Math Notes

Synthetic Division

A fast method for dividing polynomials by linear factors of the form $x - c$.

What is Synthetic Division?

Synthetic division is a shorthand method for dividing a polynomial by a linear factor of the form $x - c$. It is faster and involves fewer calculations than traditional long division, using only the coefficients of the polynomial.

Key Insight: Synthetic division only works when the divisor is a linear binomial with a leading coefficient of 1 (in the form $x - c$ or $x + c$). For other divisors, you must use polynomial long division.

When Synthetic Division Works

Use synthetic division when:

Example of "fixing" a polynomial:

$2x^3 + 5x + 1$ → write coefficients as $2, 0, 5, 1$ because the $x^2$ term is missing, so its coefficient is 0.

KaTeX Template for Synthetic Division

Since KaTeX doesn't have a built-in synthetic division environment, we use arrays and vertical bars. Here's the template format:

$$c \left| \begin{array}{cccc} a_0 & a_1 & a_2 & a_3 \\ & b_0 & b_1 & b_2 \\ \hline a_0 & c_0 & c_1 & \text{remainder} \end{array} \right.$$

Where $c$ is the root, $a_i$ are the original coefficients, and the bottom row contains the quotient coefficients plus the remainder.

Step-by-Step Process

To divide $P(x)$ by $x - a$:

  1. Write the "root" $a$ on the left (since $x - a = 0 \Rightarrow x = a$). For $x + 3$, the root is $-3$.
  2. Write the coefficients of $P(x)$ in a row, using zeros for missing powers.
  3. Bring down the first coefficient unchanged.
  4. Multiply that number by $a$; write the result under the next coefficient.
  5. Add down the column.
  6. Repeat multiply-add until you reach the end.
  7. Interpret the bottom row:
    • All but the last number are coefficients of the quotient
    • The last number is the remainder

Remember: If you start with a degree $n$ polynomial, the quotient has degree $n-1$.

Worked Examples

Example 1: Simple Cubic with Remainder

Divide $P(x) = 2x^3 - 3x^2 - 11x + 6$ by $x - 2$.

Solution: Divisor $x - 2$ → root $a = 2$

Step 1: Set up the coefficients

Write the root on the left and the coefficients in a row

$$2 \left| \begin{array}{rrrr} 2 & -3 & -11 & 6 \end{array} \right.$$

Step 2: Bring down the first coefficient

Bring the first number (2) straight down

$$2 \left| \begin{array}{rrrr} 2 & -3 & -11 & 6 \\ & & & \\ \hline 2 & & & \end{array} \right.$$

Step 3: Multiply and add

Multiply the bottom number (2) by the root (2): $2 \times 2 = 4$. Add to the next coefficient: $-3 + 4 = 1$

$$2 \left| \begin{array}{rrrr} 2 & -3 & -11 & 6 \\ & 4 & & \\ \hline 2 & 1 & & \end{array} \right.$$

Step 4: Repeat multiply-add

Multiply by root again: $1 \times 2 = 2$. Add to next coefficient: $-11 + 2 = -9$

$$2 \left| \begin{array}{rrrr} 2 & -3 & -11 & 6 \\ & 4 & 2 & \\ \hline 2 & 1 & -9 & \end{array} \right.$$

Step 5: Final multiply-add

Multiply by root again: $-9 \times 2 = -18$. Add to last coefficient: $6 + (-18) = -12$

$$2 \left| \begin{array}{rrrr} 2 & -3 & -11 & 6 \\ & 4 & 2 & -18 \\ \hline 2 & 1 & -9 & -12 \end{array} \right.$$

Final Results:

Bottom row: $2, 1, -9, -12$
Quotient coefficients: $2, 1, -9$ → $2x^2 + x - 9$
Remainder: $-12$

$$\frac{2x^3 - 3x^2 - 11x + 6}{x - 2} = 2x^2 + x - 9 - \frac{12}{x - 2}$$

Example 2: Zero Remainder (Factor)

Divide $P(x) = x^3 - 6x^2 + 11x - 6$ by $x - 1$.

Solution: Divisor $x - 1$ → root $a = 1$

Step 1: Set up the coefficients

Write the root on the left and the coefficients in a row

$$1 \left| \begin{array}{rrrr} 1 & -6 & 11 & -6 \end{array} \right.$$

Step 2: Bring down the first coefficient

Bring the first number (1) straight down

$$1 \left| \begin{array}{rrrr} 1 & -6 & 11 & -6 \\ & & & \\ \hline 1 & & & \end{array} \right.$$

Step 3: Multiply and add

Multiply by root: $1 \times 1 = 1$. Add to next coefficient: $-6 + 1 = -5$

$$1 \left| \begin{array}{rrrr} 1 & -6 & 11 & -6 \\ & 1 & & \\ \hline 1 & -5 & & \end{array} \right.$$

Step 4: Repeat multiply-add

Multiply by root again: $-5 \times 1 = -5$. Add to next coefficient: $11 + (-5) = 6$

$$1 \left| \begin{array}{rrrr} 1 & -6 & 11 & -6 \\ & 1 & -5 & \\ \hline 1 & -5 & 6 & \end{array} \right.$$

Step 5: Final multiply-add

Multiply by root again: $6 \times 1 = 6$. Add to last coefficient: $-6 + 6 = 0$

$$1 \left| \begin{array}{rrrr} 1 & -6 & 11 & -6 \\ & 1 & -5 & 6 \\ \hline 1 & -5 & 6 & 0 \end{array} \right.$$

Final Results:

Bottom row: $1, -5, 6, 0$
Quotient coefficients: $1, -5, 6$ → $x^2 - 5x + 6$
Remainder: $0$ ✅ (This means $x - 1$ is a factor!)

$$x^3 - 6x^2 + 11x - 6 = (x - 1)(x^2 - 5x + 6)$$

Example 3: Missing Term Requires Zero

Divide $P(x) = 3x^3 + 5x - 4$ by $x + 2$.

Solution: Divisor $x + 2$ → root $a = -2$

⚠️ Important!

The polynomial $3x^3 + 5x - 4$ is missing the $x^2$ term, so we need to include a 0 as a placeholder coefficient.

Step 1: Set up the coefficients

Write the root on the left and the coefficients in a row (include 0 for missing $x^2$ term)

$$-2 \left| \begin{array}{rrrr} 3 & 0 & 5 & -4 \end{array} \right.$$

Step 2: Bring down the first coefficient

Bring the first number (3) straight down

$$-2 \left| \begin{array}{rrrr} 3 & 0 & 5 & -4 \\ & & & \\ \hline 3 & & & \end{array} \right.$$

Step 3: Multiply and add

Multiply by root: $3 \times (-2) = -6$. Add to next coefficient: $0 + (-6) = -6$

$$-2 \left| \begin{array}{rrrr} 3 & 0 & 5 & -4 \\ & -6 & & \\ \hline 3 & -6 & & \end{array} \right.$$

Step 4: Repeat multiply-add

Multiply by root again: $-6 \times (-2) = 12$. Add to next coefficient: $5 + 12 = 17$

$$-2 \left| \begin{array}{rrrr} 3 & 0 & 5 & -4 \\ & -6 & 12 & \\ \hline 3 & -6 & 17 & \end{array} \right.$$

Step 5: Final multiply-add

Multiply by root again: $17 \times (-2) = -34$. Add to last coefficient: $-4 + (-34) = -38$

$$-2 \left| \begin{array}{rrrr} 3 & 0 & 5 & -4 \\ & -6 & 12 & -34 \\ \hline 3 & -6 & 17 & -38 \end{array} \right.$$

Final Results:

Bottom row: $3, -6, 17, -38$
Quotient coefficients: $3, -6, 17$ → $3x^2 - 6x + 17$
Remainder: $-38$

$$\frac{3x^3 + 5x - 4}{x + 2} = 3x^2 - 6x + 17 - \frac{38}{x + 2}$$

Key Uses and Theorems

Finding Zeros and Factoring

Synthetic division is often used to find the roots (zeros) of a polynomial. If the remainder after division by $x - c$ is zero, then $x - c$ is a factor of the polynomial, and $c$ is a zero of the function.

Remainder Theorem

This theorem states that if a polynomial $f(x)$ is divided by $x - c$, the remainder is equal to $f(c)$. Synthetic division provides a quick way to evaluate $f(c)$.

Practice Problems

Try these with the same steps. You can check your answers using long division or graphing tools.

Problem 1:

$(2x^3 - 7x^2 + x + 6) \div (x - 3)$

Problem 2:

$(x^4 - 2x^3 + 5x - 3) \div (x + 1)$ (remember the $x^2$ placeholder)

Problem 3:

$(6x^3 + 5x^2 - 7) \div (x - 1)$ (placeholder for $x$-term)