🎯⭐ INTERACTIVE LESSON

Polynomial Division and Remainder Theorem

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Polynomial Division and Remainder Theorem - Complete Interactive Lesson

Part 1: Long Division of Polynomials

➗ Polynomial Division & the Remainder Theorem

Part 1 of 7 — Long Division of Polynomials

Topics in This Part

Section
The Division Algorithm
Setting Up Long Division
Carrying Out the Steps
Handling Missing Terms

🔑 Key Concept: Dividing polynomials works just like the long division you learned for whole numbers. Every division produces a quotient and a remainder, and that single idea unlocks the Remainder Theorem, the Factor Theorem, and how to find a polynomial's zeros.

The Division Algorithm

When you divide one polynomial $p(x)$ (the dividend) by another $d(x)$ (the divisor), you get a quotient $q(x)$ and a remainder $r(x)$ that satisfy:

$p(x) = d(x)\cdot q(x) + r(x)$

where the remainder has lower degree than the divisor. With $17 \div 5$ you write $17 = 5\cdot 3 + 2$ — quotient $3$ , remainder $2$ . Polynomials follow the exact same rule:

$\underbrace{x^2 + 5x + 6}_{\text{dividend}} = \underbrace{(x+2)}_{\text{divisor}}\cdot\underbrace{(x+3)}_{\text{quotient}} + \underbrace{0}_{\text{remainder}}$

💡 If the remainder is $0$ , the divisor divides the dividend evenly — the divisor is a factor. That observation becomes the Factor Theorem in Part 4.

Vocabulary Check 🔽

For the statement $2x^3 - 3x^2 + 4x - 5 = (x-2)(2x^2 + x + 6) + 7$ , identify each piece.

Carrying Out Long Division

Let's divide $2x^3 - 3x^2 + 4x - 5$ by $x - 2$ step by step. Each round: the leading terms, , then .

Concept Check 🎯

Watch Out: Missing Terms

If the dividend skips a power of $x$ , insert a placeholder with coefficient $0$ so your columns line up. To divide $x^3 - 7x - 6$ by $x + 1$ , write the dividend as .

Long Division Drill 🧮

Carry out each division and report what's asked.

1) Divide $x^2 + 5x + 6$ by $x + 2$ . Enter the remainder. 2) Divide $2 x^{3}$ by . Enter the . Divide by . Enter the (form , e.g. ).

Part 2: Synthetic Division (the Shortcut)

➗ Polynomial Division & the Remainder Theorem

Part 2 of 7 — Synthetic Division (the Shortcut)

🔑 The Idea: When the divisor is linear and monic — that is, of the form $x - c$ — you can skip the bulky long division and use a fast number-only process called synthetic division.

Setting Up Synthetic Division

To divide $p(x)$ by $x - c$ :

Part 3: The Remainder Theorem

➗ Polynomial Division & the Remainder Theorem

Part 3 of 7 — The Remainder Theorem

🔑 The Theorem: When you divide a polynomial $p(x)$ by $x - c$ , the remainder equals $p(c)$ . So you can evaluate a polynomial just by reading off a synthetic-division remainder — no plugging-in required.

Why It Works

The Division Algorithm says where is a constant (the divisor is degree , so the remainder has degree ).

Part 4: The Factor Theorem

➗ Polynomial Division & the Remainder Theorem

Part 4 of 7 — The Factor Theorem

🔑 The Theorem: $x - c$ is a factor of $p(x)$ if and only if $p(c) = 0$ . A zero remainder and a factor are the same thing said two ways.

Part 5: Finding All the Zeros

➗ Polynomial Division & the Remainder Theorem

Part 5 of 7 — Finding All the Zeros

🔑 The Strategy: Find one zero, divide it out, and repeat on the smaller quotient until you reach a quadratic you can factor or solve. This "depress-and-conquer" loop finds every real zero of a polynomial.

The Depress-and-Conquer Loop

To find all zeros of a polynomial $p(x)$ :

Find one zero $c$ — test small integers ( $\pm 1, \pm 2, \dots$ ) until .

Part 6: Rational Roots & Slant Asymptotes

➗ Polynomial Division & the Remainder Theorem

Part 6 of 7 — Rational Roots & Slant Asymptotes

🔑 Two payoffs: The Rational Root Theorem narrows your zero hunt to a short list of fractions, and polynomial division reveals the slant (oblique) asymptote of a rational function.

The Rational Root Theorem

If a polynomial with integer coefficients $p(x) = a_n x^n + \cdots + a_0$ has a rational zero (in lowest terms), then:

Part 7: Mixed Practice & Mastery Check

➗ Polynomial Division & the Remainder Theorem

Part 7 of 7 — Mixed Practice & Mastery Check

You can now (1) divide polynomials long-hand and synthetically, (2) apply the Remainder Theorem, (3) apply the Factor Theorem, (4) find every zero, and (5) read off rational-root candidates and slant asymptotes. Let's put it all together.

Quick Reference

Goal	Key move
Divide by $x - c$ fast	synthetic division with $c$
Evaluate $p(c)$

Step 1. $\dfrac{2x^3}{x} = 2x^2$ . Multiply: $2x^2(x-2) = 2x^3 - 4x^2$ . Subtract: $(2x^3 - 3x^2) - (2x^3 - 4x^2) = x^2$

Step 2. Bring down $+4x$ . Now $\dfrac{x^2}{x} = x$ . Multiply: $x(x-2) = x^2 - 2x$ . Subtract: $(x^2 + 4x) - (x^2 - 2x) = 6x$

Step 3. Bring down $-5$ . Now $\dfrac{6x}{x} = 6$ . Multiply: $6(x-2) = 6x - 12$ . Subtract: $(6x - 5) - (6x - 12) = 7$

The remainder is $7$ , so: $2x^3 - 3x^2 + 4x - 5 = (x-2)(2x^2 + x + 6) + 7$

✅ Check: $(x-2)(2x^2+x+6) = 2x^3 + x^2 + 6x - 4x^2 - 2x - 12 = 2x^3 - 3x^2 + 4x - 12$ , and $+7$ gives $2x^3 - 3x^2 + 4x - 5$ . ✓

x^3 + 0x^2 - 7x - 6

Carrying out the division gives: $x^3 - 7x - 6 = (x+1)(x^2 - x - 6) + 0$

The remainder is $0$ , so $x+1$ is a factor — and notice $x^2 - x - 6 = (x-3)(x+2)$ .

⚠️ Most common setup error: forgetting a zero placeholder. Skipping the missing $0x^2$ term shifts every column and corrupts the whole division.

2 x^{3} - 3 x^{2} + 4 x - 5

Write $c$ in the box. Beware the sign: dividing by $x - 2$ uses $c = 2$ ; dividing by $x + 3$ uses $c = -3$ (because $x+3 = x-(-3)$ ).
List the coefficients of $p(x)$ in order, including $0$ for any missing power.
Bring down the first coefficient. Then repeatedly: multiply by $c$ , write the result under the next coefficient, and add.

Worked Example: $(x^3 - 4x^2 + 6x - 4) \div (x - 2)$

Here $c = 2$ and the coefficients are $1, -4, 6, -4$ :

\begin{array}{c|cccc} 2 & 1 & -4 & 6 & -4 \\ & & 2 & -4 & 4 \\ \hline & 1 & -2 & 2 & \;\;0 \end{array}

The bottom row gives the quotient coefficients $1, -2, 2$ and the final number is the remainder: $x^3 - 4x^2 + 6x - 4 = (x-2)(x^2 - 2x + 2) + 0$

💡 The quotient's degree is always one less than the dividend's. A cubic divided by a linear term leaves a quadratic quotient.

Concept Check 🎯

A Second Example with a Negative $c$

Divide $2x^3 + 3x^2 - 8x + 3$ by $x + 3$ . Here $c = -3$ , coefficients $2, 3, -8, 3$ :

\begin{array}{c|cccc} -3 & 2 & 3 & -8 & 3 \\ & & -6 & 9 & -3 \\ \hline & 2 & -3 & 1 & \;\;0 \end{array}

Reading the bottom row: quotient $2x^2 - 3x + 1$ , remainder $0$ .

$2x^3 + 3x^2 - 8x + 3 = (x+3)(2x^2 - 3x + 1)$

✅ Because the remainder is $0$ , $x+3$ is a factor — and $2x^2 - 3x + 1 = (2x-1)(x-1)$ , which fully factors the cubic. We'll exploit this in Part 5.

Synthetic Division Drill 🧮

Use synthetic division (remember $0$ placeholders for missing powers!).

1) $(x^3 - 4x^2 + 6x - 4) \div (x - 2)$ . Enter the remainder. 2) $(2x^3 + 3x^2 - 8x + 3) \div (x + 3)$ . Enter the constant term of the quotient. 3) $(x^4 - 16) \div (x - 2)$ . Enter the remainder.

Order the Process 🔽

Synthetic-dividing $x^3 - 4x^2 + 6x - 4$ by $x - 2$ . Fill in each stage.

p(x) = (x - c)\,q(x) + r,

p (x) = (x - c) q (x) + r,

Now substitute $x = c$ : $p(c) = (c - c)\,q(c) + r = 0\cdot q(c) + r = r.$

The $(x-c)$ factor vanishes, leaving exactly the remainder. That's the whole proof.

$\boxed{\;p(c) = \text{remainder when } p(x) \text{ is divided by } x - c\;}$

Earlier we divided $x^3 - 4x^2 + 6x - 4$ by $x - 2$ and got remainder $0$ . The theorem predicts $p(2) = 0$ . Check directly: $p(2) = 8 - 16 + 12 - 4 = 0 \;✓$

Concept Check 🎯

Synthetic Substitution

Combining synthetic division with the Remainder Theorem gives a slick way to evaluate $p(c)$ . To find $p(-1)$ for $p(x) = 2x^3 - 3x^2 + x - 5$ , synthetic-divide by $x + 1$ (so $c = -1$ ):

\begin{array}{c|cccc} -1 & 2 & -3 & 1 & -5 \\ & & -2 & 5 & -6 \\ \hline & 2 & -5 & 6 & -11 \end{array}

The remainder is $-11$ , so $p(-1) = -11$ .

✅ Direct check: $p(-1) = 2(-1)^3 - 3(-1)^2 + (-1) - 5 = -2 - 3 - 1 - 5 = -11$ ✓

This "synthetic substitution" is often faster than plugging in, especially for high-degree polynomials.

Remainder Theorem Drill 🧮

Use the Remainder Theorem (synthetic substitution or direct evaluation).

1) $p(x) = x^3 - 4x^2 + 6x - 4$ . Find the remainder when divided by $x - 2$ , i.e. $p(2)$ . 2) $p(x) = 2x^3 - 3x^2 + x - 5$ . Find $p(-1)$ . 3) $p(x) = x^4 - 3x^2 + 2x - 7$ . Find $p(3)$ .

Read the Remainders 🔽

A polynomial $p(x)$ gives these remainders. Use the Remainder Theorem.

Divide by	Remainder
$x - 1$	$0$
$x - 4$	$9$
$x + 2$	$-3$

From Remainder to Factor

The Factor Theorem is just the Remainder Theorem with remainder $= 0$ :

$p(c) = 0 \;\Longleftrightarrow\; \text{remainder} = 0 \;\Longleftrightarrow\; (x - c) \text{ divides } p(x) \text{ evenly}.$

This links three ideas that AP Precalculus treats as interchangeable:

Statement	Equivalent to
$p(c) = 0$	$c$ is a zero (root) of $p$
$(x - c)$ is a

💡 So "find the zeros," "factor the polynomial," and "find the $x$ -intercepts" are three phrasings of the same task.

Concept Check 🎯

Using It to Test Candidates

Is $x - 1$ a factor of $p(x) = x^3 - 7x + 6$ ? Test $p(1)$ : $p(1) = 1 - 7 + 6 = 0 \;✓$ Yes — so $(x-1)$ is a factor. Synthetic-dividing by $x - 1$ peels it off:

\begin{array}{c|cccc} 1 & 1 & 0 & -7 & 6 \\ & & 1 & 1 & -6 \\ \hline & 1 & 1 & -6 & \;\;0 \end{array}

So $p(x) = (x-1)(x^2 + x - 6) = (x-1)(x+3)(x-2)$ , since .

⚠️ Sign trap: A factor $(x - c)$ corresponds to a zero $x = +c$ . The factor $(x+3)$ comes from the zero , not .

Match Factor ↔ Zero 🔽

Factor Theorem Drill 🧮

For $p(x) = x^3 - 7x + 6$ , evaluate to test each candidate. Enter the value.

1) $p(1) = \,?$ (Is $x-1$ a factor?) 2) $p(-3) = \,?$ (Is $x+3$ a factor?) 3) $p(0) = \,?$ (Is $x$ a factor?)

Divide

p(x)

(x - c)

using synthetic division. The quotient has degree one lower.

Repeat on the quotient. Once it's a quadratic, factor it or use the quadratic formula.

Example: $p(x) = x^3 - 4x^2 + x + 6$

Test $x = -1$ : $\;p(-1) = -1 - 4 - 1 + 6 = 0$ ✓. Divide by $x + 1$ :

\begin{array}{c|cccc} -1 & 1 & -4 & 1 & 6 \\ & & -1 & 5 & -6 \\ \hline & 1 & -5 & 6 & \;\;0 \end{array}

So $p(x) = (x+1)(x^2 - 5x + 6) = (x+1)(x-2)(x-3)$ .

The zeros are $x = -1,\; 2,\; 3$ .

Concept Check 🎯

A Cubic with a Fractional Zero

Find all zeros of $p(x) = 2x^3 + x^2 - 13x + 6$ .

Test $x = 2$ : $\;p(2) = 16 + 4 - 26 + 6 = 0$ ✓. Divide by :

\begin{array}{c|cccc} 2 & 2 & 1 & -13 & 6 \\ & & 4 & 10 & -6 \\ \hline & 2 & 5 & -3 & \;\;0 \end{array}

The quotient is $2x^2 + 5x - 3 = (2x - 1)(x + 3)$ . Setting each factor to :

The zeros are $x = 2,\; \tfrac{1}{2},\; -3$ .

💡 Not every zero is an integer. The factor $(2x - 1)$ produces the fractional zero $x = \tfrac{1}{2}$ — Part 6's Rational Root Theorem tells you exactly which fractions to test.

From Factors to Zeros 🔽

A polynomial fully factors as $p(x) = (x + 1)(x - 2)(2x - 1)$ .

Find the Zeros Drill 🧮

1) $p(x) = x^3 - 4x^2 + x + 6$ factors as $(x+1)(x-2)(x-3)$ . Enter the smallest zero. 2) Same polynomial: enter the largest zero. 3) $p(x) = 2x^3 + x^2 - 13x + 6$ has a zero at $x = \tfrac{1}{2}$ from the factor $(2x-1)$ . Enter that zero as a fraction (e.g. 1/2).

$p$ (the numerator) divides the constant term $a_0$ , and
$q$ (the denominator) divides the leading coefficient $a_n$ .

So every possible rational zero looks like $\pm\dfrac{\text{factor of } a_0}{\text{factor of } a_n}$ .

Example: $p(x) = 2x^3 + 3x^2 - 8x + 3$

Factors of constant $a_0 = 3$ : $\;\pm 1, \pm 3$
Factors of leading $a_n = 2$ : $\;\pm 1, \pm 2$
Candidates $\dfrac{p}{q}$ : $\;\pm 1,\; \pm 3,\; \pm\dfrac{1}{2},\; \pm\dfrac{3}{2}$

Testing finds the actual zeros $x = 1,\; \tfrac{1}{2},\; -3$ (matching Part 2's factorization $(x+3)(2x-1)(x-1)$ ).

💡 The theorem doesn't say a rational zero exists — it just gives you the only fractions worth testing before reaching for the quadratic formula or technology.

Concept Check 🎯

Rational Root Drill 🧮

For $p(x) = 2x^3 + 3x^2 - 8x + 3$ (constant $3$ , leading coefficient $2$ ):

1) How many factors does the constant $3$ have (positive only: count $1$ and $3$ )? 2) Test $x = 1$ : enter $p(1)$ . 3) Test : enter .

Slant Asymptotes from Division

When a rational function $\dfrac{N(x)}{D(x)}$ has numerator degree exactly one more than the denominator, its graph approaches a slant (oblique) line asymptote. Division finds it: the quotient is the asymptote, and the remainder term vanishes as $x \to \pm\infty$ .

Example: $f(x) = \dfrac{x^2 + 3x + 5}{x + 1}$

Divide $x^2 + 3x + 5$ by $x + 1$ (synthetic, $c = -1$ ): quotient , remainder . So

As $x \to \pm\infty$ , the fraction $\dfrac{3}{x+1} \to 0$ , so the graph hugs the line .

✅ Slant asymptote: $y = x + 2$ . The remainder $3$ only controls how the curve approaches the line, not the line itself.

Find the Slant Asymptote 🔽

Each rational function is rewritten as quotient $+ \dfrac{\text{remainder}}{\text{divisor}}$ . Pick the slant asymptote.

⚠️ Three habits that prevent most errors: use $0$ placeholders for missing powers, flip the sign of $c$ (divisor $x+3 \Rightarrow c=-3$ ), and remember a factor $(x-c)$ matches the zero $x = +c$ .

Mixed Practice 🎯

Synthesis Drill 🧮

$p(x) = x^3 - 6x^2 + 11x - 6$ .

1) Compute $p(1)$ . (Is $x - 1$ a factor?) 2) Dividing by $x - 1$ gives quotient $x^2 - 5x + 6$ . Enter its zero. Compute to confirm is a zero.

Exit Quiz ✅

Answer all three to finish the lesson.

x^2 + x - 6 = (x+3)(x-2)

2x - 1 = 0 \Rightarrow x = \tfrac{1}{2}, \qquad x + 3 = 0 \Rightarrow x = -3.

f(x) = x + 2 + \frac{3}{x + 1}.

Polynomial Division and Remainder Theorem

Polynomial Division and Remainder Theorem - Complete Interactive Lesson

Part 1: Long Division of Polynomials

➗ Polynomial Division & the Remainder Theorem

Topics in This Part

The Division Algorithm

Carrying Out Long Division

Watch Out: Missing Terms

Part 2: Synthetic Division (the Shortcut)

➗ Polynomial Division & the Remainder Theorem

Setting Up Synthetic Division

Part 3: The Remainder Theorem

➗ Polynomial Division & the Remainder Theorem

Why It Works

Part 4: The Factor Theorem

➗ Polynomial Division & the Remainder Theorem

Part 5: Finding All the Zeros

➗ Polynomial Division & the Remainder Theorem

The Depress-and-Conquer Loop

Part 6: Rational Roots & Slant Asymptotes

➗ Polynomial Division & the Remainder Theorem

The Rational Root Theorem

Part 7: Mixed Practice & Mastery Check

➗ Polynomial Division & the Remainder Theorem

Quick Reference

Worked Example: (x3−4x2+6x−4)÷(x−2)(x^3 - 4x^2 + 6x - 4) \div (x - 2)(x3−4x2+6x−4)÷(x−2)

A Second Example with a Negative ccc

Example

Synthetic Substitution

From Remainder to Factor

Using It to Test Candidates

Example: p(x)=x3−4x2+x+6p(x) = x^3 - 4x^2 + x + 6p(x)=x3−4x2+x+6

A Cubic with a Fractional Zero

Example: p(x)=2x3+3x2−8x+3p(x) = 2x^3 + 3x^2 - 8x + 3p(x)=2x3+3x2−8x+3

Slant Asymptotes from Division

Example: f(x)=x2+3x+5x+1f(x) = \dfrac{x^2 + 3x + 5}{x + 1}f(x)=x+1x2+3x+5​

Worked Example: $(x^3 - 4x^2 + 6x - 4) \div (x - 2)$

A Second Example with a Negative $c$

Example: $p(x) = x^3 - 4x^2 + x + 6$

Example: $p(x) = 2x^3 + 3x^2 - 8x + 3$

Example: $f(x) = \dfrac{x^2 + 3x + 5}{x + 1}$