x1)(x - 1) is factor, then p(1)=0p(1) = 0

Step 2: Calculate p(1)p(1) p(1)=(1)2+(1)+k=0p(1) = (1)^2 + (1) + k = 0 1+1+k=01 + 1 + k = 0 k=2k = -2

Answer: k=2k = -2


Factorizing Using Factor Theorem

Steps:

  1. List possible factors (from constant term)
  2. Test each using Factor Theorem
  3. When you find one that works, divide to get quotient
  4. Factor the quotient if possible

Example: Factorize x32x2x+2x^3 - 2x^2 - x + 2

Step 1: Possible zeroes: ±1, ±2

Step 2: Test x=1x = 1: p(1)=121+2=0p(1) = 1 - 2 - 1 + 2 = 0 ✓ Found one!

Step 3: (x1)(x - 1) is a factor

Step 4: Divide to get quotient: x2x2x^2 - x - 2

Step 5: Factor quotient: (x2)(x+1)(x - 2)(x + 1)

Final Answer: (x1)(x2)(x+1)(x - 1)(x - 2)(x + 1)

Visualizer
Step-by-Step Solution
1
Is (x+2) a factor of x³+3x²+5x+6?
We need to check using Factor Theorem
2
Find zero: x + 2 = 0 → x = -2
The value to substitute
3
p(-2) = (-2)³ + 3(-2)² + 5(-2) + 6
Substitute x = -2
4
= -8 + 12 - 10 + 6 = 0
Calculate step by step
5
Since p(-2) = 0, YES its a factor!
Factor theorem confirmed!

Interactive Visualization