Synthetic Division

➗ Synthetic Division

Part 1 of 5 — Setting Up the Shortcut

Topics in This Part

🔑 Key Concept: Synthetic division is a fast, paper-saving shortcut for dividing a polynomial by a linear factor $x - c$. No variables, no long-division scaffolding — just the coefficients and one number.

What Synthetic Division Does

When you divide a polynomial $P(x)$ by $x - c$, you get a quotient and a remainder:

$P(x) = (x - c)\cdot Q(x) + r$

Long division works, but it's slow. Synthetic division does the exact same job using only:

1. The coefficients of $P(x)$, and
2. The single number $c$ (the root of the divisor).

Long division vs. synthetic division

⚠️ Limitation: Synthetic division only works when you divide by a linear factor of the form $x - c$ (leading coefficient $1$). For $2x - 3$ you must first rewrite it — we cover that in Part 4.

Step 1 — List the Coefficients

Write the coefficients of the polynomial in order of descending degree. The crucial rule:

🔑 Insert a $0$ for every missing term. A skipped power of $x$ throws off the entire alignment.

Example

The first polynomial above has every degree from $3$ down to $0$. The second skips $x^2$, so we hold its place with a $0$. The third skips $x^3$, $x^2$, and $x^1$.

Concept Check 🎯

Step 2 — Find the Divisor's Root

Synthetic division uses the root of the divisor — the value of $x$ that makes $x - c = 0$.

$x - c = 0 \;\Longrightarrow\; x = c$

💡 Watch the sign flip! When you divide by $x - 3$, you use $\mathbf{+3}$. When you divide by $x + 3$, rewrite it as $x - (-3)$ and use $\mathbf{-3}$.

Find the Root 🧮

Enter the number $c$ you would place in the synthetic-division box for each divisor.

1) Divide by $x - 4 \;\Rightarrow\; c = \,?$ 2) Divide by $x + 6 \;\Rightarrow\; c = \,?$ 3) Divide by $x + 1 \;\Rightarrow\; c = \,?$

Build the Setup 🔽

You're about to divide $2x^3 - 7x + 4$ by $x + 3$. Choose the correct setup pieces before running the algorithm.

You're Set Up

You now have the two ingredients synthetic division needs:

• a tidy coefficient list (with $0$s for any gaps), and
• the root $c$ of the divisor $x - c$.

In Part 2 we run the actual algorithm — bring down, multiply, add — and read off a quotient and remainder.

➗ Synthetic Division

Part 2 of 5 — The Algorithm: Bring Down, Multiply, Add

🔑 The Rhythm: Synthetic division is just three moves repeated across the row — bring down the first number, then loop: multiply by $c$, write it under the next coefficient, and add.

The Three Moves

Divide $x^3 - 4x^2 + 5x - 2$ by $x - 2$. Root $c = 2$; coefficients $1,\ -4,\ 5,\ -2$.

Set up. Put $c=2$ on the left and the coefficients across the top.

1. Bring down the leading coefficient $1$.

2. Multiply $1 \cdot 2 = 2$; write it under the next coefficient $-4$.

3. Add the column: $-4 + 2 = -2$.

Repeat across:

• $-2 \cdot 2 = -4$; add to $5$: $\;5 + (-4) = 1$
• $1 \cdot 2 = 2$; add to $-2$: $\;-2 + 2 = 0$

$\begin{array}{c|cccc} 2 & 1 & -4 & 5 & -2 \\ & & 2 & -4 & 2 \\ \hline & 1 & -2 & 1 & 0 \end{array}$

The bottom row $1,\ -2,\ 1,\ \boxed{0}$ holds the answer.

Reading the Bottom Row

The last number in the bottom row is the remainder. Everything before it gives the quotient, with degree one less than the original polynomial.

For our example the bottom row was $1,\ -2,\ 1,\ 0$:

• Remainder $= \boxed{0}$
• Quotient coefficients $= 1,\ -2,\ 1$ → starts at $x^2$ (one degree below the cubic):

$Q(x) = x^2 - 2x + 1$

So $\;x^3 - 4x^2 + 5x - 2 = (x - 2)(x^2 - 2x + 1)$.

✅ Check by multiplying: $(x-2)(x^2-2x+1) = x^3 -2x^2 + x -2x^2 +4x -2 = x^3 - 4x^2 + 5x - 2$ ✓

Concept Check 🎯

Run the Algorithm 🧮

Divide $x^3 + 2x^2 - 5x - 6$ by $x - 2$ (so $c = 2$; coefficients $1,\ 2,\ -5,\ -6$).

Fill in the bottom row in order.

1) First bottom-row entry (brought down) $= \,?$ 2) Second entry $= \,?$ 3) Third entry $= \,?$ 4) Remainder (last entry) $= \,?$

Name the Quotient & Remainder 🧮

A synthetic division of the cubic $x^3 - 7x + 6$ by $x - 1$ produces the bottom row $1,\ 1,\ -6,\ 0$.

1) What is the remainder? $\,?$ 2) The quotient is $x^2 + \,?\,x - 6$. Enter the middle coefficient. $\,?$

➗ Synthetic Division

Part 3 of 5 — The Remainder & Factor Theorems

🔑 The Payoff: That last number in the bottom row isn't just a leftover. It's $P(c)$ — the value of the whole polynomial at $x = c$. This is the Remainder Theorem, and it makes synthetic division a lightning-fast evaluator.

The Remainder Theorem

🔑 Remainder Theorem: When $P(x)$ is divided by $x - c$, the remainder equals $P(c)$.

So instead of plugging $c$ into a messy polynomial, you can synthetic-divide and just read the remainder.

Example: evaluate $P(x) = x^3 - 2x^2 + 3x - 5$ at $x = 2$

$\begin{array}{c|cccc} 2 & 1 & -2 & 3 & -5 \\ & & 2 & 0 & 6 \\ \hline & 1 & 0 & 3 & 1 \end{array}$

The remainder is $\boxed{1}$, so $P(2) = 1$.

✅ Check by substitution: $P(2) = 8 - 8 + 6 - 5 = 1$ ✓

Evaluate with the Remainder Theorem 🧮

Use synthetic division to find each value.

1) $P(x) = x^3 + 4x^2 - x + 2$. Find $P(1)$. 2) $P(x) = 2x^3 - 3x^2 + x - 4$. Find $P(2)$.

The Factor Theorem

The Remainder Theorem has a powerful special case:

🔑 Factor Theorem: $(x - c)$ is a factor of $P(x)$ if and only if $P(c) = 0$ — i.e., the remainder is $0$.

A remainder of $0$ means the division came out clean: $c$ is a root (zero) of the polynomial, and $x - c$ divides it evenly.

💡 This turns synthetic division into a factor test: try a candidate $c$, and a remainder of $0$ confirms a factor instantly.

Concept Check 🎯

Factor or Not? 🔽

For each test, decide what the remainder tells you. (You can compute the remainder by synthetic division, or note that it equals $P(c)$.)

➗ Synthetic Division

Part 4 of 5 — Factoring Polynomials & Special Divisors

🔑 Putting It to Work: Find one root, synthetic-divide to peel it off, and you're left with a smaller polynomial that's far easier to factor. Repeat until the polynomial is fully broken down.

Factoring with Synthetic Division

Example: factor $x^3 - 4x^2 + x + 6$

Step 1 — find a root. Test small integers. $P(2) = 8 - 16 + 2 + 6 = 0$, so $x - 2$ is a factor.

Step 2 — divide out $x - 2$ (coefficients $1,\ -4,\ 1,\ 6$, root $c = 2$):

$\begin{array}{c|cccc} 2 & 1 & -4 & 1 & 6 \\ & & 2 & -4 & -6 \\ \hline & 1 & -2 & -3 & 0 \end{array}$

Quotient: $x^2 - 2x - 3$.

Step 3 — factor the quotient. $x^2 - 2x - 3 = (x - 3)(x + 1)$.

$x^3 - 4x^2 + x + 6 = (x - 2)(x - 3)(x + 1)$

The roots are $x = 2,\ 3,\ -1$.

Concept Check 🎯

When the Divisor Isn't $x - c$

Divisor with a leading coefficient: $2x - 3$

Synthetic division needs a monic linear divisor ($x - c$). Factor out the leading coefficient first:

$2x - 3 = 2\left(x - \tfrac{3}{2}\right)$

Synthetic-divide by $x - \tfrac{3}{2}$ (use $c = \tfrac{3}{2}$), then divide the quotient by $2$ to account for the factored-out constant.

A negative root: dividing by $x + 5$

Rewrite $x + 5 = x - (-5)$ and use $c = -5$. The algorithm is identical — just carry the negative carefully through every multiply step.

⚠️ Two common slips: (1) forgetting to use the negative root for $x + c$, and (2) forgetting the place-holding $0$s for missing terms.

Divide by a Binomial with a Negative Root 🧮

Divide $x^3 + 6x^2 + 11x + 6$ by $x + 1$.

First, the root: $x + 1 = 0 \Rightarrow c = -1$. Coefficients $1,\ 6,\ 11,\ 6$. Fill in the bottom row.

1) First entry (brought down) $= \,?$ 2) Second entry $= \,?$ 3) Third entry $= \,?$ 4) Remainder $= \,?$

Complete the Factoring 🔽

From the drill above, $x^3 + 6x^2 + 11x + 6 = (x+1)\cdot Q(x)$ where $Q(x) = x^2 + 5x + 6$.

➗ Synthetic Division

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) set up coefficients and roots, (2) run the bring-down-multiply-add algorithm, (3) use the Remainder and Factor Theorems, and (4) factor polynomials completely. Time to put it together.

Quick Reference

⚠️ Remember: divide by $x + c$ ⟹ use $c$ negative, and never skip a place-holding $0$.

Mixed Practice 🎯

Full Division 🧮

Divide $2x^3 + 3x^2 - 11x - 6$ by $x - 2$ (root $c = 2$; coefficients $2,\ 3,\ -11,\ -6$).

1) First bottom-row entry $= \,?$ 2) Second entry $= \,?$ 3) Third entry $= \,?$ 4) Remainder $= \,?$

Exit Quiz ✅

Answer all three to finish the lesson.