Dividing polynomials works the same way as long division with numbers: divide, multiply, subtract, bring down, repeat. Synthetic division is a shortcut for one special case, dividing by x − c.
Long division
Here is (2x³ + 3x² − 5x + 6) ÷ (x + 3), one round at a time.
Example Divide 2x³ + 3x² − 5x + 6 by x + 3
- Divide the first terms: 2x³ ÷ x = 2x². Write 2x² on top.
- Multiply: 2x²(x + 3) = 2x³ + 6x². Subtract from the first two terms: (2x³ + 3x²) − (2x³ + 6x²) = −3x².
- Bring down −5x. Now you have −3x² − 5x.
- Divide: −3x² ÷ x = −3x. Multiply: −3x(x + 3) = −3x² − 9x. Subtract: 4x.
- Bring down 6. Now you have 4x + 6.
- Divide: 4x ÷ x = 4. Multiply: 4(x + 3) = 4x + 12. Subtract: −6.
- The quotient is 2x² − 3x + 4 with remainder −6.
Write the answer as quotient plus remainder over divisor:
2x² − 3x + 4 − 6/(x + 3)
Subtracting only the first term
When you subtract 2x³ + 6x², the minus applies to both terms. 3x² − 6x² = −3x², not 9x². Most long division errors are sign errors in this step. Writing the subtraction as “add the opposite” helps.
Fill in missing terms
If a power is missing, put in a zero placeholder. Otherwise the columns slide out of line.
Example Divide x³ − 8 by x − 2
- Rewrite as x³ + 0x² + 0x − 8.
- x³ ÷ x = x². Multiply: x³ − 2x². Subtract: 2x². Bring down 0x.
- 2x² ÷ x = 2x. Multiply: 2x² − 4x. Subtract: 4x. Bring down −8.
- 4x ÷ x = 4. Multiply: 4x − 8. Subtract: 0.
- x³ − 8 = (x − 2)(x² + 2x + 4), remainder 0.
A remainder of 0 means the divisor is a factor. This is how you break down cubics that don’t factor by grouping (see the factoring guide).
Dividing by a quadratic
Long division works for any divisor. You stop when the remainder’s degree is lower than the divisor’s.
Example Divide x⁴ + 2x³ − x + 5 by x² + 1
- Rewrite the top as x⁴ + 2x³ + 0x² − x + 5.
- x⁴ ÷ x² = x². Subtract x²(x² + 1) = x⁴ + x²: leaves 2x³ − x² − x.
- 2x³ ÷ x² = 2x. Subtract 2x³ + 2x: leaves −x² − 3x + 5.
- −x² ÷ x² = −1. Subtract −x² − 1: leaves −3x + 6.
- −3x + 6 has degree 1, lower than 2, so stop.
- Answer: x² + 2x − 1 + (−3x + 6)/(x² + 1).
Synthetic division
When the divisor is x − c, you can drop the variables and work with coefficients only.
Example Divide 2x³ + 3x² − 5x + 6 by x + 3 synthetically
- x + 3 = x − (−3), so c = −3.
- Write the coefficients: 2, 3, −5, 6.
- Bring down the 2.
- Multiply 2 by −3 = −6. Add to 3: −3.
- Multiply −3 by −3 = 9. Add to −5: 4.
- Multiply 4 by −3 = −12. Add to 6: −6.
- The bottom row is 2, −3, 4 | −6. That is 2x² − 3x + 4, remainder −6, the same as long division.
| Step | Coefficients | 2 | 3 | −5 | 6 |
|---|---|---|---|---|---|
| Multiply by −3 | −6 | 9 | −12 | ||
| Add | 2 | −3 | 4 | −6 |
Using the wrong sign for c
For x + 3, c is −3. For x − 3, c is 3. Use the number that makes the divisor zero.
Synthetic division only works when the divisor is linear with leading coefficient 1. For 2x − 1, either divide by x − ½ and then divide the quotient by 2, or use long division.
The remainder theorem
When you divide p(x) by x − c, the remainder equals p(c). So in the example above, p(−3) should be −6:
2(−27) + 3(9) − 5(−3) + 6 = −54 + 27 + 15 + 6 = −6.
That gives you two uses:
- To evaluate a polynomial at a number, synthetic division is often faster than plugging in.
- If p(c) = 0, then x − c is a factor (the factor theorem).
Example Is x − 1 a factor of x³ − 4x² + x + 2?
p(1) = 1 − 4 + 1 + 2 = 0, so yes. Synthetic division by 1 gives x² − 3x − 2, so x³ − 4x² + x + 2 = (x − 1)(x² − 3x − 2). The quadratic can be finished with the quadratic formula.
Try a few
Divide x² + 7x + 10 by x + 2.
Show answer
x + 5, remainder 0.
Divide 3x³ − 2x² + 4x − 1 by x − 1.
Show answer
3x² + x + 5, remainder 4.
Divide x³ + 27 by x + 3.
Show answer
x² − 3x + 9, remainder 0. Use 0x² and 0x as placeholders.
Find the remainder when x⁴ − 3x² + 2 is divided by x − 2.
Show answer
6. By the remainder theorem, 16 − 12 + 2 = 6.
Is x + 2 a factor of x³ + 2x² − x − 2?
Show answer
Yes. p(−2) = −8 + 8 + 2 − 2 = 0.
Divide 2x³ + x − 3 by x² + x.
Show answer
2x − 2, remainder 3x − 3.