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Polynomial long division and synthetic division: the missing-term trap

Both methods are just bookkeeping. Most wrong answers come from the bookkeeping, not the algebra: a missing power with no 0 holding its place, a sign flipped on the wrong number, or a subtraction that only reached the first term.

Short answer

Write the dividend in descending powers and put a 0 in for every missing power. In long division, repeat four moves: divide the leading terms, multiply, subtract the whole line, bring down the next term. In synthetic division by x − c, use c itself, so x + 2 means −2. The quotient's degree is one less than the dividend's. Write any remainder as a fraction over the divisor, and check it with the remainder theorem: the remainder from dividing by x − c equals p(c).

Why it works

Dividing polynomials works like long division with numbers. Each step asks “how many times does the divisor's leading term go into what's left?”, takes that many copies of the divisor away, and repeats on what remains. When you're done, you have

  1. p(x) = d(x) · q(x) + r(x)Dividend p equals divisor d times quotient q, plus remainder r. The remainder has a lower degree than the divisor. That's what makes it the place to stop.

Divide both sides by the divisor and you get the form your answer goes in: quotient plus remainderdivisor. Columns matter because each column holds one power of x. If a power is missing and you don't hold its place with a 0, every later term slides into the wrong column, and the subtractions combine terms that aren't like terms.

Example 1: long division with a remainder

Divide 2x3 − 3x2 + 4x − 5 by x − 2.

  1. 2x3 ÷ x = 2x2Divide leading term by leading term. 2x² is the first term of the quotient.
  2. 2x2(x − 2) = 2x3 − 4x2Multiply the whole divisor by that term.
  3. (2x3 − 3x2) − (2x3 − 4x2) = x2Subtract both terms: −3x² − (−4x²) = +x². The 2x³ cancels, which is the whole point of the divide step.
  4. x2 + 4xBring down the next term and repeat.
  5. x2 ÷ x = x; x(x − 2) = x2 − 2xSecond quotient term is x.
  6. (x2 + 4x) − (x2 − 2x) = 6x4x − (−2x) = 6x. Bring down the −5.
  7. 6x ÷ x = 6; 6(x − 2) = 6x − 12Third quotient term is 6.
  8. (6x − 5) − (6x − 12) = 7−5 − (−12) = 7. A constant can't be divided by x again, so 7 is the remainder.
  9. 2x2 + x + 6 + 7x − 2Quotient, plus remainder over the divisor.

Check with the remainder theorem: p(2) = 2(8) − 3(4) + 4(2) − 5 = 7. It matches the remainder, so the arithmetic held up.

Example 2: a missing power needs a 0

Divide x3 − 5x + 3 by x + 2. There is no x2 term, so rewrite the dividend as x3 + 0x2 − 5x + 3 before you start.

  1. x3 ÷ x = x2Divide the leading terms.
  2. x2(x + 2) = x3 + 2x2Multiply the whole divisor by x².
  3. (x3 + 0x2) − (x3 + 2x2) = −2x2The 0x² gives the 2x² something to line up under. Without it you'd be tempted to subtract 2x² from −5x.
  4. −2x2 ÷ x = −2xBring down −5x first; the next quotient term is −2x.
  5. −2x(x + 2) = −2x2 − 4xMultiply the whole divisor by −2x.
  6. (−2x2 − 5x) − (−2x2 − 4x) = −x−5x − (−4x) = −x. Bring down the 3.
  7. −x ÷ x = −1; −1(x + 2) = −x − 2Last quotient term is −1.
  8. (−x + 3) − (−x − 2) = 5Remainder 5.
  9. x2 − 2x − 1 + 5x + 2A cubic divided by a linear divisor gives a quadratic quotient.

Example 3: the same problem by synthetic division

Synthetic division is a shortcut for dividing by x − c. It keeps only the coefficients. Since x + 2 = x − (−2), here c = −2. Write the coefficients 1, 0, −5, 3, with the 0 for the missing x2.

  1. −2 │   1   0  −5   3c on the left, every coefficient on the right, 0 included.
  2.    │      −2   4   2Bring the 1 down. Then repeat: multiply the last bottom number by −2, write it in the next column. 1 · (−2) = −2, (−2)(−2) = 4, (−1)(−2) = 2.
  3.    │   1  −2  −1   5Add each column: 0 + (−2) = −2, −5 + 4 = −1, 3 + 2 = 5. You add here, not subtract, because using −2 instead of +2 already flipped the sign.
  4. x2 − 2x − 1 + 5x + 2The last number is the remainder. The others are the quotient's coefficients, starting one power below the dividend: x², then x, then the constant.

Same answer as the long division, in three rows. And the remainder theorem agrees: p(−2) = (−8) − 5(−2) + 3 = 5. Synthetic division only works when the divisor is x − c. For something like 2x − 1 or x2 + 1, use long division.

Common mistakes, and the exact line they happen on

1. Skipping the 0 placeholder

  1. −2 │   1  −5   3This is the mistake. x³ − 5x + 3 has no x² term, and these coefficients describe x² − 5x + 3 instead, a different polynomial.
  2.    │   1  −7  17The arithmetic from here is fine, which is why this is hard to spot.

It reads as x − 7 remainder 17, and (x + 2)(x − 7) + 17 = x2 − 5x + 3, not the cubic you started with. A quick tell: a cubic divided by x + 2 must give a quadratic, and this gave a linear quotient. Before dividing, count: a degree-3 dividend needs 4 coefficients.

2. Using +2 for x + 2

  1.  2 │   1   0  −5   3This is the mistake. Synthetic division divides by x − c. Writing 2 in the box divides by x − 2.
  2.    │   1   2  −1   1This is a correct answer to the wrong problem.

The result, x2 + 2x − 1 remainder 1, really is what you get from dividing by x − 2. Multiply it back against x + 2 and you get x3 + 4x2 + 3x − 1, nothing like the original. Use the number that makes the divisor zero: x + 2 = 0 at x = −2.

3. Subtracting only the first term

In Example 1, the first subtraction is (2x3 − 3x2) − (2x3 − 4x2).

  1. 2x2(x − 2) = 2x3 − 4x2Correct product.
  2. −3x2 − 4x2 = −7x2This is the mistake. The minus was applied to 2x³ and then forgotten. Subtracting −4x² means adding 4x²: −3x² + 4x² = x².

Carry it through and you get 2x2 − 7x − 10 with remainder −25. The remainder theorem catches it at once: p(2) = 7, not −25. The fix is mechanical: put the product in parentheses with a minus in front, or change every sign in the product and then add.

4. Dropping the remainder, or writing it wrong

  1. (6x − 5) − (6x − 12) = 7Last step of Example 1.
  2. 2x2 + x + 6This is the mistake. The 7 is gone. (x − 2)(2x² + x + 6) = 2x³ − 3x² + 4x − 12, which is 7 short of the dividend.

The remainder is part of the answer. Write it over the divisor, + 7x − 2. Not + 7, and not 7/x: the 7 is what's left over from dividing by x − 2, so x − 2 goes underneath.

5. Reading the quotient one degree too high

  1.    │   1  −2  −1   5The bottom row from Example 3.
  2. x3 − 2x2 − x + 5This is the mistake. The powers were copied from the dividend, and the remainder was folded in as a constant term.

Dividing by x − c lowers the degree by exactly one, so a cubic dividend has a quadratic quotient: x2 − 2x − 1, with the 5 as remainder. The over-degree version can't be right: multiplying x + 2 by any cubic gives an x4 term, and the dividend has none.

Practice

Divide. Write any remainder as a fraction over the divisor.

  1. (6x2 + x − 7) ÷ (2x − 1)

    Show answer

    Long division (the divisor isn't x − c). 6x2 ÷ 2x = 3x; 3x(2x − 1) = 6x2 − 3x; subtracting leaves 4x − 7. Then 4x ÷ 2x = 2; 2(2x − 1) = 4x − 2; subtracting leaves −5. Answer: 3x + 2 − 52x − 1.

  2. (x3 − 8) ÷ (x − 2)

    Show answer

    Two placeholders: coefficients 1, 0, 0, −8 with c = 2. Bottom row 1, 2, 4, 0. Answer: x2 + 2x + 4, remainder 0, so x − 2 is a factor.

  3. (2x3 + 5x2 − x + 7) ÷ (x + 3)

    Show answer

    Synthetic with c = −3: coefficients 2, 5, −1, 7, middle row −6, 3, −6, bottom row 2, −1, 2, 1. Answer: 2x2 − x + 2 + 1x + 3. Check: p(−3) = −54 + 45 + 3 + 7 = 1.

  4. (x4 − 3x2 + 2x − 1) ÷ (x − 1)

    Show answer

    Coefficients 1, 0, −3, 2, −1 (0 for the missing x3), c = 1. Bottom row 1, 1, −2, 0, −1. Answer: x3 + x2 − 2x − 1x − 1. The 0 in the bottom row is the constant term of the quotient, so there isn't one.

  5. Without dividing, find the remainder when x3 + 4x2 − 3x + 2 is divided by x + 1.

    Show answer

    By the remainder theorem it's p(−1) = −1 + 4 + 3 + 2 = 8.

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