Polynomial Long Division

➗ Polynomial Long Division

Part 1 of 5 — Why Divide Polynomials?

Topics in This Part

🔑 Key Concept: Polynomial long division works exactly like the long division you learned for whole numbers — you just divide, multiply, subtract, and bring down. The numbers are replaced by terms like $x^2$ and $3x$, but the rhythm is identical.

The Division Vocabulary

Every division problem has the same four roles. Compare arithmetic to polynomials:

We write the result the same way too. For $17 \div 5$:

$17 = 5 \cdot 3 + 2 \qquad\Longleftrightarrow\qquad \frac{17}{5} = 3 + \frac{2}{5}$

For polynomials, $\dfrac{x^2+5x+6}{x+2} = x + 3$ with remainder $0$, because $(x+2)(x+3) = x^2+5x+6$ exactly.

💡 The remainder must always be "smaller" than the divisor. With numbers that means a smaller value; with polynomials it means a lower degree.

Concept Check 🎯

Standard Form & Placeholders

Before dividing, always write both polynomials in standard form (highest power first) and fill every gap with a zero coefficient.

If a power is missing, insert it as $0 \cdot x^n$. This keeps your columns lined up — just like writing $204$ instead of "$2\;\;4$" so the place values don't shift.

Example

$x^3 - 8 \quad\longrightarrow\quad x^3 + 0x^2 + 0x - 8$

$2x^4 - 3x + 1 \quad\longrightarrow\quad 2x^4 + 0x^3 + 0x^2 - 3x + 1$

⚠️ The #1 cause of wrong answers in polynomial division is forgetting a placeholder. A missing $x^2$ or $x$ term silently shifts everything and corrupts the result.

Standard Form & Placeholders 🔽

Choose the correct full standard form for each polynomial.

One More Idea: Degree

The degree of a polynomial is its highest exponent — $x^4 - 5$ has degree $4$, and $x - 1$ has degree $1$. Degree is what tells you when to stop dividing (you'll use it in every problem from here on).

Counting placeholders is just counting the missing powers between the top degree and the constant.

Count the Placeholders 🧮

To divide $x^4 - 5$ by $x - 1$, you first rewrite the dividend with placeholders:

$x^4 + 0x^3 + 0x^2 + 0x - 5$

1) How many placeholder terms (terms with coefficient $0$) did you have to insert? 2) What is the degree of the dividend $x^4 - 5$?

You're Set Up to Divide

You now know the vocabulary (dividend, divisor, quotient, remainder), the rule that the remainder is always lower degree than the divisor, and the must-do prep step of standard form + placeholders.

In Part 2 we run the full divide–multiply–subtract–bring-down loop on a real example.

➗ Polynomial Long Division

Part 2 of 5 — The Four-Step Loop

🔑 The Loop: Every step of polynomial long division repeats four moves — Divide, Multiply, Subtract, Bring down (D-M-S-B). Memorize this rhythm and the whole method becomes mechanical.

The Four Steps

To divide, you repeat this loop until the leftover has lower degree than the divisor:

1. Divide the leading term of the current dividend by the leading term of the divisor. This gives the next term of the quotient.
2. Multiply that quotient term by the entire divisor.
3. Subtract the product from the current dividend.
4. Bring down the next term and repeat.

⚠️ Subtracting flips signs. When you subtract a product, change the sign of every term you're subtracting. Distributing that minus sign is where most arithmetic errors happen.

Worked Example: $(x^2 + 5x + 6) \div (x + 2)$

Step 1 — Divide: Leading terms: $\dfrac{x^2}{x} = x$. First quotient term is $x$.

Step 2 — Multiply: $x(x+2) = x^2 + 2x$.

Step 3 — Subtract: $(x^2 + 5x) - (x^2 + 2x) = 3x$

Step 4 — Bring down the $+6$: now we have $3x + 6$.

Repeat — Divide: $\dfrac{3x}{x} = 3$. Next quotient term is $+3$.

Multiply: $3(x+2) = 3x + 6$.

Subtract: $(3x+6) - (3x+6) = 0$.

$\frac{x^2+5x+6}{x+2} = x + 3 \quad\text{(remainder } 0\text{)}$

✅ Check: $(x+2)(x+3) = x^2 + 5x + 6$ ✓

Track the Loop 🔽

You're dividing $(x^2 + 7x + 10) \div (x + 2)$. Choose what happens at each stage.

The Subtraction Trap

Step 3 (Subtract) trips up more students than any other. Remember: you are subtracting an entire expression, so the minus sign hits every term.

$(x^2 + 7x) - (x^2 + 2x) = x^2 + 7x - x^2 - 2x = 5x$

A reliable trick: mentally flip the sign of each term in the product, then add. Try the next check on exactly this skill.

Watch the Signs 🎯

Putting the Loop Together

You now have all four moves. The complete rhythm for one full division is:

🔑 D-M-S-B: Divide leading terms → Multiply by the whole divisor → Subtract (flip signs!) → Bring down the next term → repeat.

Run the full loop yourself on the next two problems — both divide evenly, so a leftover of $0$ confirms you did it right.

Run the Loop 🧮

Divide each. Every one comes out with remainder $0$, so enter the quotient's pieces.

1) $(x^2 + 9x + 20) \div (x + 4) = x + \,?$ 2) $(x^2 - 7x + 12) \div (x - 3) = x + \,?$ (enter the constant, including its sign)

➗ Polynomial Long Division

Part 3 of 5 — Remainders & Placeholders

🔑 Reality Check: Most divisions do not come out evenly. When a remainder is left over, we write the answer as $\text{quotient} + \dfrac{\text{remainder}}{\text{divisor}}$ — exactly like a mixed number.

Writing a Remainder

When the leftover has lower degree than the divisor, you stop — that leftover is the remainder $R$. The complete answer is:

$\frac{\text{dividend}}{\text{divisor}} = \text{quotient} + \frac{R}{\text{divisor}}$

Worked Example: $(x^2 + 4x + 7) \div (x + 1)$

Divide: $x^2 \div x = x$. Multiply: $x(x+1) = x^2 + x$. Subtract: $(x^2+4x) - (x^2+x) = 3x$. Bring down $+7$: now $3x + 7$.

Divide: $3x \div x = 3$. Multiply: $3(x+1) = 3x + 3$. Subtract: $(3x+7) - (3x+3) = 4$.

$4$ has degree $0$, lower than the divisor's degree $1$, so we stop. The remainder is $4$:

$\frac{x^2+4x+7}{x+1} = x + 3 + \frac{4}{x+1}$

✅ Check: $(x+1)(x+3) + 4 = x^2 + 4x + 3 + 4 = x^2 + 4x + 7$ ✓

Concept Check 🎯

When Terms Are Missing

Worked Example: $(x^3 - 7x - 6) \div (x - 3)$

There's no $x^2$ term, so rewrite the dividend with a placeholder: $x^3 + 0x^2 - 7x - 6$.

$\frac{x^3 - 7x - 6}{x - 3} = x^2 + 3x + 2 \quad\text{(remainder } 0\text{)}$

✅ Check: $(x-3)(x^2+3x+2) = x^3 + 3x^2 + 2x - 3x^2 - 9x - 6 = x^3 - 7x - 6$ ✓

Set Up With Placeholders 🔽

You're about to divide $2x^3 - 5$ by $x + 1$.

Knowing When to Stop

You keep looping until the leftover's degree drops below the divisor's degree. With a degree-$1$ divisor like $x + 1$, you stop as soon as the leftover is a constant — that constant is your remainder.

On the next drill, run each division all the way down and report the final constant left over.

Find the Remainder 🧮

Divide and report the remainder (a single number, with its sign).

1) $(x^2 + 2x + 5) \div (x + 1)$. Remainder $= \,?$ 2) $(x^2 - 3x + 10) \div (x - 2)$. Remainder $= \,?$

➗ Polynomial Long Division

Part 4 of 5 — The Remainder & Factor Theorems

🔑 Big Payoff: Division isn't just busywork. The Remainder Theorem lets you find a remainder without dividing, and the Factor Theorem turns division into a tool for factoring and finding roots.

The Remainder Theorem

🔑 Remainder Theorem: When a polynomial $P(x)$ is divided by $(x - c)$, the remainder equals $P(c)$.

So to find the remainder of $P(x) \div (x - c)$, just plug $c$ into $P$ — no long division needed.

Example: remainder of $(x^2 + 4x + 7) \div (x + 1)$

Here the divisor is $x - (-1)$, so $c = -1$:

$P(-1) = (-1)^2 + 4(-1) + 7 = 1 - 4 + 7 = 4$

The remainder is $4$ — matching the long division we did in Part 3. ✓

⚠️ Mind the sign of $c$. Dividing by $x + 1$ means $c = -1$, not $+1$. Set the divisor equal to zero: $x + 1 = 0 \Rightarrow x = -1$.

Remainder Theorem Shortcut 🧮

Use $P(c)$ to find each remainder. Set the divisor $= 0$ to get $c$ first.

1) $P(x) = x^2 - 5x + 3$ divided by $(x - 2)$. Remainder $= P(2) = \,?$ 2) $P(x) = x^3 + 2x - 1$ divided by $(x + 1)$. Remainder $= P(-1) = \,?$

The Factor Theorem

The Factor Theorem is the special case where the remainder is zero:

🔑 Factor Theorem: $(x - c)$ is a factor of $P(x)$ if and only if $P(c) = 0$.

In words: if dividing by $(x - c)$ leaves no remainder, then $(x - c)$ divides evenly — it's a factor — and $c$ is a root (a zero) of $P$.

Example: Is $(x - 2)$ a factor of $P(x) = x^3 - 3x^2 + 4x - 4$?

$P(2) = 8 - 12 + 8 - 4 = 0$

Yes! Since $P(2) = 0$, $(x - 2)$ is a factor. Dividing then peels it off:

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

So $P(x) = (x - 2)(x^2 - x + 2)$.

💡 This is the everyday workflow for factoring cubics: test a value with the Factor Theorem, then divide to get the leftover quadratic.

Concept Check 🎯

The Test-Then-Divide Workflow

Putting the two theorems together gives a powerful routine for factoring:

1. Test small values ($1, -1, 2, -2, \dots$) by computing $P(c)$.
2. When you find $P(c) = 0$, you've found a factor $(x - c)$.
3. Divide by $(x - c)$ to peel it off and reveal the remaining factor.

Walk through this routine in the next check.

Test for Factors 🔽

Let $P(x) = x^3 - 4x^2 + x + 6$.

➗ Polynomial Long Division

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) set up with standard form and placeholders, (2) run the divide–multiply–subtract–bring-down loop, (3) write remainders correctly, and (4) use the Remainder and Factor Theorems. Time to put it all together.

Quick Reference

⚠️ The two killers: forgetting a placeholder, and not flipping signs when you subtract. Slow down on those two and your accuracy soars.

Mixed Practice 🧮

1) $(x^2 + 6x + 8) \div (x + 2)$ has remainder $0$. The quotient is $x + \,?$ (the constant) 2) Use the Remainder Theorem: the remainder of $(x^3 - 2x^2 + x + 5) \div (x - 1)$ is $\,?$

Mixing Methods

Notice how you chose the right tool for each problem above: long division when you need the full quotient, and the Remainder Theorem when you only need the remainder. A strong solver picks the shortcut whenever the question allows it.

The next set blends both skills — read each question carefully to decide which method is fastest.

Mixed Practice 🎯

Ready for the Exit Quiz

You've covered the full arc: set up with placeholders, run the D-M-S-B loop, write remainders, and apply the Remainder and Factor Theorems.

The three questions below pull from all of it. Take your time, watch your signs, and you'll finish strong. ✅

Exit Quiz ✅

Answer all three to finish the lesson.