Polynomial Division and Remainder Theorem

➗ Polynomial Division & the Remainder Theorem

Part 1 of 7 — Long Division of Polynomials

Topics in This Part

🔑 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: divide the leading terms, multiply, then subtract.

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 $2x^3 - 3x^2 + 4x - 5$. ✓

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

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.

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 $2x^3 - 3x^2 + 4x - 5$ by $x - 2$. Enter the remainder. 3) Divide $6x^2 + 11x - 35$ by $2x + 7$. Enter the quotient (form $ax + b$, e.g. 3x-5).

➗ 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 $x - c$:

1. Write $c$ in the box. Beware the sign: dividing by $x - 2$ uses $c = 2$; dividing by $x + 3$ uses $c = -3$ (because $x+3 = x-(-3)$).
2. List the coefficients of $p(x)$ in order, including $0$ for any missing power.
3. 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 $1, -4, 6, -4$:

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 $2, 3, -8, 3$:

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.

➗ 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 $p(x) = (x - c)\,q(x) + r,$ where $r$ is a constant (the divisor $x-c$ is degree $1$, so the remainder has degree $0$).

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.

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

Example

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 $c = -1$):

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.

➗ 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.

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:

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

Concept Check 🎯

Using It to Confirm a Zero You Are Given

A graph of $p(x) = x^3 - 7x + 6$ appears to cross the $x$-axis at $x = 1$. Is $x - 1$ a factor? Check $p(1)$: $p(1) = 1 - 7 + 6 = 0 \;✓$ Yes — so $(x-1)$ is a factor. Synthetic-dividing by $x - 1$ peels it off:

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

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

Match Factor ↔ Zero 🔽

Factor Theorem Drill 🧮

A graph of $p(x) = x^3 - 7x + 6$ shows three $x$-intercepts, at $x = -3$, $x = 1$, and $x = 2$. Evaluate $p$ to confirm two of them, then check the $y$-intercept.

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

➗ Polynomial Division & the Remainder Theorem

Part 5 of 7 — Finding All the Zeros

🔑 The Strategy: Start from a zero you are given (by a graph, a table, a known factor, or the problem statement), divide it out, and repeat on the smaller quotient until you reach a quadratic you can factor or solve.

The Depress-and-Conquer Loop

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

1. Get one zero $c$ — from a graph, a table, a given factor, or a value the problem gives you. Confirm it with $p(c) = 0$.
2. Divide $p(x)$ by $(x - c)$ using synthetic division. The quotient has degree one lower.
3. Repeat on the quotient. Once it's a quadratic, factor it or use the quadratic formula.

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

A table of values shows $p(-1) = 0$. Confirm: $\;p(-1) = -1 - 4 - 1 + 6 = 0$ ✓. Divide by $x + 1$:

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 $p(x) = 2x^3 + x^2 - 13x + 6$, given that $x = 2$ is a zero.

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

The quotient is $2x^2 + 5x - 3 = (2x - 1)(x + 3)$. Setting each factor to $0$: $2x - 1 = 0 \Rightarrow x = \tfrac{1}{2}, \qquad x + 3 = 0 \Rightarrow x = -3.$

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

💡 Not every zero is an integer. The factor $(2x - 1)$ produces the fractional zero $x = \tfrac{1}{2}$, which you only see after dividing out the zero you were given. You never need to guess candidate zeros on the AP exam: hand factoring is limited to common factors and quadratics, and any other zero is supplied.

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).

➗ Polynomial Division & the Remainder Theorem

Part 6 of 7 — Division, Rational Functions & Slant Asymptotes

🔑 Two payoffs: Division rewrites a rational function as polynomial + leftover fraction, and that rewritten form exposes the function's end behavior, including any slant (oblique) asymptote.

Rewriting a Rational Function by Division

If the polynomial $f$ is divided by the polynomial $g$, then

$f(x) = g(x)\,q(x) + r(x), \qquad \deg r < \deg g,$

where $q$ is the quotient and $r$ the remainder. Dividing both sides by $g(x)$ rewrites the rational function:

$\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}$

Example: a quadratic divisor (long division required)

Divide $2x^3 + x^2 - 3$ by $x^2 + 1$. Synthetic division only handles divisors $x - c$, so use long division:

• $2x^3 \div x^2 = 2x$; subtract $2x(x^2 + 1) = 2x^3 + 2x$ to leave $x^2 - 2x - 3$.
• $x^2 \div x^2 = 1$; subtract $1(x^2 + 1)$ to leave $-2x - 4$.

The remainder $-2x - 4$ has degree $1 < 2$, so stop:

$\frac{2x^3 + x^2 - 3}{x^2 + 1} = 2x + 1 + \frac{-2x - 4}{x^2 + 1}$

💡 Why it matters: the leftover fraction has a smaller-degree numerator, so it shrinks toward $0$ as $x \to \pm\infty$. The quotient $q(x)$ therefore controls the end behavior. Here the graph approaches the line $y = 2x + 1$.

Concept Check 🎯

Division Drill 🧮

Dividing $2x^2 + 3x + 4$ by $x - 1$ gives $\dfrac{2x^2 + 3x + 4}{x - 1} = ax + b + \dfrac{c}{x - 1}$.

1) Enter $a$, the coefficient of $x$ in the quotient. 2) Enter $b$, the constant in the quotient. 3) Enter the remainder $c$.

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 $x + 2$, remainder $3$. So $f(x) = x + 2 + \frac{3}{x + 1}.$

As $x \to \pm\infty$, the fraction $\dfrac{3}{x+1} \to 0$, so the graph hugs the line $y = x + 2$.

✅ 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.

➗ 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 once one is known, and (5) rewrite rational functions to read off end behavior and slant asymptotes. Let's put it all together.

Quick Reference

⚠️ 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 larger zero. 3) Compute $p(3)$ to confirm $x = 3$ is a zero.

Exit Quiz ✅

Answer all three to finish the lesson.