Basic Synthetic Division

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

Identify the divisor value: $(x - 2)$ means $c = 2$ = $c = 2$

Write coefficients in order: $x^3 - 6x^2 + 11x - 6$ has coefficients: $1, -6, 11, -6$ = $1 \quad -6 \quad 11 \quad -6$

Set up the synthetic division: $$\begin{array}{c|cccc} 2 & 1 & -6 & 11 & -6 \\ & & & & \\ \hline & & & & \end{array}$$ = Ready to calculate

Bring down the first coefficient: $$\begin{array}{c|cccc} 2 & 1 & -6 & 11 & -6 \\ & & & & \\ \hline & 1 & & & \end{array}$$ = First coefficient: $1$

Multiply and add for each column: $2 \times 1 = 2$, then $-6 + 2 = -4$ $2 \times (-4) = -8$, then $11 + (-8) = 3$ $2 \times 3 = 6$, then $-6 + 6 = 0$ = $$\begin{array}{c|cccc} 2 & 1 & -6 & 11 & -6 \\ & & 2 & -8 & 6 \\ \hline & 1 & -4 & 3 & 0 \end{array}$$

Write the quotient and remainder: Bottom row: $1, -4, 3$ are quotient coefficients; $0$ is remainder = Quotient: $x^2 - 4x + 3$, Remainder: $0$

Answer: $(x^3 - 6x^2 + 11x - 6) \div (x - 2) = x^2 - 4x + 3$ Since the remainder is $0$, $(x - 2)$ is a factor of the polynomial.

Synthetic Division with Missing Terms

Divide $(x^4 - 16)$ by $(x - 2)$

Identify missing terms: $x^4 - 16 = x^4 + 0x^3 + 0x^2 + 0x - 16$ = Coefficients: $1, 0, 0, 0, -16$

Set up with c = 2: $$\begin{array}{c|ccccc} 2 & 1 & 0 & 0 & 0 & -16 \\ & & & & & \\ \hline & & & & & \end{array}$$ = Include zeros for missing terms

Perform synthetic division: Bring down $1$ $2 \times 1 = 2$, $0 + 2 = 2$ $2 \times 2 = 4$, $0 + 4 = 4$ $2 \times 4 = 8$, $0 + 8 = 8$ $2 \times 8 = 16$, $-16 + 16 = 0$ = $$\begin{array}{c|ccccc} 2 & 1 & 0 & 0 & 0 & -16 \\ & & 2 & 4 & 8 & 16 \\ \hline & 1 & 2 & 4 & 8 & 0 \end{array}$$

Write the result: Quotient coefficients: $1, 2, 4, 8$ = Quotient: $x^3 + 2x^2 + 4x + 8$

Answer: $(x^4 - 16) \div (x - 2) = x^3 + 2x^2 + 4x + 8$ with remainder $0$

Synthetic Division with (x + c)

Divide $(2x^3 + 5x^2 - x - 6)$ by $(x + 2)$

Convert divisor to standard form: $(x + 2) = (x - (-2))$, so $c = -2$ = $c = -2$

Write coefficients: $2x^3 + 5x^2 - x - 6$ has coefficients: $2, 5, -1, -6$ = All terms present

Perform synthetic division with $c = -2$: Bring down $2$ $(-2) \times 2 = -4$, $5 + (-4) = 1$ $(-2) \times 1 = -2$, $-1 + (-2) = -3$ $(-2) \times (-3) = 6$, $-6 + 6 = 0$ = $$\begin{array}{c|cccc} -2 & 2 & 5 & -1 & -6 \\ & & -4 & -2 & 6 \\ \hline & 2 & 1 & -3 & 0 \end{array}$$

Interpret the result: Quotient: $2x^2 + x - 3$, Remainder: $0$ = $(x + 2)$ is a factor!

Answer: $(2x^3 + 5x^2 - x - 6) \div (x + 2) = 2x^2 + x - 3$ Note: When dividing by $(x + c)$, use $c = -c$ in synthetic division.

Finding a Non-Zero Remainder

Divide $(3x^3 - 2x^2 + 5x - 1)$ by $(x - 1)$

Set up with c = 1: Coefficients: $3, -2, 5, -1$ = $$\begin{array}{c|cccc} 1 & 3 & -2 & 5 & -1 \end{array}$$

Perform synthetic division: Bring down $3$ $1 \times 3 = 3$, $-2 + 3 = 1$ $1 \times 1 = 1$, $5 + 1 = 6$ $1 \times 6 = 6$, $-1 + 6 = 5$ = $$\begin{array}{c|cccc} 1 & 3 & -2 & 5 & -1 \\ & & 3 & 1 & 6 \\ \hline & 3 & 1 & 6 & 5 \end{array}$$

Write the complete answer: Quotient: $3x^2 + x + 6$, Remainder: $5$ = Non-zero remainder

Answer: $(3x^3 - 2x^2 + 5x - 1) \div (x - 1) = 3x^2 + x + 6 + \frac{5}{x - 1}$ By the Remainder Theorem: $P(1) = 3(1)^3 - 2(1)^2 + 5(1) - 1 = 3 - 2 + 5 - 1 = 5$ (equals the remainder!)

Using Synthetic Division to Test Roots

Is $x = 3$ a root of $P(x) = x^3 - 7x + 6$?

Set up synthetic division: Coefficients of $x^3 + 0x^2 - 7x + 6$: $1, 0, -7, 6$ Test $c = 3$ = Include $0$ for missing $x^2$ term

Perform synthetic division: $$\begin{array}{c|cccc} 3 & 1 & 0 & -7 & 6 \\ & & 3 & 9 & 6 \\ \hline & 1 & 3 & 2 & 12 \end{array}$$ = Remainder: $12$

Interpret the result: Remainder $= 12 \neq 0$ = $x = 3$ is NOT a root

Answer: No, $x = 3$ is not a root of $P(x) = x^3 - 7x + 6$ because the remainder is $12$, not $0$. A value is a root only if the remainder equals zero.

Mistake: Forgetting to include zeros for missing terms

Why: Every power of $x$ from highest to constant must have a coefficient. Missing terms have coefficient $0$.

Correct: For $x^4 - 16$, write: $1, 0, 0, 0, -16$ (zeros for $x^3$, $x^2$, and $x$)

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

Why: Synthetic division uses the root form $(x - c)$. When dividing by $(x + 2)$, you need $c = -2$.

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

Mistake: Adding when you should multiply, or vice versa

Why: The pattern is: multiply by $c$, then add to the next coefficient.

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

Mistake: Writing the quotient with the wrong degree

Why: The quotient has one degree less than the original polynomial.

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

Engineering: Transfer Functions

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

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

Computer Science: Algorithm Optimization

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

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

Q: Can I use synthetic division for any divisor?

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

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

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

Q: How does the Remainder Theorem connect to synthetic division?

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