How do I handle division when both the dividend and divisor have missing terms?

Write every polynomial in standard form, then insert a zero for each missing term. For the divisor, do the same if it has gaps, like x² - 1 becoming x² + 0x - 1. This keeps the columns aligned so subtraction works correctly.

What does the remainder actually mean when it is not zero?

The remainder is the polynomial left over after the division, and its degree is always strictly less than the divisor's degree. It is written as a fraction over the divisor, not as a decimal. In rational functions, that remainder term often becomes negligible for large x, which is why the quotient line becomes the oblique asymptote when the numerator's degree is one more than the denominator's.

Can I use synthetic division for a divisor like 2x - 3?

Not directly, because synthetic division only works for divisors of the form (x - c). For 2x - 3, you can factor out the 2, divide by (x - 3/2) using synthetic division, then divide the quotient by 2. That extra step usually makes long division simpler for non-monic divisors.

Why does my answer look different from the calculator's?

Equivalent forms can look different if terms are ordered differently or if a remainder is written with a different sign. Always verify by multiplying the quotient by the divisor and adding the remainder; if it matches the dividend, your answer is correct regardless of appearance.

How do I divide when the divisor has a degree greater than the dividend?

The quotient is zero, and the remainder is the entire dividend. For example, (x + 1) ÷ (x² + 1) has quotient 0 and remainder x + 1, written as 0 + (x + 1)/(x² + 1). The division algorithm stops immediately because the divisor's degree is higher.

Polynomial Long Division Examples and Practice Problems

Polynomial Long Division Examples: The Algorithm That Never Skips a Beat

If you are here, you have likely stared at a polynomial long division problem and wondered where the quotient goes or what to do with a remainder that is not a number. These polynomial long division examples are solved the same way every time: divide the leading term, multiply the whole divisor, subtract, and bring down the next term. The trap is not the arithmetic; it is the alignment. Miss a zero placeholder and the columns shift, turning a clean division into a mess. This resource covers every type of problem encountered in an Algebra 2 or Precalculus course, from a linear divisor with no remainder to a quadratic divisor with missing terms. Each example is checked with a calculator, and the practice set at the end hides the answers until you are ready. Work the problems in order, and the pattern becomes as familiar as long division with digits.

Linear Divisor, No Remainder: The Clean Slate

Start with the case where everything divides evenly. Take (x³ + 2x² - 5x - 6) ÷ (x + 1). Write the dividend inside the long division bracket and the divisor outside, both in standard form. Divide x³ by x to get x². Multiply (x + 1) by x² to get x³ + x², and subtract: the x³ cancels, leaving x². Bring down the -5x. Divide x² by x to get x, multiply (x + 1) by x to get x² + x, subtract to get -6x. Bring down the -6. Divide -6x by x to get -6, multiply (x + 1) by -6 to get -6x - 6, subtract to get zero. The quotient is x² + x - 6, and the remainder is zero.

Verify with the calculator: multiply (x² + x - 6) by (x + 1). Distribute to get x³ + x² + x² + x - 6x - 6, which simplifies to x³ + 2x² - 5x - 6, the original dividend. That multiplication check is the whole game: quotient times divisor plus remainder equals dividend. When the remainder is zero, the divisor is a factor, and the quotient is the other factor. This is the case that makes the Remainder Theorem feel obvious: if you evaluate the dividend at x = -1, you get zero because the remainder is zero.

Linear Divisor with Remainder: The Fraction Is Not Optional

Not every division comes out even. Divide (2x³ - 3x² + 4x - 5) ÷ (x - 2). Set up the bracket, divide 2x³ by x to get 2x², multiply (x - 2) by 2x² to get 2x³ - 4x², subtract to get x². Bring down the 4x. Divide x² by x to get x, multiply (x - 2) by x to get x² - 2x, subtract to get 6x. Bring down the -5. Divide 6x by x to get 6, multiply (x - 2) by 6 to get 6x - 12, subtract to get 7. The quotient is 2x² + x + 6, and the remainder is 7.

The answer is written as 2x² + x + 6 + 7/(x - 2). That fraction over the divisor is the only correct form. Writing the remainder as a decimal, like 7.5, is wrong because the remainder is a constant here only because the divisor is linear; with higher-degree divisors it stays a polynomial. Check it: multiply (2x² + x + 6) by (x - 2) to get 2x³ - 4x² + x² - 2x + 6x - 12, which is 2x³ - 3x² + 4x - 12. Add the remainder 7 to get 2x³ - 3x² + 4x - 5, matching the dividend. The Remainder Theorem applies here: evaluate the dividend at x = 2, and you get 2(8) - 3(4) + 4(2) - 5 = 16 - 12 + 8 - 5 = 7, the same remainder.

Divisor with Leading Coefficient Not 1: Scaling Each Step

When the divisor has a leading coefficient other than 1, the division works the same but each step involves a fraction or a careful scaling. Divide (6x³ - 7x² + 2x - 1) ÷ (2x - 1). Divide 6x³ by 2x to get 3x². Multiply (2x - 1) by 3x² to get 6x³ - 3x², subtract to get -4x². Bring down the 2x. Divide -4x² by 2x to get -2x. Multiply (2x - 1) by -2x to get -4x² + 2x, subtract to get 0x. Bring down the -1. Divide 0x by 2x to get 0, multiply (2x - 1) by 0 to get 0, subtract to get -1. The quotient is 3x² - 2x, and the remainder is -1.

The answer is 3x² - 2x - 1/(2x - 1). The leading coefficient of the divisor means the quotient terms often have fractional coefficients if the dividend does not divide cleanly. Some students try to use synthetic division here, but synthetic division only works for divisors of the form (x - c). To use synthetic division on 2x - 1, you would first factor out the 2, divide by (x - 1/2), and then divide the quotient by 2. That extra step is usually more work than just doing long division. Check the answer: multiply (3x² - 2x) by (2x - 1) to get 6x³ - 3x² - 4x² + 2x, which is 6x³ - 7x² + 2x. Add the remainder -1 to get 6x³ - 7x² + 2x - 1, matching the dividend.

Quadratic Divisor: When the Divisor Has Three Terms

Divide (x⁴ - 3x³ + 2x² - 5x + 1) ÷ (x² - 2x + 1). Set up the bracket. Divide x⁴ by x² to get x². Multiply (x² - 2x + 1) by x² to get x⁴ - 2x³ + x², subtract to get -x³ + x². Bring down the -5x. Divide -x³ by x² to get -x. Multiply (x² - 2x + 1) by -x to get -x³ + 2x² - x, subtract to get -x² - 4x. Bring down the 1. Divide -x² by x² to get -1. Multiply (x² - 2x + 1) by -1 to get -x² + 2x - 1, subtract to get -6x + 2. The quotient is x² - x - 1, and the remainder is -6x + 2.

This is where students often slip: they try to keep dividing because they see an x term, but the divisor has a higher degree, so the division stops. The answer is x² - x - 1 + (-6x + 2)/(x² - 2x + 1). Add the remainder -6x + 2 to get x⁴ - 3x³ + 2x² - 5x + 1, the dividend. This kind of problem matters for oblique asymptotes later, because when the numerator's degree is exactly one more than the denominator's, the quotient line is the asymptote.

Missing Terms: The Zero Placeholder That Saves the Columns

When a polynomial has a missing term, like x³ - 5x + 2, there is no x² term. Skipping it is the most common error in polynomial long division. Rewrite the dividend as x³ + 0x² - 5x + 2, putting a zero in the x² column. The quotient is x² + 2x - 1, and the remainder is zero.

The zero placeholder is not optional. Without it, the terms shift, and the subtraction step produces wrong coefficients. For a divisor like x² + 1, where the dividend might be x⁴ + 3x² - 2, you would write x⁴ + 0x³ + 3x² + 0x - 2, adding two placeholders. The calculator handles this automatically, but when you work by hand, write out every term from the highest degree down, filling zeros for gaps. This keeps the columns aligned so subtraction works correctly.

Practice Problems: Answers Hidden Until You Try

Work these on paper before checking. Cover the answers below the line, solve each, then reveal to verify.

Divide (x³ + 4x² - 3x - 12) ÷ (x + 3).
Divide (2x⁴ - 5x³ + 3x² - x + 7) ÷ (x - 1).
Divide (9x³ - 6x² + 3x - 2) ÷ (3x - 1).
Divide (x⁵ - 1) ÷ (x - 1).

Answers: 1. x² + x - 6, remainder 6. 2. 2x³ - 3x² + 0x - 1, remainder 6. 3. 3x² - x + 2/3, remainder -4/3. 4. x⁴ + x³ + x² + x + 1, remainder 0.

Polynomial Long Division Examples