9709 P3 §3.1 — Polynomial Division

💡 Lead In

Quick Recall

1. Use long division to compute $847 \div 21$. What is the quotient and remainder?

2. Expand $(x + 2)(2x^2 - 5x + 5)$. What do you get?

3. If $847 = 21 \times 40 + 7$, identify the dividend, divisor, quotient, and remainder.

The Wall

You just expanded $(x + 2)(2x^2 - 5x + 5)$ to get $2x^3 - x^2 - 5x + 10$ — multiplying polynomials is straightforward. Now try going backwards: divide $2x^3 - x^2 - 5x + 10$ by $(x + 2)$. The layout looks like long division, the steps feel like long division — but the "digits" are now powers of $x$. Where do you even start?

The Need

We need to extend the long division algorithm from numbers to polynomials, so we can factorise and simplify algebraic fractions.

📖 Key Concepts

The division identity: For any polynomial $f(x)$ divided by a non-zero divisor $d(x)$: $$f(x) = d(x) \cdot Q(x) + R$$ where $Q(x)$ is the quotient and $R$ is the remainder (with $\deg(R) < \deg(d)$, or $R = 0$).

$b$: -1

$c$: 1

$r$: 2

Green: dividend $f(x)$. Indigo (dashed): divisor $d(x) = x - b$. Amber dot: the point $(b, r)$ where $f(b) = r$.

Activity 1 Demonstration — Step-by-Step Long Division

Divide $\dfrac{x^3 - 3x^2 - x + 3}{x - 1}$ using long division. Write each step clearly — divide, multiply, subtract, bring down. Then verify your result by multiplying the quotient back by $(x - 1)$.

The long division algorithm (4 steps, repeated):

Divide the leading term of the current dividend by the leading term of the divisor.
Multiply the divisor by this new term of the quotient.
Subtract this product from the current dividend.
Bring down the next term. Repeat until no terms remain.

Step 0 of 11 — Setup

Click Next to walk through the 4-step algorithm: Divide → Multiply → Subtract → Bring down (repeat)

Division by $(ax + b)$ where $a \neq 1$: Divide the leading term by $ax$ (not just $x$). For example, $\dfrac{2x^4}{2x} = x^3$. The factor $a$ must be carried through each division step.

Division by a quadratic (e.g. $x^2 + 1$): The process is identical, but each quotient term comes from dividing by $x^2$. Continue until the remaining polynomial has degree less than 2 — that remainder is the final $R$.

Checking your answer: Multiply $Q(x)$ by the divisor, then add $R$. You should recover $f(x)$ exactly. If not, an arithmetic error occurred during subtraction.

Activity 2 · Group Pattern Detective — The Remainder Files

Three case files are open. In each one: study the evidence with your partner, agree on the pattern, then vote for the correct deduction. A case cracked on the first vote earns 3 clues (second vote 2, later votes 1) — close all three files to solve the mystery of the remainder.

File 1 File 2 File 3

Clues collected: 0 / 9

Case File 1 The Vanishing Remainder — in every file below, $f(x) \div (x - a)$ leaves $R = 0$. What does the graph of $y = f(x)$ do at $x = a$?

Evidence A

$x^3 - 3x^2 - x + 3 \div (x - 1)$

$Q(x) = x^2 - 2x - 3$, $R = 0$

Evidence B

$x^2 - 5x + 6 \div (x - 2)$

$Q(x) = x - 3$, $R = 0$

Evidence C

$2x^3 - x^2 - 5x + 10 \div (x + 2)$

$Q(x) = 2x^2 - 5x + 5$, $R = 0$

Partner brief: in each file, follow the dashed line to the amber dot. Where does it sit? Agree on your deduction before voting.

Case closed! In every file the amber dot sits on the x-axis: $R = 0$ means $f(a) = 0$, so the curve passes through the point $(a, 0)$ — an x-intercept. And since $f(x) = (x - a)Q(x)$ with nothing left over, $(x - a)$ is a factor of $f(x)$.

Case File 2 The Stubborn Remainder — now every division leaves $R \neq 0$. Can the divisor $(x - a)$ still be a factor of $f(x)$?

Evidence D

$x^2 + x + 1 \div (x - 1)$

$Q(x) = x + 2$, $R = 3$

Evidence E

$x^2 - 4 \div (x - 1)$

$Q(x) = x + 1$, $R = -3$

Evidence F

$x^3 - 3x^2 + 5 \div (x - 2)$

$Q(x) = x^2 - x - 2$, $R = 1$

Partner brief: try to rewrite $f(x) = (x - a)Q(x) + R$ so that $(x - a)$ becomes a factor. What blocks you? Also watch where each amber dot sits this time…

Case closed! With $R \neq 0$ there is always a leftover: $f(x) = (x - a)Q(x) + R$ can never be rearranged into $(x - a) \times$ (a polynomial). So $(x - a)$ is a factor if and only if $R = 0$. One more clue for the final file: each amber dot now sits exactly at height $R$ above $x = a$ — the points $(1, 3)$, $(1, -3)$ and $(2, 1)$.

Case File 3 The Instant Check — one pattern links all six files. Which deduction below both fits the evidence table and checks a division answer in five seconds?

The division identity, once more: $f(x) = (x - a) \cdot Q(x) + R$

File	Divisor	$a$	$R$	$f(a)$	$Q(a)$	$f(0)$
A	$x - 1$	1	0	0	−4	3
B	$x - 2$	2	0	0	−1	6
C	$x + 2$	−2	0	0	23	10
D	$x - 1$	1	3	3	3	1
E	$x - 1$	1	−3	−3	2	−4
F	$x - 2$	2	1	1	0	5

Partner brief: test each candidate deduction against every row of the table — only one of them survives all six rows.

Elementary! Substitute $x = a$ into the division identity: $f(a) = (a - a)Q(a) + R = R$. The remainder is simply the value of $f$ at the root of the divisor — in every file. So the five-second check: evaluate $f(a)$ and compare it with the claimed remainder. (The full check — multiply back $(x - a)Q(x) + R$ and confirm you recover $f(x)$ — still works too.) Next lesson gives your discovery its official name: the Remainder Theorem.

CASE CLOSED

The value link: $f(a) = R$ — the remainder is just $f$ evaluated at the root of the divisor.

The factor link: $(x - a)$ is a factor of $f(x)$ $\iff$ $R = 0$.

The graph link: $R = 0$ $\iff$ the curve passes through the point $(a, 0)$.

Two ways to check any division claim: instant — evaluate $f(a)$ and compare with the claimed remainder; full — multiply back $(x - a)Q(x) + R$ and confirm you recover $f(x)$.

Next lesson — the Factor and Remainder Theorems — formalises exactly what you discovered today.

✏️ Worked Examples

Guided Example 1 [Group] — Linear Divisor with $a \neq 1$

Exercise 1A Q23

Divide $4x^4 + 4x^3 - x^2 + 7x - 4$ by $2x - 1$.

Hint

The divisor has a leading coefficient of $2$. Divide each leading term by $2x$ (not just $x$). Start with $\dfrac{4x^4}{2x} = 2x^3$.

(4x⁴ + 4x³ − x² + 7x − 4) ÷ (2x − 1) = 2x³ + 3x² + x + 4, remainder 0

Final Answer

$$\text{Quotient} = 2x^3 + 3x^2 + x + 4, \quad \text{Remainder} = 0$$

Guided Example 2 [Individual] — Quadratic Divisor

Exercise 1A Q24

Divide $2x^4 + 2x^3 + 5x^2 + 2x + 3$ by $x^2 + 1$.

Hint

The divisor is quadratic, so divide the leading term by $x^2$ (not $x$). Start with $\dfrac{2x^4}{x^2} = 2x^2$, so $a = 2$. Continue until the remaining polynomial has degree less than 2.

(2x⁴ + 2x³ + 5x² + 2x + 3) ÷ (x² + 1) = 2x² + 2x + 3, remainder 0

Final Answer

$$\text{Quotient} = 2x^2 + 2x + 3, \quad \text{Remainder} = 0$$

🎯 Polynomial Division Concept Quiz — Quick Check

Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!

Score: 0 / 4

Q1 (Objective 1) — First step of long division

When dividing $x^3 - 6x^2 + 11x - 6$ by $x - 1$ using long division, what is the first term of the quotient?

$x$

$x^2$

$-6x^2$

$x^3$

The first quotient term comes from dividing the leading term of the dividend by the leading term of the divisor: $\dfrac{x^3}{x} = x^2$. This becomes the first term of the quotient polynomial.

Q2 (Objective 2) — Divisor with leading coefficient $\neq 1$

When dividing $4x^3 + 2x^2 - x$ by $2x - 1$, the first division step gives:

$\dfrac{4x^3}{x} = 4x^2$

$\dfrac{4x^3}{2x} = 2x^2$

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

$\dfrac{4x^3}{2} = 2x^3$

When the divisor is $(ax + b)$ with $a \neq 1$, you must divide by the full leading term $ax$ (not just $x$). Here $\dfrac{4x^3}{2x} = 2x^2$. The factor $2$ must be carried through each division step.

Q3 (Objective 3) — When the remainder is zero

After dividing $f(x)$ by $d(x)$ using long division, the final subtraction yields $0$. Which conclusion is correct?

The quotient $Q(x)$ must equal $1$

The division was performed incorrectly

$d(x)$ is a factor of $f(x)$, and the remainder $R = 0$

$f(x)$ must be a constant polynomial

When the final subtraction gives $0$, the remainder $R = 0$. By the division identity $f(x) = d(x) \cdot Q(x) + R$, this means $f(x) = d(x) \cdot Q(x)$, so $d(x)$ divides $f(x)$ exactly — i.e. $d(x)$ is a factor of $f(x)$. The quotient $Q(x)$ can be any polynomial; it need not be $1$.

Q4 (Objective 4) — Finding an unknown coefficient

When $2x^3 + 5x^2 + 4x + 3$ is divided by $2x + 1$, the quotient is $x^2 + bx + 1$ with remainder $2$. Use the division identity to find $b$.

$b = 1$

$b = 3$

$b = 4$

$b = 2$

Using $f(x) = d(x) \cdot Q(x) + R$: $(2x+1)(x^2 + bx + 1) + 2 = 2x^3 + (2b+1)x^2 + (b+2)x + 3$. Matching the $x^2$ coefficient with $5x^2$: $2b + 1 = 5 \Rightarrow b = 2$. (Check: the $x$ coefficient gives $2 + 2 = 4$ ✓ and the constant gives $3$ ✓.)

🎉

You scored 0 / 4

✏️ More Practice

Exercise 1A Q25 Challenging

Find the quotient when $x^4 + 3x^3 - 8x^2 - 27x - 9$ is divided by $x^2 - 9$.

Hint 1

Divide each leading term by $x^2$. Note that $x^2 - 9 = (x-3)(x+3)$ is a difference of squares.

(x⁴ + 3x³ − 8x² − 27x − 9) ÷ (x² − 9) = x² + 3x + 1, remainder 0

Final Answer

$$\text{Quotient} = x^2 + 3x + 1, \quad \text{Remainder} = 0$$

Exercise 1A Q28 Advanced

Divide $x^4 + 3x^3 + x^2 - 2$ by $x^2 + x + 1$.

Hint 1

The divisor $x^2 + x + 1$ has discriminant $1 - 4 = -3 < 0$, so it does not factorise. Use long division directly, dividing each leading term by $x^2$.

(x⁴ + 3x³ + x² − 2) ÷ (x² + x + 1) = x² + 2x − 2, remainder 0

Final Answer

$$\text{Quotient} = x^2 + 2x - 2, \quad \text{Remainder} = 0$$

🎫

Exit Ticket

Check the learning objectives you feel confident about:

I can divide a polynomial by a linear divisor using long division (divide, multiply, subtract, bring down). I can divide by a divisor of the form $(ax + b)$ where $a \neq 1$, carrying the leading coefficient through each step. I can divide a polynomial by a quadratic divisor and identify the quotient and remainder. I can verify a division result by multiplying the quotient by the divisor and checking I recover the original polynomial.