Synthetic Division Calculator

Divide a Polynomial by x − c Step by Step

Synthetic Division Formula

Formula

\[ P(x) = (x - c) \cdot Q(x) + R \] \[ b_0 = a_0, \quad b_i = a_i + c \cdot b_{i-1} \;\; (i = 1, \dots, n), \quad R = b_n \]

Where:

\(a_0, a_1, \dots, a_n\) — coefficients of \(P(x)\) written from the highest power down to the constant term
\(c\) — the root of the divisor, i.e. the number for which the divisor equals \(x - c\)
\(b_0, b_1, \dots, b_{n-1}\) — the coefficients of the quotient \(Q(x)\)
\(R\) — the remainder; each step multiplies the previous result by \(c\) and adds the next coefficient

The formula expresses the division algorithm: bring down the leading coefficient, then repeatedly multiply by \(c\) and add the next coefficient. All results except the last form the quotient, and the last one is the remainder. By the remainder theorem, \(R = P(c)\).

Synthetic Division – Calculation Example

Example: divide \(P(x) = 2x^3 - 6x^2 + 2x - 1\) by \(x - 3\), so \(c = 3\). The coefficients are \(2, -6, 2, -1\).

\[ b_0 = 2 \]

\[ b_1 = -6 + 3 \cdot 2 = 0, \quad b_2 = 2 + 3 \cdot 0 = 2 \]

\[ R = -1 + 3 \cdot 2 = 5 \]

Result: \(Q(x) = 2x^2 + 0x + 2 = 2x^2 + 2\) and remainder \(R = 5\), so \(2x^3 - 6x^2 + 2x - 1 = (x - 3)(2x^2 + 2) + 5\). Check: \(P(3) = 54 - 54 + 6 - 1 = 5 = R\).

How to use the calculator

Choose how to enter the polynomial: Polynomial text (e.g. 2x^3 - 6x^2 + 2x - 1) or Coefficients (e.g. 2, -6, 2, -1).
Fill in the Polynomial P(x) field or the Coefficients of P(x) field, depending on the mode you chose. Write powers with ^, list coefficients from the highest power down, and use 0 for any missing power.
Select the Divisor form: x − c or x + c. For x + c the calculator internally divides by \(x - (-c)\).
Enter the Value of c (any real number, e.g. 3).
Read the results: the quotient \(Q(x)\) and remainder \(R\), the synthetic division table, the identity \(P(x) = (x - c)\cdot Q(x) + R\), the root check, and the detailed steps with a remainder-theorem verification.

What is synthetic division?

Synthetic division is a shortcut for dividing a polynomial by a linear divisor of the form \(x - c\). It replaces the full long-division layout with a compact table that uses only the coefficients of the polynomial, so no variables or powers have to be written or aligned. For linear divisors it produces exactly the same quotient and remainder as long division — just much faster and with fewer opportunities for sign errors.

The method relies on the division identity \(P(x) = (x - c) \cdot Q(x) + R\), where the quotient \(Q(x)\) has degree one less than \(P(x)\) and the remainder \(R\) is a constant. Writing the coefficients of \(P(x)\) in a row and repeating “multiply by \(c\), add the next coefficient” fills in the coefficients of \(Q(x)\) plus the remainder \(R\) at the end. A nice by-product is the remainder theorem: the last value in the bottom row equals \(P(c)\), the polynomial evaluated at \(c\).

Interpreting the result is straightforward. The bottom row gives the quotient, and the shaded final cell gives the remainder. If the remainder is 0, then \(c\) is a root of \(P(x)\) and \(x - c\) is a factor (the factor theorem); the calculator reports this automatically. If the remainder is not 0, the divisor does not divide the polynomial evenly — the “left-over” is exactly \(R\).

Synthetic division is used constantly when factoring polynomials, finding rational roots, sketching polynomial graphs, and partial fraction decomposition. Once one root is found, you can apply the quotient again to hunt for further roots, lowering the degree each time.

Limitations and practical tips

Linear divisors only. Synthetic division works only for divisors of the form \(x - c\). To divide by a quadratic or higher-degree polynomial, such as \(x^2 - 5x + 6\), use the Polynomial Long Division Calculator.
Convert the divisor first. If the divisor is \(x + c\), rewrite it as \(x - (-c)\) and use the negated value as the root: dividing by \(x + 3\) means \(c = -3\). The calculator does this automatically when you select the x + c form.
Include zeros for missing powers. When entering coefficients, every power from the highest down to the constant must appear, so a polynomial like \(x^4 - 5x^2 + 7\) has coefficients 1, 0, −5, 0, 7. Entering the polynomial as text fills the gaps automatically.
Divisors with a leading coefficient. If the divisor is \(ax - b\) with \(a \neq 1\), divide the whole divisor by \(a\) to get \(x - b/a\), run synthetic division with \(c = b/a\), and remember that the true quotient and remainder are the results divided by \(a\).
Degrees up to 30. The calculator accepts polynomials of degree 30 or less with real coefficients; roots and factors of polynomials often go hand in hand with the Factor Calculator when testing rational-root candidates, and the quotient from synthetic division can be checked further with the Multiplying Binomials Calculator for simple cases.

Frequently Asked Questions

How do I do synthetic division by hand?

Write the root \(c\) of the divisor in a corner, then list the coefficients of \(P(x)\) from the highest power down in a row. Bring down the first coefficient. Then repeat two moves across the row: multiply the last number you wrote by \(c\), write the product in the next column, and add it to the coefficient above. Every sum is one coefficient of the quotient \(Q(x)\); the final sum is the remainder \(R\).

Why do I need to write zeros for missing powers?

The table works positionally: each column corresponds to one power of \(x\). If a term is missing, its column must still exist or every later coefficient shifts into the wrong power and the whole result is wrong. For example, \(x^4 - 5x^2 + 7\) must be entered as the list 1, 0, −5, 0, 7 — not as 1, −5, 7. This is the single most common mistake when doing synthetic division manually.

How do I divide by x + c instead of x − c?

Rewrite the divisor so it looks like \(x - c\). Since \(x + c = x - (-c)\), you use the negated value as the root: dividing by \(x + 3\) means setting \(c = -3\) in the table. The signs of the products in the middle row flip compared with \(x - 3\), which is easy to get wrong by hand — the calculator handles this automatically when you choose the x + c form.

Can synthetic division handle divisors like 2x − 6 or 3x + 1?

Only after a small adjustment, because the shortcut requires a leading coefficient of 1. Factor out the leading coefficient: \(2x - 6 = 2(x - 3)\), so divide by \(x - 3\) with \(c = 3\), then divide the resulting quotient and remainder by 2 to get the true answer. In general, for \(ax - b\) use \(c = b/a\) and remember the correction factor \(a\).

What does the remainder tell me, and how can I check my answer?

By the remainder theorem, \(R\) equals \(P(c)\) — the polynomial evaluated at the root of the divisor. If \(R = 0\), then \(c\) is a root and \(x - c\) is a full factor of \(P(x)\); otherwise the division leaves a leftover constant. A quick check: plug your \(c\) directly into \(P(x)\), or expand \((x - c)\cdot Q(x) + R\) and confirm you get back the original polynomial.

How is the quotient's degree related to the dividend's degree?

When a degree-\(n\) polynomial is divided by a linear divisor, the quotient always has degree \(n - 1\). So the bottom row of the table (without the remainder cell) has exactly one fewer entry than the top row. If your quotient comes out with the same degree as the dividend, you almost certainly dropped a zero or put an extra coefficient in the top row.

When should I use synthetic division instead of long division?

Use synthetic division whenever the divisor is linear, because it skips all the variable bookkeeping of long division and runs in one compact pass. Use long division when the divisor has degree 2 or more — for example \(x^2 - 5x + 6\) — because the table has nowhere to put the extra terms. Synthetic division is also the fastest hand method for testing whether a candidate rational root actually works.