Polynomial Division

➗ Polynomial Division

Part 1 of 5 — The Big Picture & Dividing by a Monomial

Topics in This Part

🔑 Key Concept: Polynomial division answers the question "how many times does one polynomial fit inside another?" — just like $\dfrac{17}{5} = 3$ remainder $2$, we'll write $\dfrac{2x^2 + 7x + 3}{x + 3}$ as a quotient plus a remainder.

The Division Vocabulary

Every division problem — with numbers or polynomials — uses the same four words:

$\underbrace{17}_{\text{dividend}} \div \underbrace{5}_{\text{divisor}} = \underbrace{3}_{\text{quotient}} \; \text{remainder } \underbrace{2}_{\text{remainder}}$

The Division Algorithm ties them together. For polynomials:

$\text{dividend} = (\text{divisor})(\text{quotient}) + \text{remainder}$

or, written as fractions:

$\frac{P(x)}{D(x)} = Q(x) + \frac{R(x)}{D(x)}$

🔑 Key Rule: The remainder must have a smaller degree than the divisor. You keep dividing until the leftover is "too small" to divide further — exactly like long division of numbers.

Concept Check 🎯

Dividing by a Monomial

The easiest case: a monomial (one term) divisor. Split the fraction term-by-term and use the quotient rule for exponents, $\dfrac{x^m}{x^n} = x^{m-n}$.

Worked Example: $\dfrac{6x^3 + 9x^2 - 3x}{3x}$

Divide each term by $3x$:

$\frac{6x^3}{3x} + \frac{9x^2}{3x} - \frac{3x}{3x} = 2x^2 + 3x - 1$

⚠️ Most common mistake: dividing only the first term by the divisor. You must divide every term — $\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c}$, not just $\dfrac{a}{c} + b$.

Worked Example: $\dfrac{8x^4 - 12x^2}{4x^2}$

$\frac{8x^4}{4x^2} - \frac{12x^2}{4x^2} = 2x^2 - 3$

Divide by a Monomial 🧮

Simplify each quotient. Enter the result as a polynomial (e.g. 2x^2+3x-1).

1) $\dfrac{10x^3 + 15x^2}{5x^2} = \,?$ 2) $\dfrac{12x^4 - 8x^3 + 4x^2}{4x^2} = \,?$

Match the Vocabulary 🔽

Consider $\dfrac{6x^2 + 11x + 4}{2x + 1} = 3x + 4$ (remainder $0$). Identify each part.

Looking Ahead

Monomial division is easy because nothing "carries over." But what about dividing by a binomial like $x + 3$? That needs the workhorse of this lesson: polynomial long division, coming up in Part 2.

💡 Preview: Every polynomial division can be checked the same way you check number division — multiply the quotient by the divisor and add the remainder. You should land back on the original dividend.

➗ Polynomial Division

Part 2 of 5 — Long Division

🔑 The Idea: Polynomial long division mirrors the long division you learned for numbers. Four repeating steps — Divide, Multiply, Subtract, Bring down — until the remainder's degree drops below the divisor's.

The Four-Step Loop

To compute $P(x) \div D(x)$:

1. Divide the leading term of the current dividend by the leading term of the divisor → next quotient term.
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.

Stop when the remaining polynomial's degree is less than the divisor's degree.

⚠️ Setup rule: Write both polynomials in descending powers of $x$, and insert a $0$ placeholder for any missing power (e.g. write $x^3 + 0x^2 - 5x + 1$). Skipping a placeholder is the #1 source of long-division errors.

Worked Example: $(2x^2 + 7x + 3) \div (x + 3)$

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

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

Step 3 — Subtract: $(2x^2 + 7x) - (2x^2 + 6x) = x$

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

Repeat — Divide: $\dfrac{x}{x} = 1$. Multiply: $1(x+3) = x + 3$. Subtract: $(x+3)-(x+3) = 0$.

$2x^2 + 7x + 3 = (x + 3)(2x + 1) + 0$

So the quotient is $2x + 1$ with remainder $0$.

✅ Check: $(x+3)(2x+1) = 2x^2 + x + 6x + 3 = 2x^2 + 7x + 3$ ✓

Trace the Steps 🔽

You're dividing $x^2 + 5x + 6$ by $x + 2$. Choose what happens at each stage.

When the Remainder Isn't Zero

Not every division comes out even. Consider $(x^2 + 3x + 5) \div (x + 1)$.

Divide: $\dfrac{x^2}{x} = x$. Multiply: $x(x+1) = x^2 + x$. Subtract: $(x^2 + 3x) - (x^2 + x) = 2x$. Bring down $+5$: now $2x + 5$.

Divide: $\dfrac{2x}{x} = 2$. Multiply: $2(x+1) = 2x + 2$. Subtract: $(2x+5)-(2x+2) = 3$.

The leftover $3$ has degree $0 <$ degree of $x+1$, so we stop:

$\frac{x^2 + 3x + 5}{x + 1} = x + 2 + \frac{3}{x + 1}$

💡 Read it like a number: $\dfrac{17}{5} = 3 + \dfrac{2}{5}$. Same shape — quotient plus (remainder over divisor).

Concept Check 🎯

Long Division Practice 🧮

Find the quotient $Q(x)$ and remainder $R$ for each.

1) $(x^2 + 7x + 10) \div (x + 2)$: quotient $= \,?$ , remainder $= \,?$ 2) $(x^2 + 4x + 7) \div (x + 1)$: quotient $= \,?$ , remainder $= \,?$

➗ Polynomial Division

Part 3 of 5 — Synthetic Division

🔑 The Shortcut: When the divisor is a simple linear $x - c$, synthetic division does the whole job with just the coefficients — no $x$'s, no rewriting. It's faster and cleaner than long division.

Setting Up Synthetic Division

Synthetic division works only when the divisor has the form $x - c$ (degree 1, leading coefficient 1).

⚠️ Sign flip: For divisor $x - c$ you use $+c$; for divisor $x + c$ you use $-c$. Dividing by $x + 3$ means $c = -3$. Always solve $x - c = 0$ to find the number you carry in.

The procedure (for $x - c$):

1. Write $c$ in the box, then list the dividend's coefficients (with $0$ placeholders).
2. Bring down the first coefficient.
3. Multiply it by $c$, write the product under the next coefficient.
4. Add the column. Repeat multiply-and-add across the row.
5. The last number is the remainder; the rest are the quotient's coefficients (degree one less than the dividend).

Worked Example: $(2x^3 - 3x^2 - 11x + 6) \div (x - 3)$

Divisor $x - 3 \Rightarrow c = 3$. Coefficients: $2,\; -3,\; -11,\; 6$.

$\begin{array}{c|rrrr} 3 & 2 & -3 & -11 & 6 \\ & & 6 & 9 & -6 \\ \hline & 2 & 3 & -2 & 0 \end{array}$

• Bring down $2$. Multiply $2\cdot 3 = 6$, add to $-3$ → $3$.
• Multiply $3\cdot 3 = 9$, add to $-11$ → $-2$.
• Multiply $-2\cdot 3 = -6$, add to $6$ → $0$ (the remainder).

The bottom row $2,\;3,\;-2$ gives the quotient (start one degree below the cubic, so degree 2):

$2x^3 - 3x^2 - 11x + 6 = (x - 3)(2x^2 + 3x - 2) + 0$

✅ Check: $(x-3)(2x^2+3x-2) = 2x^3+3x^2-2x -6x^2-9x+6 = 2x^3 - 3x^2 - 11x + 6$ ✓

Concept Check 🎯

Don't Forget the Placeholders!

If a power is missing, you must include a $0$ coefficient. Divide $x^3 - 8$ by $x - 2$:

Write the coefficients as $1,\; 0,\; 0,\; -8$ (for $x^3 + 0x^2 + 0x - 8$), with $c = 2$:

$\begin{array}{c|rrrr} 2 & 1 & 0 & 0 & -8 \\ & & 2 & 4 & 8 \\ \hline & 1 & 2 & 4 & 0 \end{array}$

So $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ — a perfect difference-of-cubes factorization!

⚠️ If you had written only $1, -8$, you'd get nonsense. The $0$ placeholders keep every column aligned with its power of $x$.

Synthetic Division Practice 🧮

1) Divide $x^2 - 5x + 6$ by $x - 2$. The quotient is $x + a$ — enter $a$. Then enter the remainder. 2) Divide $x^3 + 2x^2 - 5x - 6$ by $x - 2$. Enter the remainder.

Fill the Synthetic Row 🔽

Divide $x^3 + 4x^2 + x - 6$ by $x - 1$ ($c = 1$). Coefficients: $1,\;4,\;1,\;-6$. Complete the bottom row left to right.

➗ Polynomial Division

Part 4 of 5 — The Remainder & Factor Theorems

🔑 Big Payoff: Division isn't just bookkeeping — it reveals where a polynomial equals zero. Two theorems turn a single division (or even a single substitution) into deep information about roots and factors.

The Remainder Theorem

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

This means you can find a remainder without dividing at all — just plug in $c$.

Example

Find the remainder when $P(x) = x^3 - 4x^2 + 2x + 5$ is divided by $x - 1$.

Here $c = 1$:

$P(1) = (1)^3 - 4(1)^2 + 2(1) + 5 = 1 - 4 + 2 + 5 = 4$

So the remainder is $4$ — no long division needed.

💡 This is why synthetic division's last entry is the remainder: that final add is secretly computing $P(c)$.

Use the Remainder Theorem 🧮

Use $P(c)$ — not long division — to find each remainder.

1) $P(x) = x^2 + 3x - 4$ divided by $x - 2$. Remainder $= P(2) = \,?$ 2) $P(x) = 2x^3 - x + 1$ 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 other words, $c$ is a root (zero) of the polynomial exactly when $x - c$ divides it evenly.

Example

Is $x - 3$ a factor of $P(x) = x^3 - 2x^2 - 5x + 6$?

$P(3) = 27 - 18 - 15 + 6 = 0 \;\checkmark$

Yes! Since $P(3) = 0$, $x - 3$ is a factor. Dividing confirms it:

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

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

Factor or Not? 🔽

For $P(x) = x^3 - 7x + 6$, decide each statement. (Compute $P(c)$ in your head.)

Concept Check 🎯

➗ Polynomial Division

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) divide by a monomial, (2) run polynomial long division, (3) use synthetic division for $x - c$, and (4) apply the Remainder and Factor Theorems. Let's put it all together.

Quick Reference

⚠️ Two habits that prevent most errors: write polynomials in descending order with $0$ placeholders, and remember the sign flip — divisor $x + c$ uses $-c$ in synthetic division.

Mixed Practice 🎯

One More Long Division 🧮

Divide $2x^2 + 5x - 1$ by $x + 3$ and express as $Q(x) + \dfrac{R}{x+3}$.

1) Quotient $Q(x) = \,?$ (enter as e.g. 2x-1 ) 2) Remainder $R = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.