Synthetic Division Calculator

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Math Input
Synthetic division of x^3 - 4x + 5 by x - 2
Divide 2x^4 + 3x^3 - x + 7 by x + 1
Synthetic division of x^5 - 3x^2 + 2 by x - 3
Use synthetic division to evaluate p(2) for p(x) = x^4 - 2x^3 + x - 1

What is Synthetic Division?

Synthetic division is a shortcut for dividing a polynomial p(x)p(x) by a linear factor xkx - k. It's faster than long division and produces the same quotient and remainder, just with less writing.

Given p(x)=anxn+an1xn1++a0p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 divided by xkx - k, synthetic division produces:

p(x)=(xk)q(x)+rp(x) = (x - k) q(x) + r

where q(x)q(x) is the quotient (degree n1n - 1) and rr is the constant remainder.

Key uses:

  1. Quick polynomial division when the divisor is a linear xkx - k.
  2. Evaluate p(k)p(k) — by the Remainder Theorem, p(k)=rp(k) = r, so the remainder is exactly the function value.
  3. Factor polynomials — if r=0r = 0, then (xk)(x - k) is a factor and q(x)q(x) tells you the cofactor.
  4. Find rational roots combined with the Rational Roots Theorem.

How to Perform Synthetic Division

Setup

To divide p(x)=anxn+an1xn1++a0p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0 by xkx - k:

  1. Write the divisor's zero kk on the left.
  2. List the coefficients of p(x)p(x) on the right, including zeros for any missing terms.

Algorithm

  1. Bring down the first coefficient (ana_n) unchanged.
  2. Multiply by kk and write the result under the next coefficient (an1a_{n-1}).
  3. Add the column. Write the sum on the bottom row.
  4. Repeat: multiply that sum by kk, write under the next coefficient, add.
  5. Continue until you finish all coefficients.

Reading the Result

The bottom row contains:

  • The first nn entries: coefficients of the quotient q(x)q(x) (in descending order of degree).
  • The last entry: the remainder rr.

Example: (x34x+5)÷(x2)(x^3 - 4x + 5) \div (x - 2)

Coefficients of x3+0x24x+5x^3 + 0x^2 - 4x + 5: [1,0,4,5][1, 0, -4, 5]. Divisor zero: k=2k = 2.

 2 |  1   0  -4   5
   |      2   4   0
   |________________
      1   2   0   5

Quotient: x2+2x+0=x2+2xx^2 + 2x + 0 = x^2 + 2x. Remainder: 55.

So x34x+5=(x2)(x2+2x)+5x^3 - 4x + 5 = (x - 2)(x^2 + 2x) + 5.

Connection to the Remainder Theorem

The remainder rr in p(x)=(xk)q(x)+rp(x) = (x - k)q(x) + r equals p(k)p(k). Setting x=kx = k:

p(k)=(kk)q(k)+r=rp(k) = (k - k) q(k) + r = r

So synthetic division is a quick way to evaluate p(k)p(k) without plugging in.

Factor Theorem

A corollary: (xk)(x - k) is a factor of p(x)p(x) iff p(k)=0p(k) = 0 iff the synthetic-division remainder is 00.

Common Mistakes to Avoid

  • Missing zero placeholders: For p(x)=x34x+5p(x) = x^3 - 4x + 5, you must include a 00 for the missing x2x^2 term. Otherwise the columns misalign.
  • Sign error on kk: To divide by x2x - 2, use k=2k = 2 (the zero of the divisor). To divide by x+3x + 3, use k=3k = -3.
  • Cannot use directly for axkax - k divisors: Synthetic division as taught works for xkx - k (leading coefficient 1). For axkax - k, factor out aa first or use polynomial long division.
  • Forgetting to drop the first coefficient: The first step is always 'bring down ana_n' — multiply nothing yet.
  • Misreading the quotient: The bottom row's first nn entries are coefficients, and the degree drops by 1. A degree-4 polynomial divided by xkx - k gives a degree-3 quotient.

Examples

Step 1: Coefficients with placeholder for x2x^2: [1,0,4,5][1, 0, -4, 5]. k=2k = 2
Step 2: Bring down 1
Step 3: Multiply: 12=21 \cdot 2 = 2. Add to 00: 22
Step 4: Multiply: 22=42 \cdot 2 = 4. Add to 4-4: 00
Step 5: Multiply: 02=00 \cdot 2 = 0. Add to 55: 55 (remainder)
Step 6: Bottom row: [1,2,0,5][1, 2, 0, 5]
Answer: Quotient x2+2xx^2 + 2x, remainder 55

Step 1: Coefficients: [1,2,0,1,1][1, -2, 0, 1, -1]. k=3k = 3
Step 2: Bring down 1
Step 3: 13=31 \cdot 3 = 3, add to 2-2: 11
Step 4: 13=31 \cdot 3 = 3, add to 00: 33
Step 5: 33=93 \cdot 3 = 9, add to 11: 1010
Step 6: 103=3010 \cdot 3 = 30, add to 1-1: 2929
Step 7: Remainder =29= 29, so p(3)=29p(3) = 29
Answer: p(3)=29p(3) = 29

Step 1: Divide by x+1x + 1, so k=1k = -1. Coefficients: [1,2,1,2][1, 2, -1, -2]
Step 2: Bring down 1
Step 3: 1(1)=11 \cdot (-1) = -1, add to 2: 1
Step 4: 1(1)=11 \cdot (-1) = -1, add to 1-1: 2-2
Step 5: 2(1)=2-2 \cdot (-1) = 2, add to 2-2: 00 (remainder)
Step 6: Since remainder is 0, (x+1)(x + 1) is a factor and the quotient is x2+x2x^2 + x - 2
Answer: (x+1)(x + 1) is a factor; p(x)=(x+1)(x2+x2)p(x) = (x + 1)(x^2 + x - 2)

Frequently Asked Questions

When the divisor is a linear polynomial of the form x - k. For divisors like x² + 1 or 2x - 3 with non-unit leading coefficient, you need polynomial long division or you must factor out the leading coefficient first.

If you divide a polynomial p(x) by (x - k), the remainder equals p(k). This is why synthetic division is also a fast way to evaluate a polynomial at a specific number.

(x - k) is a factor of p(x) if and only if p(k) = 0 — equivalently, if and only if the synthetic-division remainder is zero. This is the key tool for factoring higher-degree polynomials.

Insert zeros as placeholders for any missing degree. For p(x) = x⁴ + 3x - 2, write coefficients as [1, 0, 0, 3, -2]. Skipping a zero shifts every subsequent column and gives wrong results.

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