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Polynomial Division and Remainder Theorem

Understanding polynomial long division, synthetic division, and the Remainder and Factor Theorems

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Polynomial Division and Remainder Theorem

Why Divide Polynomials?

Polynomial division helps us:

  • Factor polynomials
  • Find zeros of polynomials
  • Simplify rational expressions
  • Analyze polynomial behavior

Long Division of Polynomials

The process is similar to long division with numbers.

Steps for Polynomial Long Division

  1. Arrange both polynomials in descending order of powers
  2. Divide the leading term of the dividend by the leading term of the divisor
  3. Multiply the result by the divisor
  4. Subtract from the dividend
  5. Bring down the next term
  6. Repeat until the degree of the remainder is less than the degree of the divisor

Example Format

dividenddivisor=quotient+remainderdivisor\frac{\text{dividend}}{\text{divisor}} = \text{quotient} + \frac{\text{remainder}}{\text{divisor}}

Synthetic Division

Synthetic division is a shortcut for dividing by linear factors of the form (x−c)(x - c).

When to Use Synthetic Division

✓ Divisor is (x−c)(x - c) (linear with leading coefficient 1) ✗ Cannot use for divisors like (2x−3)(2x - 3) or (x2+1)(x^2 + 1)

Steps for Synthetic Division

  1. Write cc (from x−cx - c) outside the box
  2. Write coefficients of the dividend inside
  3. Bring down the first coefficient
  4. Multiply by cc, add to next coefficient
  5. Repeat across all coefficients
  6. Last number is the remainder

Example Setup

Dividing f(x)f(x) by (x−c)(x - c):

c & a_n & a_{n-1} & \cdots & a_0 \\ & & ca_n & \cdots & \\ \hline & a_n & b_{n-1} & \cdots & r \end{array}$$ The bottom row gives quotient coefficients and remainder $r$. ## The Remainder Theorem **Remainder Theorem**: When a polynomial $f(x)$ is divided by $(x - c)$, the remainder is $f(c)$. $$f(x) = (x - c) \cdot q(x) + r$$ where $r = f(c)$ ### Why It Matters - Quick way to find remainders without full division - Just evaluate $f(c)$! ## The Factor Theorem **Factor Theorem**: $(x - c)$ is a factor of $f(x)$ if and only if $f(c) = 0$. In other words: - If $f(c) = 0$, then $(x - c)$ divides evenly into $f(x)$ - If the remainder is 0, then $c$ is a zero of $f(x)$ ### Applications 1. **Testing for factors**: Check if $f(c) = 0$ 2. **Finding zeros**: If $(x - c)$ is a factor, then $c$ is a zero 3. **Factoring**: Use known zeros to write factored form ## Rational Zero Theorem If $f(x) = a_nx^n + ... + a_0$ has integer coefficients, then any rational zero $\frac{p}{q}$ must have: - $p$ divides the constant term $a_0$ - $q$ divides the leading coefficient $a_n$ This gives us a list of **possible** rational zeros to test. ## Complete Factorization Strategy 1. Use Rational Zero Theorem to list possible zeros 2. Test candidates using synthetic division or direct evaluation 3. Once you find a zero $c$, factor out $(x - c)$ 4. Repeat on the quotient polynomial 5. Factor completely over the real numbers

📚 Practice Problems

1Problem 1medium

❓ Question:

Use synthetic division to divide f(x)=2x3−5x2+x+2f(x) = 2x^3 - 5x^2 + x + 2 by (x−2)(x - 2).

💡 Show Solution

Solution:

Step 1: Set up synthetic division with c=2c = 2.

The coefficients of f(x)f(x) are: 2,−5,1,22, -5, 1, 2

2 & 2 & -5 & 1 & 2 \\ & & 4 & -2 & -2 \\ \hline & 2 & -1 & -1 & 0 \end{array}$$ Step 2: Perform the operations. - Bring down 2 - Multiply: $2 \times 2 = 4$, add: $-5 + 4 = -1$ - Multiply: $2 \times (-1) = -2$, add: $1 + (-2) = -1$ - Multiply: $2 \times (-1) = -2$, add: $2 + (-2) = 0$ Step 3: Interpret the result. The bottom row gives: quotient coefficients $2, -1, -1$ and remainder $0$. **Quotient**: $2x^2 - x - 1$ **Remainder**: $0$ Therefore: $$f(x) = (x - 2)(2x^2 - x - 1)$$ Since the remainder is 0, $(x - 2)$ is a factor and $x = 2$ is a zero of $f(x)$. **Answer:** Quotient: $2x^2 - x - 1$, Remainder: $0$

2Problem 2easy

❓ Question:

Use the Remainder Theorem to find the remainder when f(x)=x4−3x3+2x−5f(x) = x^4 - 3x^3 + 2x - 5 is divided by (x+1)(x + 1).

💡 Show Solution

Solution:

The Remainder Theorem states that the remainder when dividing f(x)f(x) by (x−c)(x - c) is f(c)f(c).

Step 1: Identify cc.

We're dividing by (x+1)=(x−(−1))(x + 1) = (x - (-1)), so c=−1c = -1.

Step 2: Evaluate f(−1)f(-1).

f(−1)=(−1)4−3(−1)3+2(−1)−5f(-1) = (-1)^4 - 3(-1)^3 + 2(-1) - 5

=1−3(−1)−2−5= 1 - 3(-1) - 2 - 5

=1+3−2−5= 1 + 3 - 2 - 5

=−3= -3

Answer: The remainder is −3-3.

Note: We found this without doing any division!

3Problem 3hard

❓ Question:

Find all rational zeros of f(x)=2x3−x2−13x−6f(x) = 2x^3 - x^2 - 13x - 6 and factor completely.

💡 Show Solution

Solution:

Step 1: List possible rational zeros using the Rational Zero Theorem.

Factors of constant term (-6): ±1,±2,±3,±6\pm 1, \pm 2, \pm 3, \pm 6 Factors of leading coefficient (2): ±1,±2\pm 1, \pm 2

Possible rational zeros: ±1,±2,±3,±6,±12,±32\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}

Step 2: Test candidates.

Try x=−1x = -1: f(−1)=2(−1)3−(−1)2−13(−1)−6f(-1) = 2(-1)^3 - (-1)^2 - 13(-1) - 6 =−2−1+13−6=4≠0= -2 - 1 + 13 - 6 = 4 \neq 0 ✗

Try x=2x = 2: f(2)=2(2)3−(2)2−13(2)−6f(2) = 2(2)^3 - (2)^2 - 13(2) - 6 =16−4−26−6=−20≠0= 16 - 4 - 26 - 6 = -20 \neq 0 ✗

Try x=−2x = -2: f(−2)=2(−2)3−(−2)2−13(−2)−6f(-2) = 2(-2)^3 - (-2)^2 - 13(-2) - 6 =−16−4+26−6=0= -16 - 4 + 26 - 6 = 0 ✓

Step 3: Use synthetic division with x=−2x = -2.

-2 & 2 & -1 & -13 & -6 \\ & & -4 & 10 & 6 \\ \hline & 2 & -5 & -3 & 0 \end{array}$$ Quotient: $2x^2 - 5x - 3$ Step 4: Factor the quotient. $$2x^2 - 5x - 3 = (2x + 1)(x - 3)$$ Step 5: Write complete factorization. $$f(x) = (x + 2)(2x + 1)(x - 3)$$ **Zeros**: $x = -2, x = -\frac{1}{2}, x = 3$ **Answer:** Zeros are $-2, -\frac{1}{2}, 3$; Factored form: $(x + 2)(2x + 1)(x - 3)$

4Problem 4medium

❓ Question:

Use synthetic division to divide: (2x³ - 5x² + 3x - 7) ÷ (x - 2)

💡 Show Solution

Step 1: Set up synthetic division with divisor x - 2 (use 2): 2 | 2 -5 3 -7

Step 2: Bring down the first coefficient: 2 | 2 -5 3 -7 ↓ 2

Step 3: Multiply and add: 2 | 2 -5 3 -7 4 -2 2 ――――――――――――― 2 -1 1 -5

Step 4: Write the result: Quotient: 2x² - x + 1 Remainder: -5

Step 5: Express as quotient + remainder/divisor: (2x³ - 5x² + 3x - 7) = (x - 2)(2x² - x + 1) - 5

Answer: Quotient: 2x² - x + 1, Remainder: -5

5Problem 5hard

❓ Question:

Use the Remainder Theorem to find the remainder when f(x) = x⁴ - 3x³ + 2x - 5 is divided by (x + 1).

💡 Show Solution

Step 1: Recall the Remainder Theorem: When f(x) is divided by (x - c), the remainder is f(c)

Step 2: Identify c from (x + 1): x + 1 = x - (-1) So c = -1

Step 3: Evaluate f(-1): f(-1) = (-1)⁴ - 3(-1)³ + 2(-1) - 5 = 1 - 3(-1) - 2 - 5 = 1 + 3 - 2 - 5 = -3

Step 4: Conclusion: By the Remainder Theorem, the remainder is -3

Step 5: Verify with synthetic division if desired: -1 | 1 -3 0 2 -5 -1 4 -4 2 ―――――――――――――――――― 1 -4 4 -2 -3 ✓

Answer: -3

Explain using:

⚠️ Common Mistakes: Polynomial Division and Remainder Theorem

Avoid these 4 frequent errors

🌍 Real-World Applications: Polynomial Division and Remainder Theorem

See how this math is used in the real world

📝 Worked Example: Related Rates — Expanding Circle

Problem:

A stone is dropped into a still pond, creating a circular ripple. The radius of the ripple is increasing at a rate of 22 cm/s. How fast is the area of the circle increasing when the radius is 1010 cm?

2Write the relationship between variables
3Differentiate both sides with respect to time
4Substitute known values
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❓ Frequently Asked Questions

What is Polynomial Division and Remainder Theorem?▾
Understanding polynomial long division, synthetic division, and the Remainder and Factor Theorems
How can I study Polynomial Division and Remainder Theorem effectively?▾
Start by reading the study notes and working through the examples on this page. Then use the flashcards to test your recall. Practice with the 5 problems provided, checking solutions as you go. Regular review and active practice are key to retention.
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Polynomial Division and Remainder Theorem is part of the AP Precalculus course on Study Mondo, specifically in the Polynomial and Rational Functions section. You can explore the full course for more related topics and practice resources.
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Yes, this page includes 5 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.