Lesson 2.3: Polynomial Division & Remainder Theorem

Master Polynomial Division & Advanced Factoring

Dive deep into polynomial division techniques! Learn long division (A-APR.2), remainder theorem, factor theorem, and systematic approaches to factoring high-degree polynomials. Master the art of polynomial manipulation and zero finding.

Learning Objectives (A-APR.2, A-SSE.3)

Perform polynomial long division (A-APR.2)

Apply remainder theorem and factor theorem

Factor high-degree polynomials systematically

Use rational root theorem for zero finding

Solve polynomial equations by factoring

Verify solutions and check work

Core Concepts & Theoretical Foundation

Polynomial Long Division (A-APR.2)

Systematic division process:

1. Arrange terms: Both dividend and divisor in descending powers

2. Divide leading terms: Divide highest degree terms to get first quotient term

3. Multiply and subtract: Multiply divisor by quotient term, subtract from dividend

4. Bring down next term: Continue with next term of dividend

5. Repeat process: Continue until remainder degree is less than divisor degree

6. Result format: $\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}$

Remainder & Factor Theorems

Powerful tools for polynomial analysis:

• Remainder Theorem: If polynomial $f(x)$ is divided by $(x-a)$ , then remainder = $f(a)$

• Factor Theorem: If $f(a) = 0$ , then $(x-a)$ is a factor of $f(x)$

• Applications:

- Find remainders without performing division

- Test if $(x-a)$ is a factor by evaluating $f(a)$

- Find zeros by testing potential factors

• Verification: Always check that $f(a) = 0$ when claiming $(x-a)$ is a factor

High-Degree Polynomial Factoring (A-SSE.3)

Systematic factoring approach:

1. Factor out GCF: Remove greatest common factor from all terms

2. Apply rational root theorem: Test possible rational roots

3. Use factor theorem: If $f(a) = 0$ , then $(x-a)$ is a factor

4. Perform division: Divide by found factor to reduce degree

5. Repeat process: Continue factoring the reduced polynomial

6. Check completeness: Ensure all factors are prime (irreducible)

• Common patterns: Difference of squares, sum/difference of cubes, perfect squares

Detailed Worked Examples

Example 1: Polynomial Long Division (A-APR.2)

Divide $x^3 - 5x^2 + 7x - 3$ by $x - 2$ and verify using remainder theorem.

Step 1: Set up long division

$x - 2 \overline{)x^3 - 5x^2 + 7x - 3}$

Step 2: Divide leading terms

$\frac{x^3}{x} = x^2$ (first term of quotient)

Multiply: $x^2 \cdot (x - 2) = x^3 - 2x^2$

Subtract: $(x^3 - 5x^2) - (x^3 - 2x^2) = -3x^2$

Step 3: Continue division

Bring down +7x: $-3x^2 + 7x$

$\frac{-3x^2}{x} = -3x$ (second term of quotient)

Multiply: $-3x \cdot (x - 2) = -3x^2 + 6x$

Subtract: $(-3x^2 + 7x) - (-3x^2 + 6x) = x$

Step 4: Final step

Bring down -3: $x - 3$

$\frac{x}{x} = 1$ (third term of quotient)

Multiply: $1 \cdot (x - 2) = x - 2$

Subtract: $(x - 3) - (x - 2) = -1$ (remainder)

Step 5: Result

Quotient: $x^2 - 3x + 1$

Remainder: -1

Verification: $x^3 - 5x^2 + 7x - 3 = (x - 2)(x^2 - 3x + 1) - 1$

Step 6: Remainder theorem verification

Evaluate $f(2) = 2^3 - 5(2)^2 + 7(2) - 3 = 8 - 20 + 14 - 3 = -1$

Remainder theorem confirms: remainder = f(2) = -1 ✓

Key Insight: The remainder theorem provides a quick way to find remainders without performing the full division process. This is especially useful for checking work or finding specific remainder values.

Example 2: Factor Theorem Application

Factor the polynomial $f(x) = x^3 - 6x^2 + 11x - 6$ completely and solve $f(x) = 0$ .

Step 1: Apply rational root theorem

Possible rational roots: $\pm\frac{\text{factors of 6}}{\text{factors of 1}} = \pm1, \pm2, \pm3, \pm6$

Step 2: Test potential roots

$f(1) = 1 - 6 + 11 - 6 = 0$ → $(x-1)$ is a factor ✓

$f(2) = 8 - 24 + 22 - 6 = 0$ → $(x-2)$ is a factor ✓

$f(3) = 27 - 54 + 33 - 6 = 0$ → $(x-3)$ is a factor ✓

Step 3: Perform division to verify

Divide $f(x)$ by $(x-1)$ :

$\frac{x^3 - 6x^2 + 11x - 6}{x - 1} = x^2 - 5x + 6$

Factor the quadratic: $x^2 - 5x + 6 = (x-2)(x-3)$

Step 4: Complete factorization

$f(x) = (x-1)(x-2)(x-3)$

Step 5: Solve f(x) = 0

$(x-1)(x-2)(x-3) = 0$

Solutions: $x = 1, x = 2, x = 3$

Step 6: Verification

Check: $f(1) = f(2) = f(3) = 0$ ✓

All three roots are confirmed

Factorization Success: The polynomial factors completely into three linear factors, giving us three distinct real roots. This is a perfect example of how the factor theorem helps identify all factors systematically.

Example 3: Complex Polynomial Factoring

Factor $g(x) = x^4 - 5x^3 + 5x^2 + 5x - 6$ completely.

Step 1: Test rational roots

Possible roots: $\pm1, \pm2, \pm3, \pm6$

$g(1) = 1 - 5 + 5 + 5 - 6 = 0$ → $(x-1)$ is a factor ✓

$g(-1) = 1 + 5 + 5 - 5 - 6 = 0$ → $(x+1)$ is a factor ✓

Step 2: Divide by first factor

Divide by $(x-1)$ : $\frac{g(x)}{x-1} = x^3 - 4x^2 + x + 6$

Step 3: Continue factoring

Divide $x^3 - 4x^2 + x + 6$ by $(x+1)$ :

$\frac{x^3 - 4x^2 + x + 6}{x+1} = x^2 - 5x + 6$

Step 4: Factor remaining quadratic

$x^2 - 5x + 6 = (x-2)(x-3)$

Step 5: Complete factorization

$g(x) = (x-1)(x+1)(x-2)(x-3)$

This can also be written as: $g(x) = (x^2-1)(x-2)(x-3)$

Step 6: Verify all factors

Check: $g(1) = g(-1) = g(2) = g(3) = 0$ ✓

All four roots are confirmed

Systematic Approach: By testing rational roots and using polynomial division, we systematically reduced the 4th-degree polynomial to a product of linear factors. This method works for any polynomial with rational roots.

Example 4: Polynomial Equation Solving

Solve the equation $2x^3 - 7x^2 + 7x - 2 = 0$ .

Step 1: Apply rational root theorem

Possible rational roots: $\pm\frac{\text{factors of 2}}{\text{factors of 2}} = \pm1, \pm2, \pm\frac{1}{2}$

Step 2: Test potential roots

$f(1) = 2 - 7 + 7 - 2 = 0$ → $(x-1)$ is a factor ✓

$f(2) = 16 - 28 + 14 - 2 = 0$ → $(x-2)$ is a factor ✓

$f(\frac{1}{2}) = 2(\frac{1}{8}) - 7(\frac{1}{4}) + 7(\frac{1}{2}) - 2 = \frac{1}{4} - \frac{7}{4} + \frac{7}{2} - 2 = 0$ → $(x-\frac{1}{2})$ is a factor ✓

Step 3: Factor completely

Since we have three factors, the polynomial factors as:

$2x^3 - 7x^2 + 7x - 2 = 2(x-1)(x-2)(x-\frac{1}{2})$

Or equivalently: $2x^3 - 7x^2 + 7x - 2 = (x-1)(x-2)(2x-1)$

Step 4: Solve the equation

$(x-1)(x-2)(2x-1) = 0$

Solutions: $x = 1, x = 2, x = \frac{1}{2}$

Step 5: Verification

Check each solution:

• $x = 1: 2(1)^3 - 7(1)^2 + 7(1) - 2 = 0$ ✓

• $x = 2: 2(8) - 7(4) + 7(2) - 2 = 0$ ✓

• $x = \frac{1}{2}: 2(\frac{1}{8}) - 7(\frac{1}{4}) + 7(\frac{1}{2}) - 2 = 0$ ✓

Complete Solution: The cubic equation has three real roots: 1, 2, and 1/2. The factor theorem helped us identify all factors, leading to a complete factorization and solution set.

Advanced Techniques & Problem-Solving Strategies

Synthetic Division

Alternative method for dividing by linear factors:

• Setup: Write coefficients in a row, with the zero of the divisor on the left

• Process: Bring down first coefficient, multiply by zero, add to next coefficient

• Result: Last number is remainder, others are coefficients of quotient

• Advantage: Faster than long division for linear divisors

• Example: Dividing by $(x-2)$ , use zero = 2

Descartes' Rule of Signs

Estimating number of positive and negative roots:

• Positive roots: Count sign changes in f(x)

• Negative roots: Count sign changes in f(-x)

• Limitation: Gives maximum possible roots, not exact count

• Use: Helps narrow down which rational roots to test

Polynomial Identities

Useful factoring patterns:

• Difference of squares: $a^2 - b^2 = (a-b)(a+b)$

• Sum of cubes: $a^3 + b^3 = (a+b)(a^2-ab+b^2)$

• Difference of cubes: $a^3 - b^3 = (a-b)(a^2+ab+b^2)$

• Perfect squares: $a^2 \pm 2ab + b^2 = (a \pm b)^2$

Common Pitfalls & Error Prevention

Pitfall 1: Sign Errors in Long Division

Error: Forgetting to change signs when subtracting in long division.

Solution: Always distribute the negative sign when subtracting.

Pitfall 2: Incorrect Remainder Theorem Application

Error: Using the wrong value when applying remainder theorem.

Solution: For divisor $(x-a)$ , evaluate $f(a)$ , not $f(-a)$ .

Pitfall 3: Incomplete Factoring

Error: Stopping factorization before all factors are found.

Solution: Continue factoring until all factors are prime (irreducible).

Pitfall 4: Missing Rational Roots

Error: Not testing all possible rational roots from the rational root theorem.

Solution: Systematically test all possible rational roots.

Pitfall 5: Forgetting to Check Work

Error: Not verifying that found factors actually work.

Solution: Always substitute back to verify factorization is correct.

Comprehensive Practice Problems

Problem 1: Long Division

Divide $3x^3 - 2x^2 + 5x - 1$ by $x + 2$ .

Show Solution

Quotient: $3x^2 - 8x + 21$

Remainder: -43

Verification: $f(-2) = 3(-8) - 2(4) + 5(-2) - 1 = -43$

Problem 2: Factor Theorem

Show that $(x-3)$ is a factor of $x^3 - 7x^2 + 15x - 9$ .

Show Solution

Evaluate f(3): $27 - 63 + 45 - 9 = 0$

Since f(3) = 0: $(x-3)$ is a factor

Problem 3: Complete Factoring

Factor $x^4 - 10x^3 + 35x^2 - 50x + 24$ completely.

Show Solution

Test rational roots: x = 1, 2, 3, 4 all work

Complete factorization: $(x-1)(x-2)(x-3)(x-4)$

Problem 4: Equation Solving

Solve $x^3 - 8x^2 + 19x - 12 = 0$ .

Show Solution

Factor: $(x-1)(x-3)(x-4) = 0$

Solutions: x = 1, 3, 4

← Previous Lesson Next Lesson: Polynomial Applications →