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Teacher Guide: Synthetic Division

Learn a faster method for dividing polynomials by linear factors using synthetic division.

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10 questions on Polynomials. Students join with a name, you see everyone's score.

For Teachers

Learning Objectives
  • Perform synthetic division to divide polynomials by linear binomials
  • Set up synthetic division correctly, including zeros for missing terms
  • Interpret the result as quotient and remainder
  • Apply the Remainder Theorem to evaluate polynomials
  • Use synthetic division to test potential roots of polynomials
Prerequisites
  • • Understanding of polynomial terminology (degree, coefficient, terms)
  • • Polynomial long division
  • • Adding, subtracting, and multiplying integers
  • • Evaluating polynomials at specific values
Discussion Starters
  • 1. Why is synthetic division faster than polynomial long division?
  • 2. How can you tell if (x−c)(x - c) is a factor just from the synthetic division result?
  • 3. What happens if you try synthetic division with a divisor like (2x−1)(2x - 1)? Why doesn't it work directly?
  • 4. How does the Remainder Theorem let you evaluate polynomials without substituting?
Common Misconceptions

Thinking synthetic division works for any polynomial divisor

Remediation: Emphasize that it ONLY works for (x−c)(x - c) form where the coefficient of xx is 11. For (2x−1)(2x - 1), you'd need to factor out the 22 first or use long division.

Confusing which sign to use for cc

Remediation: Connect to finding roots: 'What value of xx makes the divisor zero?' For (x−3)(x - 3), it's x=3x = 3. For (x+3)(x + 3), it's x=−3x = -3.

Forgetting that the quotient's degree is one less than the dividend's

Remediation: Use the counting principle: a cubic divided by linear gives quadratic. The bottom row has one fewer number than coefficients in the dividend.

Differentiation Ideas

For Struggling Students:

  • • Use a synthetic division template with labeled columns
  • • Practice with polynomials that have all terms present first
  • • Color-code the multiply and add steps

For On-Level Students:

  • • Include problems with missing terms
  • • Practice with both (x−c)(x - c) and (x+c)(x + c) divisors
  • • Apply Remainder Theorem to evaluate polynomials

For Advanced Students:

  • • Use synthetic division to completely factor polynomials
  • • Explore connection to Horner's method for efficient evaluation
  • • Investigate what modifications allow division by (ax−b)(ax - b)
Standards Alignment
  • A-APR.D.6 (CCSS.MATH.CONTENT.HSA.APR.D.6)

    Rewrite simple rational expressions in different forms; write a(x)/b(x) in the form q(x) + r(x)/b(x)

  • A-APR.B.2 (CCSS.MATH.CONTENT.HSA.APR.B.2)

    Know and apply the Remainder Theorem

Lesson Resources
  • visualStep-by-Step Synthetic Division Tool

    Interactive tool showing each step of the algorithm

  • activityRoot Testing Challenge

    Use synthetic division to find roots of polynomials

  • worksheetPractice Problems

    Graduated difficulty from basic to complex divisions

Lesson Content

Everything students see: definition, examples, common mistakes, applications. Tap to open.

Definition

Synthetic division is a shortcut method for dividing a polynomial by a linear binomial of the form (x−c)(x - c).
Instead of using long division, we work only with the coefficients and perform simple arithmetic operations.
When to use synthetic division:
  • The divisor must be linear: (x−c)(x - c) or (x+c)(x + c)
  • The coefficient of xx in the divisor must be 11
The setup: To divide P(x)P(x) by (x−c)(x - c):
  1. 1.Write the value of cc (the root) on the left
  2. 2.Write the coefficients of P(x)P(x) in order
  3. 3.Include 00 for any missing terms
canan−1⋯a0\begin{array}{c|cccc} c & a_n & a_{n-1} & \cdots & a_0 \\ \end{array}

Worked Examples

Divide (x3−6x2+11x−6)(x^3 - 6x^2 + 11x - 6) by (x−2)(x - 2)
1

Identify the divisor value

(x−2)(x - 2) means c=2c = 2
2

Write coefficients in order

x3−6x2+11x−6x^3 - 6x^2 + 11x - 6 has coefficients: 1,−6,11,−61, -6, 11, -6 → 1−611−61 \quad -6 \quad 11 \quad -6
3

Set up the synthetic division

21−611−6\begin{array}{c|cccc} 2 & 1 & -6 & 11 & -6 \\ & & & & \\ \hline & & & & \end{array}
→ Ready to calculate
4

Bring down the first coefficient

21−611−61\begin{array}{c|cccc} 2 & 1 & -6 & 11 & -6 \\ & & & & \\ \hline & 1 & & & \end{array}
→ First coefficient: 11
5

Multiply and add for each column

2×1=22 \times 1 = 2, then −6+2=−4-6 + 2 = -4 2×(−4)=−82 \times (-4) = -8, then 11+(−8)=311 + (-8) = 3 2×3=62 \times 3 = 6, then −6+6=0-6 + 6 = 0 →
21−611−62−861−430\begin{array}{c|cccc} 2 & 1 & -6 & 11 & -6 \\ & & 2 & -8 & 6 \\ \hline & 1 & -4 & 3 & 0 \end{array}
6

Write the quotient and remainder

Bottom row: 1,−4,31, -4, 3 are quotient coefficients; 00 is remainder → Quotient: x2−4x+3x^2 - 4x + 3, Remainder: 00

Common Mistakes

Forgetting to include zeros for missing terms

Why it's wrong: Every power of xx from highest to constant must have a coefficient. Missing terms have coefficient 00.

Correct: For x4−16x^4 - 16, write: 1,0,0,0,−161, 0, 0, 0, -16 (zeros for x3x^3, x2x^2, and xx)

Using the wrong sign for c when dividing by (x+c)(x + c)

Why it's wrong: Synthetic division uses the root form (x−c)(x - c). When dividing by (x+2)(x + 2), you need c=−2c = -2.

Correct: (x+2)=(x−(−2))(x + 2) = (x - (-2)), so use c=−2c = -2 in the synthetic division setup

Adding when you should multiply, or vice versa

Why it's wrong: The pattern is: multiply by cc, then add to the next coefficient.

Correct: Always: bring down first coefficient, then repeat (multiply by cc, add to next coefficient)

Writing the quotient with the wrong degree

Why it's wrong: The quotient has one degree less than the original polynomial.

Correct: If dividing a cubic (x3x^3) by linear, the quotient is quadratic (x2x^2)

Why It Matters

Synthetic division is a powerful tool that saves time and reduces errors:
  • Speed: Much faster than polynomial long division
  • Simplicity: Only uses basic arithmetic (addition and multiplication)
  • Factoring: Helps find roots of polynomials quickly
  • Remainder Theorem: If you divide P(x)P(x) by (x−c)(x - c), the remainder equals P(c)P(c)
Real applications:
  • Engineering: Analyzing transfer functions in control systems
  • Computer graphics: Polynomial curve calculations
  • Finance: Modeling growth patterns with polynomial functions

Real World Applications

Engineering: Transfer Functions

Engineers use polynomial division when analyzing control systems. The transfer function of a system is often a ratio of polynomials.

Example:

If a system has transfer function x3−6x2+11x−6x−2\frac{x^3 - 6x^2 + 11x - 6}{x - 2}, synthetic division simplifies it to x2−4x+3x^2 - 4x + 3.

1Try It Yourself

A filter circuit has polynomial response P(x)=x3+2x2−5x−6P(x) = x^3 + 2x^2 - 5x - 6. You need to factor it by testing if x=2x = 2 is a root.

Use synthetic division to test if (x−2)(x - 2) is a factor.

Step 1: Write the mathematical expression

Set up: coefficients are 1,2,−5,−61, 2, -5, -6 with c=2c = 2

Computer Science: Algorithm Optimization

Evaluating polynomials efficiently uses ideas from synthetic division. Horner's method is essentially synthetic division for function evaluation.

Example:

To find P(5)P(5) for P(x)=2x3−x2+3x−7P(x) = 2x^3 - x^2 + 3x - 7, use synthetic division with c=5c = 5. The remainder equals P(5)=213P(5) = 213.

2Try It Yourself

You need to evaluate P(x)=x3−4x+2P(x) = x^3 - 4x + 2 at x=3x = 3 for a graphics calculation.

Use synthetic division to find P(3)P(3).

Step 1: Write the mathematical expression

Coefficients: 1,0,−4,21, 0, -4, 2 (remember the missing x2x^2 term!)

Key Takeaways

  • 1Synthetic division is a shortcut for dividing polynomials by (x−c)(x - c)
  • 2Use only the coefficients, including 00 for missing terms
  • 3Pattern: bring down, multiply by cc, add, repeat
  • 4The last number is the remainder; other numbers are quotient coefficients
  • 5For (x+c)(x + c), use c=−cc = -c in the setup
  • 6Remainder Theorem: the remainder equals P(c)P(c)

Frequently Asked Questions

Can I use synthetic division for any divisor?

No, synthetic division only works when dividing by a linear binomial (x−c)(x - c) where the coefficient of xx is 11. For other divisors like (2x−3)(2x - 3) or (x2+1)(x^2 + 1), use polynomial long division.

Why do I use c=−2c = -2 when dividing by (x+2)(x + 2)?

Synthetic division is based on the root form (x−c)(x - c). Since (x+2)=(x−(−2))(x + 2) = (x - (-2)), the value of cc is −2-2. Think of it as finding what makes the divisor equal zero: x+2=0x + 2 = 0 means x=−2x = -2.

How does the Remainder Theorem connect to synthetic division?

When you divide P(x)P(x) by (x−c)(x - c), the remainder you get equals P(c)P(c) - the value of the polynomial at x=cx = c. This means you can use synthetic division to quickly evaluate polynomials!

Glossary

Synthetic division
A shortcut method for dividing a polynomial by a linear binomial (x−c)(x - c) using only coefficients
Divisor
The polynomial you are dividing by; in synthetic division, it must be of the form (x−c)(x - c)
Quotient
The result of the division (excluding the remainder)
Remainder
The amount left over after division; equals P(c)P(c) by the Remainder Theorem
Remainder Theorem
States that when P(x)P(x) is divided by (x−c)(x - c), the remainder equals P(c)P(c)
Factor Theorem
States that (x−c)(x - c) is a factor of P(x)P(x) if and only if P(c)=0P(c) = 0

Formula Card

Synthetic Division Setup

canan−1⋯a0\begin{array}{c|ccc} c & a_n & a_{n-1} & \cdots & a_0 \end{array}

Write $c$ on left, coefficients in descending order on right

The Algorithm

Bring down →\to Multiply by cc →\to Add →\to Repeat

Continue until all coefficients are processed

Result Interpretation

P(x)=(x−c)⋅Q(x)+RP(x) = (x-c) \cdot Q(x) + R

Last number is remainder $R$; others form quotient $Q(x)$

Remainder Theorem

P(c)=RP(c) = R

The remainder equals the polynomial evaluated at $c$

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