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.
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.
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 $R(x)$.
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: $R(x) = 0$
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
Skills
- Perform polynomial long division step by step
- State the quotient and remainder for any division
- Verify the result by expanding $D(x) \cdot Q(x) + R(x)$
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:
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.
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 $R(x)$?
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:
- Divide the leading term of the current dividend by the leading term of the divisor to get the next term of $Q(x)$.
- Multiply the entire divisor by this new term.
- 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 $R(x)$.
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.
Worked examples · 3 in a row, reveal as you go
Divide $P(x) = x^3 - 2x^2 + 3x - 4$ by $D(x) = x - 1$.
Divide $P(x) = 2x^3 + 3x^2 - x + 5$ by $D(x) = x + 2$.
When $P(x)$ is divided by $x + 1$, the quotient is $x^2 + 3x - 2$ and the remainder is 5. Find $P(x)$.
Fill the gap: When $x^3 + 2x - 1$ is divided by $x - 1$, the remainder equals .
Misconceptions to fix · the 3 traps that cost marks
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.
Activities · practice with the ideas
Divide $x^3 - 3x^2 + x - 2$ by $x - 2$. State $Q(x)$ and $R(x)$.
Divide $x^3 + 1$ by $x + 1$. (Hint: insert placeholder terms first.)
When $P(x)$ is divided by $(x - 3)$, the quotient is $2x^2 + x - 1$ and the remainder is 4. Find $P(x)$.
Divide $3x^3 - x^2 + 4x - 1$ by $x^2 + 1$. State quotient and remainder.
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?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Divide $x^3 + 2x^2 - 5x + 6$ by $x - 2$. State the quotient and remainder. (3 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)
Q3. Divide $3x^3 - x^2 + 4x - 1$ by $x^2 + 1$. (3 marks)
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].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering polynomial division questions. Lighter alternative to the boss.
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