What Synthetic Division Is
Synthetic division is a compact way to divide a polynomial by a linear factor $x - c$, working only with the coefficients of the dividend and the single number $c$. It returns the same quotient and remainder as long division of polynomials, but with far less writing because the variables are dropped and every subtraction becomes an addition.
The one condition: the divisor must be linear with leading coefficient 1, meaning it has the form $x - c$. If you can read off $c$ and list the coefficients in order, you can run the method.
Steps of Synthetic Division
Every synthetic division follows the same short loop. Read it as a way of evaluating the polynomial, not as motion to memorise.
Write the polynomial in standard form and insert a 0 for any missing power, so each column stands for one degree of $x$.
Find $c$ from the divisor. Set $x - c = 0$; for a divisor $x + 2$ that gives $c = -2$.
List the dividend's coefficients in a row and place $c$ to the left.
Bring the first coefficient straight down into the bottom row.
Multiply that bottom number by $c$ and write the product under the next coefficient.
Add the column and write the sum in the bottom row.
Repeat the multiply-and-add step across every remaining column.
Split the bottom row: the last number is the remainder, and the numbers before it are the quotient coefficients, one degree lower than the dividend.
The two habits that keep the method honest are inserting zeros in step 1 and taking $c$ from $x - c$ in step 2. Everything else is one multiplication and one addition per column.
Examples of Synthetic Division of Polynomials
The examples build from a clean linear divisor to a missing-term case, a wrong-path check, a zero-remainder factor test, and a non-monic divisor.
Example 1
Divide $x^2 + x - 2$ by $x + 2$.
Write the divisor as $x - c$, so $x + 2 = x - (-2)$ and $c = -2$.
List the coefficients: $1, 1, -2$.
Bring down the first coefficient.
$$1$$
Multiply by $c = -2$ and add to the next coefficient.
$$1 \times (-2) = -2$$ $$1 + (-2) = -1$$
Multiply again and add to the last coefficient.
$$-1 \times (-2) = 2$$ $$-2 + 2 = 0$$
The bottom row is $1, -1, 0$. The last number is the remainder, 0. The rest are the quotient coefficients, one degree lower.
$$\text{Quotient} = x - 1, \quad \text{Remainder} = 0$$
Example 2
Divide $2x^3 + 3x^2 - 4x + 1$ by $x - 1$.
Here $c = 1$ and the coefficients are $2, 3, -4, 1$.
Bring down the 2.
$$2$$
Multiply and add across.
$$2 \times 1 = 2, \quad 3 + 2 = 5$$ $$5 \times 1 = 5, \quad -4 + 5 = 1$$ $$1 \times 1 = 1, \quad 1 + 1 = 2$$
The bottom row is $2, 5, 1, 2$.
$$\text{Quotient} = 2x^2 + 5x + 1, \quad \text{Remainder} = 2$$
Example 3
Divide $x^3 - 7x + 6$ by $x - 2$.
The $x^2$ term is missing. This is where a first instinct goes wrong.
The wrong path: listing only the visible coefficients $1, -7, 6$ and running the method.
$$1, \quad 1 \times 2 = 2, \quad -7 + 2 = -5$$ $$-5 \times 2 = -10, \quad 6 + (-10) = -4$$
This gives a remainder of $-4$, suggesting $x - 2$ is not a factor. But $2^3 - 7(2) + 6 = 8 - 14 + 6 = 0$, so the remainder should be 0. The result is obviously wrong.
The rescue: every power must have a slot. Insert a 0 for the missing $x^2$ term, so the coefficients are $1, 0, -7, 6$.
$$1, \quad 1 \times 2 = 2, \quad 0 + 2 = 2$$ $$2 \times 2 = 4, \quad -7 + 4 = -3$$ $$-3 \times 2 = -6, \quad 6 + (-6) = 0$$
The bottom row is $1, 2, -3, 0$.
$$\text{Quotient} = x^2 + 2x - 3, \quad \text{Remainder} = 0$$
Example 4
Use synthetic division to test whether $x - 3$ is a factor of $x^3 - 4x^2 + x + 6$.
Set $c = 3$ with coefficients $1, -4, 1, 6$.
$$1, \quad 1 \times 3 = 3, \quad -4 + 3 = -1$$ $$-1 \times 3 = -3, \quad 1 + (-3) = -2$$ $$-2 \times 3 = -6, \quad 6 + (-6) = 0$$
The remainder is 0, so $x - 3$ is a factor.
$$x^3 - 4x^2 + x + 6 = (x - 3)(x^2 - x - 2)$$
This is the Factor Theorem in action: a zero remainder means the divisor is a factor.
Example 5
Divide $x^4 - 16$ by $x - 2$.
Both middle terms and the linear term are missing, so fill them with zeros: $1, 0, 0, 0, -16$, with $c = 2$.
$$1, \quad 1 \times 2 = 2, \quad 0 + 2 = 2$$ $$2 \times 2 = 4, \quad 0 + 4 = 4$$ $$4 \times 2 = 8, \quad 0 + 8 = 8$$ $$8 \times 2 = 16, \quad -16 + 16 = 0$$
The bottom row is $1, 2, 4, 8, 0$.
$$\text{Quotient} = x^3 + 2x^2 + 4x + 8, \quad \text{Remainder} = 0$$
Example 6
Divide $6x^2 + 7x - 20$ by $2x + 5$.
The divisor is not of the form $x - c$; its leading coefficient is 2. First find the zero of the divisor.
$$2x + 5 = 0$$ $$x = -\frac{5}{2}$$
Use $c = -\dfrac{5}{2}$ with coefficients $6, 7, -20$.
$$6, \quad 6 \times \left(-\tfrac{5}{2}\right) = -15, \quad 7 + (-15) = -8$$ $$-8 \times \left(-\tfrac{5}{2}\right) = 20, \quad -20 + 20 = 0$$
The bottom row is $6, -8, 0$, but because the divisor was $2x + 5$ rather than $x + \tfrac{5}{2}$, divide the quotient coefficients by 2.
$$\text{Quotient} = \frac{6x - 8}{2} = 3x - 4, \quad \text{Remainder} = 0$$
Synthetic Division vs Long Division
Both methods return the identical quotient and remainder; they differ only in how much you write and when each one is allowed.
Feature | Synthetic division | Long division of polynomials |
|---|---|---|
Divisor allowed | Linear only, form $x - c$ | Any degree |
What you write | Coefficients and $c$ | Full terms with variables |
Core operation | Multiply, then add | Multiply, then subtract |
Space and speed | One row, fast | Several lines, slower |
Missing terms | Placeholder 0 required | Placeholder 0 required |
The takeaway: reach for synthetic division whenever the divisor is a single linear factor $x - c$, and fall back to long division the moment the divisor has degree 2 or higher.
Why It Works — "The remainder equals the polynomial at $c$"
Synthetic division is not a trick; it is bookkeeping for the Remainder Theorem. When you divide $p(x)$ by $x - c$, the remainder is exactly $p(c)$. Each row of the synthetic table is one step of evaluating $p(c)$ by nested multiplication.
The historical root is Horner's method, published by William George Horner in 1819 for evaluating polynomials efficiently, though Chinese and Persian mathematicians used the same nesting centuries earlier.
Where it is going: the same nested-evaluation idea powers how calculators and computers evaluate polynomials today, because it minimises the number of multiplications. You can read the outline of Horner's method on Wikipedia.
The payoff for algebra: once a factor is found by synthetic division, the leftover quotient is a lower-degree polynomial that is easier to solve, which is how many polynomial equations get cracked open one root at a time.
At Bhanzu, trainers teach synthetic division alongside the Remainder Theorem so students see the row of numbers as evaluation, not memorised motion.
Common Mistakes With Synthetic Division of a Polynomial
Mistake 1: Forgetting to insert zeros for missing terms
Where it slips in: dividing a polynomial like $x^3 - 7x + 6$ that skips a power.
Don't do this: listing only $1, -7, 6$ and running the method, which shifts every coefficient into the wrong column.
The correct way: write a 0 for each missing power first, so the coefficients are $1, 0, -7, 6$. This is the habit that fixes most wrong answers in synthetic division.
Mistake 2: Using the wrong sign for $c$
Where it slips in: reading the divisor $x + 2$ and using $c = 2$.
Don't do this: the second-guessing student often flips the sign back and forth mid-problem.
The correct way: the divisor is $x - c$, so $x + 2 = x - (-2)$ gives $c = -2$. Set $x - c = 0$ and solve to get $c$ every time.
Mistake 3: Reading the last number as a quotient coefficient
Where it slips in: at the final step, when the rusher writes the whole bottom row as the answer.
Don't do this: treating $1, -1, 0$ as the quotient $x^2 - x + 0$.
The correct way: the last number is the remainder; only the numbers before it are quotient coefficients, and the quotient is one degree lower than the dividend.
Conclusion
Synthetic division divides a polynomial by a linear factor $x - c$ using only coefficients and the number $c$.
Bring down, multiply by $c$, add, and repeat; the last number is the remainder and the rest are the quotient, one degree lower.
Insert a 0 for every missing power before you start.
A remainder of 0 means $x - c$ is a factor, linking the method to the Factor Theorem and Remainder Theorem.
For a non-monic divisor such as $2x + 5$, use its zero as $c$ and divide the quotient coefficients afterward.
A Practical Next Step
Now try dividing $x^3 - 6x^2 + 11x - 6$ by $x - 1$, then by $x - 2$, and watch each zero remainder peel off a factor. If you lose track of a column, come back to Example 3 and check that every missing power has its 0. To build this skill with a live trainer who ties the coefficient row back to the Remainder Theorem, explore Bhanzu's algebra tutor, help with algebra, or math classes online.
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