Unit 5 · Lesson 5 of 5

Dividing polynomials

About 25 minutes · 5 steps · one idea: divide, multiply, subtract, bring down

You'll be able to: divide a polynomial by a monomial or by a binomial, and write any remainder correctly.

  1. 1 Warm-up
  2. 2 The idea
  3. 3 Examples
  4. 4 Practice
  5. 5 Wrap-up

Step 1 of 5Warm-up

Two quick questions from earlier lessons. Answer from memory first. Pulling it out of your head is what makes it stick.

Multiply out: \((x - 5)^2\).

First squared: \(x^2\). Twice the product: \(2 \cdot x \cdot 5 = 10x\), with a minus. Last squared: \(+25\).

Answer: \(x^2 - 10x + 25\).

Simplify \(\dfrac{12x^5}{3x^2}\).

Divide the numbers: \(12 \div 3 = 4\). Same base, so subtract the exponents: \(x^{5-2} = x^3\).

Answer: \(4x^3\).

Step 2 of 5The idea

By a monomial: split the fraction. Divide each term on top by the bottom, using the quotient rule for exponents.

By a binomial: use long division, the same steps as long division with numbers. The answer on top is the quotient. What is left at the end is the remainder.

Monomial. \(\dfrac{a + b}{m} = \dfrac{a}{m} + \dfrac{b}{m}\).

Long division loop. (1) Divide the first term by the divisor's first term. (2) Multiply that result by the whole divisor. (3) Subtract (flip every sign). (4) Bring down the next term. Repeat until nothing is left to bring down.

Missing powers. Before you start, write the top in standard form and fill gaps with 0 terms: \(x^3 - 8\) becomes \(x^3 + 0x^2 + 0x - 8\).

Remainder. Answer \(=\) quotient \(+ \dfrac{\text{remainder}}{\text{divisor}}\). Check: divisor \(\times\) quotient \(+\) remainder \(=\) the top.

Why this shows up on the placement test: dividing polynomials is a standard "simplify" item, and it comes back when you simplify rational expressions.

Step 3 of 5Worked examples

Example 1: \(\dfrac{12x^3 - 8x^2 + 4x}{4x}\)

  1. Split: \(\dfrac{12x^3}{4x} - \dfrac{8x^2}{4x} + \dfrac{4x}{4x}\).
  2. Each piece: \(3x^2 - 2x + 1\). \(\dfrac{4x}{4x} = 1\), not 0. Don't drop it.

Example 2: \((2x^2 + 3x - 4) \div (x + 2)\)

  1. Divide: \(2x^2 \div x = 2x\). This is the first term of the quotient.
  2. Multiply: \(2x(x + 2) = 2x^2 + 4x\).
  3. Subtract: \((2x^2 + 3x) - (2x^2 + 4x) = -x\). Bring down the \(-4\): now you have \(-x - 4\).
  4. Divide again: \(-x \div x = -1\). Multiply: \(-1(x + 2) = -x - 2\).
  5. Subtract: \((-x - 4) - (-x - 2) = -4 + 2 = -2\). Nothing left to bring down, so the remainder is \(-2\).
  6. Answer: \(2x - 1 + \dfrac{-2}{x + 2}\), which is the same as \(2x - 1 - \dfrac{2}{x + 2}\).
  7. Check: \((x + 2)(2x - 1) + (-2) = 2x^2 + 3x - 2 - 2 = 2x^2 + 3x - 4\). That is the top, so it is right.

Optional shortcut: synthetic division. It works only when the divisor looks like \(x - c\). For \(x + 2\), \(c = -2\). Write \(c\), then the coefficients of the top: 2, 3, −4.

\[\begin{array}{r|rrr} -2 & 2 & 3 & -4 \\ & & -4 & 2 \\ \hline & 2 & -1 & -2 \end{array}\]

Bring down the 2. Multiply by \(c\) and write it under the next number: \(2 \cdot (-2) = -4\). Add: \(3 + (-4) = -1\). Repeat: \(-1 \cdot (-2) = 2\), and \(-4 + 2 = -2\).

The bottom row gives the quotient's coefficients \(2, -1\) (so \(2x - 1\)) and the remainder \(-2\), the same as Example 2. If this feels like one thing too many, skip it: long division always works.

Step 4 of 5Practice

Type your answer and press Check. If you miss twice, open the worked solution. Then mark how it went. If there is a remainder, type the whole answer, like x + 2 + 3/(x - 1).

Divide: \(\dfrac{15x^4 - 10x^3 + 5x^2}{5x^2}\).

Split: \(\dfrac{15x^4}{5x^2} - \dfrac{10x^3}{5x^2} + \dfrac{5x^2}{5x^2}\).

Each piece: \(3x^2 - 2x + 1\).

Answer: \(3x^2 - 2x + 1\).

Divide: \(\dfrac{10x^3 - 6x^2 + 4x}{-2x}\).

Split: \(\dfrac{10x^3}{-2x} + \dfrac{-6x^2}{-2x} + \dfrac{4x}{-2x}\).

Each piece: \(-5x^2 + 3x - 2\). Dividing by a negative flips every sign.

Answer: \(-5x^2 + 3x - 2\).

Divide: \((x^2 + 7x + 12) \div (x + 3)\).

\(x^2 \div x = x\). Multiply: \(x(x + 3) = x^2 + 3x\). Subtract: \(7x - 3x = 4x\). Bring down: \(4x + 12\).

\(4x \div x = 4\). Multiply: \(4(x + 3) = 4x + 12\). Subtract: \(0\).

Answer: \(x + 4\).

Divide: \((2x^2 - 5x - 3) \div (x - 3)\).

\(2x^2 \div x = 2x\). Multiply: \(2x(x - 3) = 2x^2 - 6x\). Subtract: \(-5x - (-6x) = x\). Bring down: \(x - 3\).

\(x \div x = 1\). Multiply: \(1(x - 3) = x - 3\). Subtract: \(0\).

Answer: \(2x + 1\).

Divide: \((x^3 - 8) \div (x - 2)\).

Fill the gaps: \(x^3 + 0x^2 + 0x - 8\).

\(x^3 \div x = x^2\). Multiply: \(x^3 - 2x^2\). Subtract: \(0x^2 - (-2x^2) = 2x^2\). Bring down: \(2x^2 + 0x\).

\(2x^2 \div x = 2x\). Multiply: \(2x^2 - 4x\). Subtract: \(4x\). Bring down: \(4x - 8\).

\(4x \div x = 4\). Multiply: \(4x - 8\). Subtract: \(0\). Answer: \(x^2 + 2x + 4\).

Divide: \((x^2 + 4x - 1) \div (x + 1)\). Write the answer as quotient \(+\) remainder over divisor.

\(x^2 \div x = x\). Multiply: \(x^2 + x\). Subtract: \(4x - x = 3x\). Bring down: \(3x - 1\).

\(3x \div x = 3\). Multiply: \(3x + 3\). Subtract: \(-1 - 3 = -4\). The remainder is \(-4\).

Answer: \(x + 3 + \dfrac{-4}{x + 1} = x + 3 - \dfrac{4}{x + 1}\).

Divide: \((2x^3 + x - 5) \div (x - 1)\). Write the answer as quotient \(+\) remainder over divisor.

Fill the gap: \(2x^3 + 0x^2 + x - 5\).

\(2x^3 \div x = 2x^2\). Multiply: \(2x^3 - 2x^2\). Subtract: \(2x^2\). Bring down: \(2x^2 + x\).

\(2x^2 \div x = 2x\). Multiply: \(2x^2 - 2x\). Subtract: \(x - (-2x) = 3x\). Bring down: \(3x - 5\).

\(3x \div x = 3\). Multiply: \(3x - 3\). Subtract: \(-5 - (-3) = -2\). The remainder is \(-2\).

Answer: \(2x^2 + 2x + 3 - \dfrac{2}{x - 1}\).

Which expression is equal to \(\dfrac{x^2 - 2x + 5}{x - 3}\)?

  1. \(x + 1 - \dfrac{8}{x - 3}\)
  2. \(x - 5 - \dfrac{10}{x - 3}\)
  3. \(x + 1 + \dfrac{2}{x - 3}\)
  4. \(x + 1 + \dfrac{8}{x - 3}\)

\(x^2 \div x = x\). Multiply: \(x^2 - 3x\). Subtract: \(-2x - (-3x) = x\). Bring down: \(x + 5\).

\(x \div x = 1\). Multiply: \(x - 3\). Subtract: \(5 - (-3) = 8\). The remainder is \(8\).

Answer: \(x + 1 + \dfrac{8}{x - 3}\), choice (d). Check: \((x - 3)(x + 1) + 8 = x^2 - 2x - 3 + 8 = x^2 - 2x + 5\).

(b) comes from not flipping the sign of \(-3x\) in the first subtraction; (a) and (c) come from sign slips in the last subtraction.

Step 5 of 5Wrap-up

You can now divide by a monomial (split into one fraction per term) and by a binomial (divide, multiply, subtract, bring down), fill missing powers with 0, and write a remainder as remainder over divisor.

Read or watch next: OpenStax Intermediate Algebra 2e, 5.4 Dividing Polynomials (free textbook, with more practice at the end of the section).

Also good: Khan Academy: Algebra 2: Dividing polynomials by x (short videos). Quick rules: Exponents and polynomials cheat sheet.

Something unclear? Ask your teacher bot a follow-up question, for example: "Why do I need to write \(0x^2\) when a power is missing?" It knows this course and can explain things another way.