The same four moves as ordinary long division, with the sign of the subtraction doing all the damage.
Find the remainder
(2x³ − 3x² + 5) ÷ (x − 2)
The remainder theorem beats long division here: the remainder is P(2) = 2(8) − 3(4) + 5 = 16 − 12 + 5 = 9. C evaluates at −2 — the divisor x − 2 is zero at x = +2. Note the missing x term; write it as 0x if you do divide properly.
There is no x term, so write 0x. Columns must line up or the subtraction goes wrong.
2x³ − 3x² + 5
as given — the x is missing
2x³ − 3x² + 0x + 5
now every power is present
It is the same reason 1005 ÷ 5 is written with the zeros in place. A gap in the columns shifts everything below it, and the remainder comes out wrong.
2x³ ÷ x = 2x²
divide the leading terms only
2x²(x − 2) = 2x³ − 4x²
multiply back
−3x² − (−4x²) = x²
subtract, then bring down 0x
x² ÷ x = x → x² − 2x
round two
0x − (−2x) = 2x, bring down 5
2x + 5
2x ÷ x = 2 → 2x − 4
round three
5 − (−4) = 9
degree 0, so we stop
Every term of the product changes sign — not just the first.
Dividing by x − 2 means you stop once the leftover has no x in it. Trying to continue into fractions of x is a sign the process already finished.
2x² + x + 2 + 9x − 2
dividend = divisor × quotient + remainder
the division algorithm
2x³ − 3x² + 5 = (x − 2)(2x² + x + 2) + 9
the same statement, cleared of fractions
Exactly like 17 ÷ 5 = 3 remainder 2, which is also 3 + ⅖, and also 17 = 5 × 3 + 2. Which form to give depends on the question — “find the quotient and remainder” wants the two pieces named separately.
Only for a divisor of the form x − c. Here c = 2.
| c = 2 | 2 | −3 | 0 | 5 |
|---|---|---|---|---|
| add | 4 | 2 | 4 | |
| result | 2 | 1 | 2 | 9 |
bring down the 2
the leading coefficient
2 × 2 = 4, −3 + 4 = 1
multiply by c, add down
1 × 2 = 2, 0 + 2 = 2
repeat
2 × 2 = 4, 5 + 4 = 9
the last one is the remainder
2x² + x + 2, r 9
degree drops by one
For x + 3 the value of c is −3, because x + 3 is x − (−3). And for 2x − 1, divide the whole thing by 2 first, or use long division — synthetic division only handles a leading coefficient of 1.
Divide 6x³ + x² − 7x + 5 by 2x + 1.
6x³ ÷ 2x = 3x² → 6x³ + 3x²
subtract: −2x², bring down −7x
−2x² ÷ 2x = −x → −2x² − x
subtract: −6x, bring down 5
−6x ÷ 2x = −3 → −6x − 3
subtract: 5 − (−3) = 8
quotient 3x² − x − 3, remainder 8
P(−½) = −¾ + ¼ + 3½ + 5 = 8 ✓
Different: every leading term is divided by 2x, not x, so the quotient’s coefficients are half the size you might expect.
Divide x⁴ − 3x² + x + 5 by x + 2, synthetically.
c = −2; row: 1 0 −3 1 5
x + 2 = x − (−2); 0 for x³
1, 0 − 2 = −2, −3 + 4 = 1, 1 − 2 = −1, 5 + 2 = 7
multiply by −2, add down
quotient x³ − 2x² + x − 1, remainder 7
P(−2) = 16 − 12 − 2 + 5 = 7 ✓
Different: c is negative, so the multiplied numbers alternate in sign — and the missing x³ still needs its 0.
With +2 the row runs 1, 2, 1, 3, 11: quotient x³ + 2x² + x + 3, remainder 11. That is the division by x − 2, a different question.
Long division works for any divisor; synthetic division does not. Divide 2x³ + x² − x + 4 by x² + x − 1.
2x³ ÷ x² = 2x → 2x³ + 2x² − 2x
divide leading terms, multiply back
subtract: −x² + x + 4
x² − 2x², −x + 2x, bring down 4
−x² ÷ x² = −1 → −x² − x + 1
round two
subtract: 2x + 3
degree 1 < 2, so stop
With a quadratic divisor you stop when the leftover drops to degree 1, not 0. A remainder like 2x + 3 is finished — do not try to divide it again.
Backwards: when P(x) is divided by x + 1, the quotient is x² − 3x + 2 and the remainder is −5. Find P(x). Use dividend = divisor × quotient + remainder from Step 4.
P(x) = (x + 1)(x² − 3x + 2) + (−5)
the division algorithm
= x³ − 3x² + 2x + x² − 3x + 2 − 5
expand
= x³ − 2x² − x − 3
P(−1) = −1 − 2 + 1 − 3 = −5 ✓
If the dividend has an unknown coefficient instead, expand divisor × quotient the same way and match it to the dividend term by term — the unknown is read off one coefficient, and the constant terms then give the remainder.
Division items ask for a quotient (use synthetic division when the divisor is x − c), a remainder (use the remainder theorem — no division at all), or work backwards from dividend = divisor × quotient + remainder. Remember placeholders for missing powers; skipping one shifts every column.
Find the remainder
(x³ + 4x² − 7) ÷ (x + 1)
x + 1 is zero at x = −1, so the remainder is P(−1) = −1 + 4 − 7 = −4. C evaluates at +1. The sign always flips: x + 1 tests −1, and x − 1 tests +1.
