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hscscience Maths Ext 1 · Y11
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Module 2 · L2 of 15 ~35 min ⚡ +95 XP available

Polynomial Division

When you divide 17 by 5 you get quotient 3 and remainder 2, and you write $17 = 5 \times 3 + 2$. The exact same idea extends to polynomials: divide $P(x)$ by $D(x)$, get a quotient $Q(x)$ and a remainder $R(x)$, and write $P(x) = D(x) \cdot Q(x) + R(x)$. Mastering this algorithm unlocks the Remainder Theorem, the Factor Theorem, and every factorisation problem in Module 2.

Today's hook, When you divide 17 by 5, you get quotient 3 and remainder 2. How might this idea extend to dividing polynomials? The same structure, dividend equals divisor times quotient plus remainder, applies, and it unlocks every factorisation technique you'll use for the rest of Module 2.
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Recall, your gut answer first

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Recall, your gut answer first
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When you divide 17 by 5, you get quotient 3 and remainder 2. We write $17 = 5 \times 3 + 2$. Without using a formula how do you think this idea might extend to dividing polynomials? Write your prediction before reading on.

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Orientation and key terms

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The two moves
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There are only two core ideas in this lesson. Lock the division algorithm ($P = DQ + R$, with $\deg R < \deg D$) and the long division process (divide leading terms, multiply back, subtract, repeat) into muscle memory.

Every polynomial division problem lives on one of two roads: state the division algorithm to express the relationship between dividend, divisor, quotient and remainder, or execute long division to actually find $Q(x)$ and $R(x)$.

Two parallel roads for polynomial division: stating the division algorithm versus performing long division.
$P(x) = D(x) \cdot Q(x) + R(x)$
Degree of remainder
The remainder $R(x)$ must have lower degree than the divisor $D(x)$. If divisor is linear (degree 1), remainder is a constant.
Exact division
If $R(x) = 0$, the divisor divides the dividend exactly, this means the divisor is a factor. This connects to the Factor Theorem in Lesson 4.
Check by expansion
Always verify: expand $D(x) \cdot Q(x) + R(x)$ and confirm it equals $P(x)$. This catches errors before you write your final answer.
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What you'll master
Know

Key facts

  • The division algorithm: $P(x) = D(x) \cdot Q(x) + R(x)$
  • The degree constraint: $\deg(R) < \deg(D)$
  • What "divides exactly" means: $R(x) = 0$
Understand

Concepts

  • Why the remainder's degree must be less than the divisor's
  • How the algorithm mirrors long division of integers
  • Why exact division implies the divisor is a factor
Can do

Skills

  • Perform polynomial long division step by step
  • State the quotient and remainder for any division
  • Verify the result by expanding $D(x) \cdot Q(x) + R(x)$
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Key terms
Dividend $P(x)$The polynomial being divided, what you start with.
Divisor $D(x)$The polynomial you are dividing by, the denominator in a fraction sense.
Quotient $Q(x)$The result of the division, the polynomial "answer".
Remainder $R(x)$What is left over after division; its degree must be strictly less than the degree of $D(x)$.
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The division algorithm

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The division algorithm
core concept

For any polynomials $P(x)$ (dividend) and $D(x)$ (non-zero divisor), there exist unique polynomials $Q(x)$ (quotient) and $R(x)$ (remainder) such that:

$$P(x) = D(x) \cdot Q(x) + R(x) \quad \text{where } \deg(R) < \deg(D)$$

This mirrors the integer division fact $17 = 5 \times 3 + 2$, where the remainder 2 is smaller than the divisor 5.

If $R(x) = 0$, then $P(x) = D(x) \cdot Q(x)$, meaning $D(x)$ divides $P(x)$ exactly the divisor is a factor of the dividend.

Uniqueness matters. The theorem guarantees there is only one pair $(Q(x), R(x))$ satisfying these conditions. This is why polynomial long division always gives you the right answer, there is no other valid answer.

Division algorithm: P(x) = D(x) Q(x) + R(x), with (R) < (D); Degree of quotient: (Q) = (P) - (D)

Pause, copy the Division Algorithm into your book: $P(x) = D(x)Q(x) + R(x)$ where $\deg R < \deg D$; degree of quotient $= \deg P - \deg D$.

Quick check: When $P(x)$ (degree 4) is divided by $D(x)$ (degree 2), what is the degree of the remainder $R(x)$?

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Long division method
core concept

We just saw that evaluating $P(a)$ substitutes a number, giving a single value. That raises a question: when we divide $P(x)$ by $(x-a)$, we get a quotient polynomial and a remainder, how is this long division process carried out, and what does the Division Algorithm state? This card answers it → the Division Algorithm states $P(x) = D(x)Q(x) + R(x)$ with $\deg R < \deg D$; execute by the "divide–multiply–subtract" loop.

The long division algorithm for polynomials follows exactly the same three-step loop as for integers:

  1. Divide the leading term of the current dividend by the leading term of the divisor to get the next term of $Q(x)$.
  2. Multiply the entire divisor by this new term.
  3. Subtract the result from the current dividend to get a new (lower-degree) dividend.

Repeat until the remaining polynomial has degree strictly less than the divisor. That remaining polynomial is $R(x)$.

Key insight. At each step you are reducing the degree by at least 1. Since degree cannot go below 0, the algorithm always terminates, just like integer long division always produces a remainder smaller than the divisor.

Step loop: divide leading terms → multiply divisor → subtract (repeat); Stop when remaining degree < degree of divisor

Pause, copy the long division loop into your book: (1) divide leading terms to get the next term of $Q$; (2) multiply the divisor by that term; (3) subtract; (4) repeat until $\deg$ of remainder $< \deg$ of divisor.

Did you get this? True or false: when dividing a degree-3 polynomial by a linear factor, the quotient always has degree 2.

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Worked examples

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · LINEAR DIVISOR

Divide $P(x) = x^3 - 2x^2 + 3x - 4$ by $D(x) = x - 1$.

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$x^3 \div x = x^2$. Multiply: $x^2(x-1) = x^3 - x^2$. Subtract: $(x^3 - 2x^2) - (x^3 - x^2) = -x^2$. Bring down $+3x$: current dividend is $-x^2 + 3x$.
Divide the leading term of the dividend by the leading term of the divisor. Subtract to eliminate the $x^3$ term.
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PROBLEM 2 · ANOTHER LINEAR DIVISOR

Divide $P(x) = 2x^3 + 3x^2 - x + 5$ by $D(x) = x + 2$.

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$2x^3 \div x = 2x^2$. Multiply: $2x^2(x+2) = 2x^3+4x^2$. Subtract: $(2x^3+3x^2) - (2x^3+4x^2) = -x^2$. Bring down $-x$.
Note: $3x^2 - 4x^2 = -x^2$, be careful with the sign when subtracting.
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PROBLEM 3 · RECONSTRUCT P(x)

When $P(x)$ is divided by $x + 1$, the quotient is $x^2 + 3x - 2$ and the remainder is 5. Find $P(x)$.

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Use the division algorithm: $P(x) = (x+1)(x^2+3x-2) + 5$.
The algorithm works in reverse: if you know $D$, $Q$, and $R$, you can reconstruct $P$.
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Fill the gap: When $x^3 + 2x - 1$ is divided by $x - 1$, the remainder equals .

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Misconceptions to fix

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Forgetting placeholder terms
If the dividend is missing a power (e.g. $x^3 + 2x - 5$ has no $x^2$ term), you must insert $0x^2$ as a placeholder before starting. Skipping this shifts all subsequent subtractions and gives a completely wrong answer.
Trap 02
Sign errors when subtracting
After multiplying, you subtract the entire product. Distribute the minus sign across every term. The most common error: subtracting $-(−x^2)$ and writing $-x^2$ instead of $+x^2$. Write subtraction brackets explicitly every time.
Trap 03
Stopping too early
You must keep dividing until the remaining polynomial has degree strictly less than the divisor. If your divisor is linear (degree 1), you stop only when you have a constant remainder. Students often stop one step too early.

Did you get this? True or false: when dividing $P(x) = x^3 + 1$ by $D(x) = x - 1$, you should insert a $0x^2 + 0x$ placeholder before beginning.

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Activities

Activities · practice with the ideas

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Divide $x^3 - 3x^2 + x - 2$ by $x - 2$. State $Q(x)$ and $R(x)$.

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Divide $x^3 + 1$ by $x + 1$. (Hint: insert placeholder terms first.)

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When $P(x)$ is divided by $(x - 3)$, the quotient is $2x^2 + x - 1$ and the remainder is 4. Find $P(x)$.

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Divide $3x^3 - x^2 + 4x - 1$ by $x^2 + 1$. State quotient and remainder.

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Explain in your own words why polynomial long division must eventually terminate (stop producing terms).

Odd one out: Three of these are correct statements about dividing $P(x) = x^3 + 2x^2 - 5x + 6$ by $D(x) = x - 2$. Which statement is WRONG?

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Revisit your thinking

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Revisit your thinking

Earlier you were asked: how might integer division extend to polynomials?

Just as $17 = 5 \times 3 + 2$ (divisor $\times$ quotient $+$ remainder), we write $P(x) = D(x) \cdot Q(x) + R(x)$ where $\deg(R) < \deg(D)$. The key similarity: the remainder is always "smaller" than the divisor (in degree for polynomials, in size for integers). The key difference: polynomial subtraction can generate negative terms with mixed signs, requiring more careful arithmetic.

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Multiple choice

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Multiple choice
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Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

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Short answer
ApplyBand 43 marks

Q1. Divide $x^3 + 2x^2 - 5x + 6$ by $x - 2$. State the quotient and remainder. (3 marks)

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ApplyBand 42 marks

Q2. When $P(x)$ is divided by $x + 1$, the quotient is $x^2 + 3x - 2$ and the remainder is 5. Find $P(x)$. (2 marks)

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AnalyseBand 53 marks

Q3. Divide $3x^3 - x^2 + 4x - 1$ by $x^2 + 1$. (3 marks)

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Comprehensive answers (click to reveal)

Activity 1: $x^3-3x^2+x-2 = (x-2)(x^2-x-1) + (-4)$. $Q(x)=x^2-x-1$, $R=-4$.

Activity 2: $x^3+1 = x^3+0x^2+0x+1$. $(x+1)(x^2-x+1) + 0 = x^3+1$. $Q(x)=x^2-x+1$, $R=0$ (exact division).

Activity 3: $P(x) = (x-3)(2x^2+x-1)+4 = 2x^3+x^2-x-6x^2-3x+3+4 = 2x^3-5x^2-4x+7$.

Activity 4: $3x^3-x^2+4x-1 = (x^2+1)(3x-1) + (x)$. $Q(x)=3x-1$, $R(x)=x$.

Activity 5: Each division step reduces the degree of the current remainder by at least 1. Since degree cannot be negative, the process must terminate in at most $\deg(P) - \deg(D) + 1$ steps.

Q1 (3 marks): Step 1: $x^3 \div x = x^2$; $x^2(x-2)=x^3-2x^2$; remainder $0x^2+2x^2-5x=2x^2-5x$ ... wait, rework: $x^3+2x^2-5x+6$ divided by $x-2$: $x^2(x-2)=x^3-2x^2$; new: $4x^2-5x$; $4x(x-2)=4x^2-8x$; new: $3x+6$; $3(x-2)=3x-6$; remainder $12$. [Checking: actually $2^3+2(4)-5(2)+6=8+8-10+6=12$. $Q(x)=x^2+4x+3$, $R=12$.] [3 marks for correct working + quotient + remainder]

Q2 (2 marks): $P(x) = (x+1)(x^2+3x-2)+5$. Expand: $x^3+3x^2-2x+x^2+3x-2+5 = x^3+4x^2+x+3$ [2 marks].

Q3 (3 marks): $3x^3 \div x^2 = 3x$. $3x(x^2+1)=3x^3+3x$. Subtract: $-x^2+x-1$. $-x^2 \div x^2 = -1$. $-1(x^2+1)=-x^2-1$. Subtract: $x$. Degree of $x$ is less than degree of $x^2+1$, stop. $Q(x)=3x-1$, $R(x)=x$ [3 marks].

Module overview · Extension 1 · Checkpoint 1 · Module quiz