Polynomial Division and Remainder Theorem

Polynomial Long Division

Similar to long division of numbers

Process:

1. Divide leading terms
2. Multiply divisor by quotient term
3. Subtract
4. Bring down next term
5. Repeat until degree of remainder < degree of divisor

Example 1: Basic Division

Divide: (x² + 5x + 6) ÷ (x + 2)

Step-by-step:

         x + 3
       ________
x + 2 | x² + 5x + 6
        x² + 2x
        ________
             3x + 6
             3x + 6
             ______
                  0

Answer: x + 3 with remainder 0

Check: (x + 2)(x + 3) = x² + 5x + 6 ✓

Example 2: With Remainder

Divide: (x³ + 2x² - 5x + 1) ÷ (x - 1)

         x² + 3x - 2
       ______________
x - 1 | x³ + 2x² - 5x + 1
        x³ - x²
        __________
            3x² - 5x
            3x² - 3x
            ________
                -2x + 1
                -2x + 2
                _______
                    -1

Answer: x² + 3x - 2 with remainder -1

Written as: x² + 3x - 2 + (-1)/(x - 1)

Example 3: Missing Terms

Divide: (x³ - 8) ÷ (x - 2)

Insert placeholders for missing terms:

x³ + 0x² + 0x - 8

Divide:

         x² + 2x + 4
       ______________
x - 2 | x³ + 0x² + 0x - 8
        x³ - 2x²
        __________
            2x² + 0x
            2x² - 4x
            ________
                 4x - 8
                 4x - 8
                 ______
                      0

Answer: x² + 2x + 4

Synthetic Division

Shortcut for dividing by linear expressions (x - c)

Faster and less error-prone than long division

Only works when divisor is (x - c)

How to Use Synthetic Division

Divide (x³ + 2x² - 5x + 1) by (x - 1):

Step 1: Write c (from x - c) and coefficients

1 | 1   2  -5   1

Step 2: Bring down first coefficient

1 | 1   2  -5   1
  |_______________
    1

Step 3: Multiply and add repeatedly

1 | 1   2  -5   1
  |     1   3  -2
  |_______________
    1   3  -2  -1

Answer: Quotient: x² + 3x - 2, Remainder: -1

Example 1: Synthetic Division

Divide: (2x³ - 3x² + 4x - 5) ÷ (x + 2)

Note: x + 2 = x - (-2), so c = -2

-2 | 2  -3   4  -5
   |    -4  14 -36
   |________________
     2  -7  18 -41

Answer: 2x² - 7x + 18 with remainder -41

Example 2: Zero Remainder

Divide: (x³ - 6x² + 11x - 6) ÷ (x - 1)

1 | 1  -6  11  -6
  |     1  -5   6
  |_______________
    1  -5   6   0

Answer: x² - 5x + 6, remainder 0

Since remainder is 0, (x - 1) is a factor!

Remainder Theorem

When polynomial P(x) is divided by (x - c), the remainder is P(c)

Use: Evaluate polynomial without dividing

Example 1: Find Remainder

Find remainder when x³ - 4x² + 3x - 2 is divided by (x - 2)

Use Remainder Theorem:

P(2) = 2³ - 4(2)² + 3(2) - 2
     = 8 - 16 + 6 - 2
     = -4

Answer: Remainder is -4

Example 2: Using Synthetic Division to Verify

Same problem: (x³ - 4x² + 3x - 2) ÷ (x - 2)

2 | 1  -4   3  -2
  |     2  -4  -2
  |_______________
    1  -2  -1  -4

Remainder: -4 (matches Remainder Theorem!)

Factor Theorem

Extension of Remainder Theorem:

(x - c) is a factor of P(x) if and only if P(c) = 0

Use: Test if (x - c) divides polynomial evenly

Example 1: Test for Factor

Is (x - 3) a factor of x³ - 2x² - 5x + 6?

Evaluate P(3):

P(3) = 3³ - 2(3)² - 5(3) + 6
     = 27 - 18 - 15 + 6
     = 0

Since P(3) = 0, yes (x - 3) is a factor!

Example 2: Find Factor

Show (x + 2) is a factor of x³ + 3x² - 4

Evaluate P(-2):

P(-2) = (-2)³ + 3(-2)² - 4
      = -8 + 12 - 4
      = 0

P(-2) = 0, so (x + 2) is a factor

Factoring Using Division

Once you find one factor, divide to find remaining factors

Example: Complete Factorization

Factor completely: x³ - 6x² + 11x - 6

Step 1: Test possible factors (try x = 1)

P(1) = 1 - 6 + 11 - 6 = 0

So (x - 1) is a factor

Step 2: Divide by (x - 1) using synthetic division

1 | 1  -6  11  -6
  |     1  -5   6
  |_______________
    1  -5   6   0

Quotient: x² - 5x + 6

Step 3: Factor quotient

x² - 5x + 6 = (x - 2)(x - 3)

Complete factorization:

x³ - 6x² + 11x - 6 = (x - 1)(x - 2)(x - 3)

Rational Root Theorem

Helps find possible rational zeros

If p/q is a rational zero of polynomial with integer coefficients:

• p is a factor of the constant term
• q is a factor of the leading coefficient

Example: Find Possible Rational Zeros

For P(x) = 2x³ - x² - 7x + 6

Constant term: 6 Factors of 6: ±1, ±2, ±3, ±6

Leading coefficient: 2 Factors of 2: ±1, ±2

Possible rational zeros:

±1, ±2, ±3, ±6, ±1/2, ±3/2

Test these to find actual zeros

Dividing by Quadratic

Can use long division for divisors of higher degree

Example: Divide by Quadratic

(x⁴ - 16) ÷ (x² - 4)

Recognize difference of squares:

x⁴ - 16 = (x²)² - 4²
        = (x² - 4)(x² + 4)

So (x⁴ - 16) ÷ (x² - 4) = x² + 4

Applications

Finding zeros: Use division to reduce degree after finding one zero

Graphing: Factor to find x-intercepts

Simplifying rational expressions: Cancel common factors

Engineering: Signal processing, control systems

Example: Find All Zeros

Find all zeros of P(x) = x³ - 3x² - 10x + 24

Test x = 2:

P(2) = 8 - 12 - 20 + 24 = 0 ✓

Divide by (x - 2):

2 | 1  -3  -10   24
  |     2   -2  -24
  |________________
    1  -1  -12    0

Factor x² - x - 12:

= (x - 4)(x + 3)

All zeros: x = 2, 4, -3

Practice