Dividing Polynomials

Q: What happens when a term divides to exactly 1?

It stays in the answer as $1$ (or $-1$). $\dfrac{3x}{3x} = 1$, so $\dfrac{6x^{3} + 3x}{3x} = 2x^{2} + 1$. Dropping that $1$ is a very common mistake.

Dividing a polynomial by a single term looks harder than it is. A problem like $\dfrac{8x^{2} + 12x}{4}$ is really two small division problems sharing one fraction bar: split it into $\dfrac{8x^{2}}{4} + \dfrac{12x}{4}$ and handle each piece on its own.

Once it's split, every piece is a job you already know from the division rule for exponents: divide the coefficients and subtract the exponents. Split, divide, done.

Split it up

When several terms on top share one denominator, the fraction splits into one fraction per term: $\dfrac{a + b + c}{m} = \dfrac{a}{m} + \dfrac{b}{m} + \dfrac{c}{m}$ .

Every term on top must get divided — that's the part students skip. In $\dfrac{8x^{2} + 12x}{4}$ , both the $8x^{2}$ and the $12x$ go over the $4$ . Dividing only the first term is the number-one error on these problems.

Then simplify each piece

For each small fraction, divide the coefficients and subtract the exponents on each variable. $\dfrac{10x^{3}}{2x} = 5x^{2}$ : coefficients $10 \div 2 = 5$ , exponents $3 - 1 = 2$ .

Keep the signs with their terms. In $\dfrac{6x^{3} - 9x^{2} + 3x}{3x}$ , the middle piece is $\dfrac{-9x^{2}}{3x} = -3x$ — the minus sign rides along.

Watch the last term when the divisor has a variable: $\dfrac{3x}{3x} = 1$ , not $0$ . A term dividing to $1$ still shows up in the answer.

Worked examples

Example 1: divide by a number

Divide $\dfrac{8x^{2} + 12x}{4}$ .

Split into one fraction per term

\dfrac{8x^{2}}{4} + \dfrac{12x}{4}

Divide the first piece

2x^{2}

Divide the second piece

3x

Put them together

2x^{2} + 3x

Answer: $2x^{2} + 3x$

Example 2: divide by a monomial with a variable

Divide $\dfrac{10x^{3} + 6x^{2}}{2x}$ .

Split into one fraction per term

\dfrac{10x^{3}}{2x} + \dfrac{6x^{2}}{2x}

First piece: divide coefficients, subtract exponents

5x^{2}

Second piece

3x

Put them together

5x^{2} + 3x

Answer: $5x^{2} + 3x$

Example 3: three terms and a term that divides to 1

Divide $\dfrac{6x^{3} - 9x^{2} + 3x}{3x}$ .

Split into one fraction per term

\dfrac{6x^{3}}{3x} - \dfrac{9x^{2}}{3x} + \dfrac{3x}{3x}

Divide each piece

2x^{2} - 3x + 1

Note the last piece:

\dfrac{3x}{3x} = 1

, not

0

Answer: $2x^{2} - 3x + 1$

Try one yourself

Common questions

Do I have to divide every term on top?

Yes — every single one. The fraction bar is a grouping symbol, so the whole numerator gets divided. Splitting into one fraction per term makes it impossible to forget a term.

What happens when a term divides to exactly 1?

It stays in the answer as $1$ (or $-1$ ). $\dfrac{3x}{3x} = 1$ , so $\dfrac{6x^{3} + 3x}{3x} = 2x^{2} + 1$ . Dropping that $1$ is a very common mistake.

What if a term on top has a smaller exponent than the divisor?

The subtraction gives a negative exponent, meaning that piece keeps a variable in its denominator — for example $\dfrac{4x}{2x^{2}} = \dfrac{2}{x}$ . In this lesson, problems are built so that doesn't happen, but the rule covers it.

How do I check my answer?

Multiply it back by the divisor. $(2x^{2} + 3x) \cdot 4 = 8x^{2} + 12x$ — you should land exactly on the original numerator. Multiplication is the reverse of division, so this catches missed terms fast.

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Dividing Polynomials

Split it up

Then simplify each piece

Worked examples

Example 1: divide by a number

Example 2: divide by a monomial with a variable

Example 3: three terms and a term that divides to 1

Try one yourself

Common questions

Keep learning

Division Rule

Multiplication Rule

Negative Exponent Rule