acalculator

How do I do synthetic division?

Type the polynomial to divide and a divisor of degree 1, such as x - 3 or 2x + 1. The synthetic division calculator shows the quotient, the remainder and each bring-down, multiply and add step.

Your numbers

Quotient
5x + 12

(5x^2 - 3x - 36) ÷ (x - 3) = 5x + 12, remainder 0.

Remainder
0
k
3
Remainder theorem
p(3) = 0
Bottom row
5, 12 | 0
Steps
The divisor x - 3 is 0 at x = 3, so k = 3; Write the coefficients of the dividend, highest power first, with 0 for a missing power: 5, -3, -36; Bring down 5; Multiply 3 × 5 = 15, then add -3 + 15 = 12; Multiply 3 × 12 = 36, then add -36 + 36 = 0; The last number, 0, is the remainder. The quotient is 5x + 12

Quotient: 5x + 12. (5x^2 - 3x - 36) ÷ (x - 3) = 5x + 12, remainder 0.

How does the synthetic division go?

How to calculate

Divides a polynomial by a linear divisor such as x − 3 or 2x + 1 with synthetic division, in exact fractions: the quotient, the remainder, p(k), and every bring-down, multiply and add step.

Example with the default inputs (Dividend 5x^2 - 3x - 36, Divisor x - 3): (5x^2 - 3x - 36) ÷ (x - 3) = 5x + 12, remainder 0.

Method: For a divisor ax + b, k = −b ÷ a. Bring down the first coefficient; then multiply by k and add to the next coefficient, to the end. The last number is the remainder; the others, divided by a, are the quotient.

  • Coefficients are exact fractions: 0.5 is 1/2.
  • The divisor has degree 1; the dividend has degree 1 to 30, in the same letter.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Dividend 5x^2 - 3x - 36, Divisor x - 3 gives Quotient 5x + 12, Remainder 0, k 3, Bottom row 5, 12 | 0, Remainder theorem p(3) = 0.Source: OpenStax, Algebra and Trigonometry 2e, §5.4 Dividing Polynomials, https://openstax.org/books/algebra-and-trigonometry-2e/pages/5-4-dividing-polynomials, Example 3 Using Synthetic Division to Divide a Second-Degree Polynomial: 5x + 12, remainder 0
  2. Dividend 4x^3 + 10x^2 - 6x - 20, Divisor x + 2 gives Quotient 4x^2 + 2x - 10, Remainder 0, k -2.Source: OpenStax, Algebra and Trigonometry 2e, §5.4 Dividing Polynomials, https://openstax.org/books/algebra-and-trigonometry-2e/pages/5-4-dividing-polynomials, Example 4 Using Synthetic Division to Divide a Third-Degree Polynomial: 4x^2 + 2x - 10
  3. Dividend -9x^4 + 10x^3 + 7x^2 - 6, Divisor x - 1 gives Quotient -9x^3 + x^2 + 8x + 8, Remainder 2, Bottom row -9, 1, 8, 8 | 2.Source: OpenStax, Algebra and Trigonometry 2e, §5.4 Dividing Polynomials, https://openstax.org/books/algebra-and-trigonometry-2e/pages/5-4-dividing-polynomials, Example 5 Using Synthetic Division to Divide a Fourth-Degree Polynomial: −9x^3 + x^2 + 8x + 8, remainder 2
  4. Dividend 6x^3 + 11x^2 - 31x + 15, Divisor 3x - 2 gives Quotient 2x^2 + 5x - 7, Remainder 1, k 2/3, Bottom row 6, 15, -21 | 1.Source: OpenStax, Algebra and Trigonometry 2e, §5.4 Dividing Polynomials, https://openstax.org/books/algebra-and-trigonometry-2e/pages/5-4-dividing-polynomials, Example 2 (by long division): 2x^2 + 5x - 7, remainder 1

How it works

Type the dividend p, a polynomial of degree 1 to 30 in one letter, and the divisor ax + b (degree 1, a ≠ 0) in the same letter. Numbers are exact fractions: a decimal such as 0.5 is read as 1/2. Typed maths such as 2(x+1)^2 is multiplied out first.

  1. k = −b ÷ a, the zero of the divisor.
  2. Write the coefficients c₀, c₁, …, cₙ of p, highest power first, with 0 for each missing power.
  3. Bring down: d₀ = c₀.
  4. For each next coefficient: dᵢ = cᵢ + k × dᵢ₋₁.
  5. The last number dₙ is the remainder; it equals p(k) (the remainder theorem).
  6. The quotient has the coefficients d₀, …, dₙ₋₁ (degree n − 1). When a ≠ 1, each is divided by a; the remainder is not.

So p = (ax + b) × quotient + remainder.

What the page shows

  • Quotient and Remainder as maths. The quotient is written highest power first, with ^ for powers, a coefficient of 1 left out, a fraction coefficient's denominator written after the term (x^2/4, 3x/2), and - for minus: -9x^3 + x^2 + 8x + 8, x^2/4 - x/8 - 7/16. The remainder is one fraction: 0, 2, -7/4.
  • k: the fraction k, such as 3, -2 or 2/3.
  • Remainder theorem: p(k) = remainder, such as p(3) = 0.
  • Bottom row: d₀, …, dₙ₋₁ joined by , , then | and the remainder, such as 5, 12 | 0 (before any division by a).
  • Steps, one per line: the value of k; the coefficients; "Bring down d₀"; for each i, "Multiply k × dᵢ₋₁ = product, then add cᵢ + product = dᵢ" (a negative number after × or + is in brackets); "Divide the bottom row (but not the remainder) by a, the leading coefficient of the divisor" when a ≠ 1; and the remainder and quotient.

Rules

  • The dividend must have degree 1 or more and at most 30; the divisor must have degree exactly 1. Both use the same single letter (any letter but e); e and π cannot appear. Otherwise the page gives no answer and says why.

Worked examples by hand

5x² − 3x − 36 ÷ (x − 3) (OpenStax Example 3). k = 3. Coefficients 5, −3, −36. Bring down 5. 3 × 5 = 15, −3 + 15 = 12. 3 × 12 = 36, −36 + 36 = 0. Bottom row 5, 12 | 0: quotient 5x + 12, remainder 0, so p(3) = 0.

4x³ + 10x² − 6x − 20 ÷ (x + 2) (Example 4). k = −2. Bring down 4. −2 × 4 = −8, 10 − 8 = 2. −2 × 2 = −4, −6 − 4 = −10. −2 × −10 = 20, −20 + 20 = 0. Quotient 4x² + 2x − 10, remainder 0.

−9x⁴ + 10x³ + 7x² − 6 ÷ (x − 1) (Example 5). k = 1. Coefficients −9, 10, 7, 0, −6. Bottom row −9, 1, 8, 8 | 2. Quotient −9x³ + x² + 8x + 8, remainder 2.

6x³ + 11x² − 31x + 15 ÷ (3x − 2) (Example 2 works it by long division). k = 2/3. Bring down 6. 2/3 × 6 = 4, 11 + 4 = 15. 2/3 × 15 = 10, −31 + 10 = −21. 2/3 × −21 = −14, 15 − 14 = 1. Bottom row 6, 15, −21 | 1. Divide by 3: quotient 2x² + 5x − 7, remainder 1.

Other questions people ask

How does synthetic division work?

Write the dividend’s coefficients, highest power first. For a divisor x − k, bring down the first coefficient, multiply it by k, add the product to the next coefficient, and repeat to the end. The last number is the remainder; the others are the quotient’s coefficients, one degree lower.

What is k for a divisor like x + 2?

k is the number that makes the divisor 0. For x + 2 that is −2, for x − 3 it is 3, and for 3x − 2 it is 2/3.

What if a power is missing from the dividend?

Write 0 for it. For −9x^4 + 10x^3 + 7x^2 − 6 the coefficients are −9, 10, 7, 0, −6; dividing by x − 1 gives −9x^3 + x^2 + 8x + 8, remainder 2.

Can I divide by 2x − 1 or 3x − 2?

Yes. Divide by x − k with k = 2/3 for 3x − 2, then divide the quotient (not the remainder) by 3. 6x^3 + 11x^2 − 31x + 15 ÷ (3x − 2) gives 2x^2 + 5x − 7, remainder 1.

What does the remainder tell me?

By the remainder theorem it is the dividend’s value at k: p(k). A remainder of 0 means x − k is a factor and k is a zero of the polynomial, as with 5x^2 − 3x − 36 and x − 3.

When can I not use synthetic division?

When the divisor has degree 2 or more, such as x^2 + 1. Use polynomial long division for those; the polynomial calculator does it.