How does polynomial long division work?
It follows the same pattern as long division with numbers. You divide, multiply, subtract, and bring down, using the leading terms to decide each piece of the quotient.
The answer has two parts. The quotient is what you get on top, and the remainder is what is left at the bottom. You can write the result as quotient + remainder/divisor.
Why do you need placeholder zeros?
If a power is missing from the dividend, write it with a 0 coefficient so every column lines up. Skipping this is the most common reason long division goes wrong.
In x³ − 2x² − 4 there is no x term. Rewrite it as x³ − 2x² + 0x − 4 before you start. The calculator does this step for you and shows it in the first line of its working.
Worked example: (x³ − 2x² − 4) ÷ (x − 3)
Here is the full divide, multiply, subtract cycle. Each line uses only the leading term of what is left.
Written as one expression, (x³ − 2x² − 4)/(x − 3) = x² + x + 3 + 5/(x − 3). The remainder 5 has degree 0, which is lower than the degree of x − 3, so the division stops there.
How do you check a polynomial division answer?
Multiply the divisor by the quotient and add the remainder. If you get the original dividend back, the division is right.
This matches the dividend, so the answer checks out. The calculator runs this same check as its last step on every problem.
The remainder also tells you something on its own. When you divide by x − c, the remainder equals the value of the polynomial at x = c. Plugging x = 3 into x³ − 2x² − 4 gives 27 − 18 − 4 = 5, the same remainder. A nonzero remainder means x − 3 is not a factor. This is the remainder theorem, and it is how you test possible roots.
How do you divide by a quadratic?
The steps are the same, but you keep going until the leftover has a degree lower than 2. That means the remainder can be a linear expression, not only a number.
For (x⁴ + 1) ÷ (x² + 1), write the dividend as x⁴ + 0x³ + 0x² + 0x + 1. Divide x⁴ by x² to get x², then subtract x⁴ + x² to leave −x² + 1. Divide −x² by x² to get −1, then subtract −x² − 1 to leave 2.
The quotient is x² − 1 and the remainder is 2. Check: (x² + 1)(x² − 1) + 2 = x⁴ − 1 + 2 = x⁴ + 1. Synthetic division cannot handle this divisor, which is why long division still matters.
What if the divisor has a leading coefficient other than 1?
Nothing changes in the method, but fractions can appear in the quotient. Each step still divides the leading term of what is left by the leading term of the divisor.
Take (2x³ − 3x² + 4x − 5) ÷ (2x − 1). The first step is 2x³ ÷ 2x = x². Subtracting 2x³ − x² leaves −2x² + 4x. Next, −2x² ÷ 2x = −x, and subtracting −2x² + x leaves 3x − 5. Then 3x ÷ 2x = 3/2, and subtracting 3x − 3/2 leaves −7/2.
So the quotient is x² − x + 3/2 and the remainder is −7/2. The calculator gives exactly this answer, written with fractions instead of rounded decimals.
When should you use the polynomial long division calculator instead of synthetic division?
Use synthetic division when the divisor is linear, like x − 3 or 2x − 1. Use long division for anything else.
| Divisor | Long division | Synthetic division |
|---|---|---|
| x − c | Works | Works and is faster |
| ax − b | Works | Works with an extra step |
| Quadratic or higher | Works | Does not work |
What mistakes happen in polynomial division?
The biggest one is subtracting only the first term. When you subtract (x³ − 3x²), you must change the sign of both terms, so −2x² − (−3x²) becomes x². Putting the subtracted line in parentheses helps.
Missing placeholder zeros and stopping too early are the other two. Keep dividing until the leftover has a lower degree than the divisor.