Polynomials — The Basics
General form · Degree · Key terminology · Identities
What is a polynomial?
A polynomial in x is an expression where every power of x is a non-negative integer and all coefficients are real numbers.
General Form
\(f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_1 x + a_0\)
n = degree · \(a_n\) = leading coefficient (≠0) · \(a_0\) = constant term
Not polynomials: \(x^{-2}\) (negative power), \(x^{1/2}=\sqrt{x}\) (fractional power), \(2^x\) (variable exponent)
Naming by degree
degree 1Linear ax + b
degree 2Quadratic ax² + bx + c
degree 3Cubic ax³ + bx² + cx + d
degree 4Quartic axâ´ + bx³ + …
Monic polynomial: leading coefficient = 1, e.g. \(x^3 - 5x + 2\).
Key vocabulary
root / zerovalue of x where f(x) = 0
coefficientmultiplier of each xâ¿ term
identity ≡true for ALL values of x
multiplicityhow many times a root repeats
Identity ≡ vs Equation =
An equation is true for specific values of x. An identity is true for every value of x.
Equation\(x^2 - 1 = 0\) — only when \(x = \pm1\)
Identity\(x^2-1 \equiv (x-1)(x+1)\) — always true
Exam habit: Use ≡ when writing factorised polynomial identities. Writing = can cost marks.
The Division Algorithm — foundation of this chapter
The Identity
\(f(x) \equiv d(x)\cdot Q(x) + R\)
f = dividend, d = divisor, Q = quotient, R = remainder
Degree Rule
deg(Q) = deg(f) − deg(d)
Linear divisor → R is constant
Quadratic divisor → R is at most linear (px + q)
Zero Remainder
If R = 0 then d(x) is a factor of f(x). This links directly to the Factor Theorem.
The Remainder Theorem
Instant remainders — no long division needed
The Theorem
When \(f(x)\) is divided by a linear expression, the remainder equals f evaluated at the root of the divisor.
\(f(x)\div(x-a)\Rightarrow\text{Remainder}=f(a)\)
\(f(x)\div(ax-b)\Rightarrow\text{Remainder}=f\!\left(\tfrac{b}{a}\right)\)
Why it works: Division gives \(f(x)=(x-a)Q(x)+R\). Substituting \(x=a\) kills the quotient term, leaving \(f(a)=R\).
Example A — divide \(2x^3-3x^2+x-5\) by \((x-2)\)
Thm
Root of \((x-2)\) is \(x=2\), so Remainder \(=f(2)\)
Sub
\(f(2)=2(8)-3(4)+2-5\)
Calc
\(=16-12+2-5\)
✓
R = 1
Example B — divide same f(x) by \((2x+1)\)
Root
\(2x+1=0\Rightarrow x=-\tfrac{1}{2}\)
Sub
\(f(-\tfrac{1}{2})=2(-\tfrac{1}{8})-3(\tfrac{1}{4})+(-\tfrac{1}{2})-5\)
Calc
\(=-\tfrac{1}{4}-\tfrac{3}{4}-\tfrac{1}{2}-5\)
✓
R = −6½
Interactive — Remainder Calculator
For \(f(x)=x^3+ax^2+bx+c\) divided by \((x-r)\)
The Factor Theorem
Special case of Remainder Theorem · Zero remainder means factor
The Theorem — both directions
The Factor Theorem is the Remainder Theorem with R = 0. It works in both directions.
\((x-a)\text{ is a factor of }f(x)\iff f(a)=0\)
\((ax-b)\text{ is a factor of }f(x)\iff f\!\left(\tfrac{b}{a}\right)=0\)
Both directions matter: if f(a) = 0 then (x−a) is a factor; if (x−a) is a factor then f(a) = 0. Examiners test both in "show that" questions.
Show \((x-3)\) is a factor of \(f(x)=x^3-5x^2+2x+6\)
Step 1
Evaluate f at the root of the proposed factor: \(x=3\)
Step 2
\(f(3)=27-45+6+6=0\)
✓
Since \(f(3)=0\), by the Factor Theorem, \((x-3)\) is a factor of \(f(x)\).
Rational Root Test — finding candidates
For a monic cubic \(x^3+px^2+qx+r\), test \(\pm\) the factors of the constant term \(r\).
Smart order: Try ±1 first (simplest arithmetic). Check if sum of all coefficients = 0 → (x−1) is a factor. If alternating sum = 0 → (x+1) is a factor.
Finding unknowns using the Factor Theorem
One unknown — find k, given \((x-2)\) is a factor of \(f(x)=x^3+kx-4\)
FT
Factor Theorem: \(f(2)=0\)
Sub
\(8+2k-4=0\)
Solve
\(2k=-4\)
✓
\(k=-2\)
Two unknowns — \((x-1)\) and \((x+2)\) are factors of \(x^3+ax^2+bx-6\)
Eqn â‘
\(f(1)=0\): \(1+a+b-6=0 \Rightarrow a+b=5\)
Eqn â‘¡
\(f(-2)=0\): \(-8+4a-2b-6=0 \Rightarrow 2a-b=7\)
â‘ +â‘¡
\(3a=12 \Rightarrow a=4\), then \(b=1\)
✓
\(a=4,\quad b=1\)
Polynomial Long Division
Visual step-through · Two methods compared · Worked examples
Visual Step-Through — click each stage to follow the working
Dividing \(2x^3+x^2-13x+6\) by \((x-2)\)
Loading…
Method 1 — Long Division (systematic)
- Write dividend in descending degree order. Insert \(0\cdot x^k\) for any missing powers.
- Divide leading term of dividend by leading term of divisor → first term of Q(x)
- Multiply that term by the entire divisor and subtract from the dividend
- Bring down the next term; repeat until deg(remainder) < deg(divisor)
- Write the identity: \(f(x)\equiv d(x)\cdot Q(x)+R\)
Sign trap: When subtracting a row, every term flips sign. Write it out — don't do it mentally.
Method 2 — Comparing Coefficients
- State the expected form: e.g. \(f(x)\equiv(x-2)(ax^2+bx+c)+R\)
- Expand the right side and collect like terms
- Equate coefficients of each power of x on both sides
- Solve the linear system — start from the highest or lowest degree for speed
Best for: when the quotient structure is obvious, or when dividing by a quadratic where R = px + q needs two substitutions.
Comparing Coefficients — Full Worked Example
Divide \(x^3+2x^2-5x-6\) by \((x-2)\)
Set up
Divisor is degree 1 → Q(x) is degree 2.
\[x^3+2x^2-5x-6\equiv(x-2)(x^2+bx+c)\]
\[x^3+2x^2-5x-6\equiv(x-2)(x^2+bx+c)\]
Expand
\(=x^3+bx^2+cx-2x^2-2bx-2c\)
\(=x^3+(b-2)x^2+(c-2b)x-2c\)
\(=x^3+(b-2)x^2+(c-2b)x-2c\)
x² coeff
\(b-2=2\Rightarrow b=4\)
constant
\(-2c=-6\Rightarrow c=3\)
verify x
\(c-2b=3-8=-5\) ✓
Result
\(x^3+2x^2-5x-6\equiv(x-2)(x^2+4x+3)\)
Factorise the quadratic: \(x^2+4x+3=(x+1)(x+3)\)
\(\equiv(x-2)(x+1)(x+3)\)
Verify: Roots are x = 2, −1, −3. Sub each into f(x) to confirm they all give 0.
Factorising Cubic Polynomials
Full strategy · Sum & difference of cubes · Repeated roots · Interactive
General strategy for any cubic
- List all integer factors of the constant term \(a_0\) — these are your candidate roots
- Test each with the Factor Theorem. The moment f(p) = 0, stop — you have a linear factor \((x-p)\)
- Divide \(f(x)\) by \((x-p)\) using long division or comparing coefficients to get the quotient quadratic Q(x)
- Factorise Q(x) by inspection, completing the square, or the quadratic formula
- Write: \(f(x)\equiv(x-p)\cdot Q(x)\)
Shortcuts: If the sum of all coefficients = 0, then x = 1 is a root. If coefficients alternate in sign, x = −1 is a root.
Fully factorise \(f(x)=x^3-6x^2+11x-6\)
Candidates
Factors of 6: ±1, ±2, ±3, ±6
Test x=1
\(f(1)=1-6+11-6=0\) ✓ → \((x-1)\) is a factor
Divide
\(f(x)=(x-1)(x^2-5x+6)\)
Factorise Q
\(x^2-5x+6=(x-2)(x-3)\)
✓
\(f(x)\equiv(x-1)(x-2)(x-3)\)
Fully factorise \(f(x)=2x^3+x^2-13x+6\)
Candidates
Factors of 6 with leading coeff 2: ±1, ±2, ±3, ±6, ±½, ±³â„â‚‚
Test x=2
\(f(2)=16+4-26+6=0\) ✓ → \((x-2)\) is a factor
Divide
\(f(x)=(x-2)(2x^2+5x-3)\)
Factorise Q
\(2x^2+5x-3=(2x-1)(x+3)\)
✓
\(f(x)\equiv(x-2)(2x-1)(x+3)\)
Sum of Cubes
\(a^3+b^3=(a+b)(a^2-ab+b^2)\)
The quadratic factor has no real roots (discriminant < 0). Cannot be factorised further over â„.
\(x^3+8=(x+2)(x^2-2x+4)\)
Difference of Cubes
\(a^3-b^3=(a-b)(a^2+ab+b^2)\)
Same structure — the quadratic factor is also irreducible over â„.
\(27x^3-1=(3x-1)(9x^2+3x+1)\)
Repeated Roots
A root has multiplicity 2 if \((x-a)^2\) is a factor. On a graph, the curve touches but doesn't cross the x-axis at that point.
\(f(x)=(x-a)^2(x-b)\)
Both \(f(a)=0\) and \(f'(a)=0\) hold at a repeated root.
Example: \(f(x)=x^3-4x^2+4x=x(x-2)^2\)
Roots
\(x=0\) (simple) and \(x=2\) (repeated, multiplicity 2)
Graph
Curve crosses x-axis at \(x=0\), touches (bounces off) at \(x=2\)
Interactive — Cubic Root Finder
\(f(x)=x^3+ax^2+bx+c\)
Adjust sliders…
Practice Questions
3 Easy · 3 Medium · 3 Hard — Factor & Remainder Theorem
Easy 1
[2] Find the remainder when \(p(x)=x^3-3x^2+x+2\) is divided by \((x-2)\).
SOLUTION
Remainder Thm
Sub \(x=2\): \(8-12+2+2=\boldsymbol{0}\) (so \(x-2\) is a factor)
Easy 2
[3] Show that \((x+1)\) is a factor of \(x^3+2x^2-x-2\).
SOLUTION
Factor Thm
Sub \(x=-1\): \(-1+2+1-2=0\) ✓ So \((x+1)\) is a factor.
Easy 3
[2] Fully factorise \(x^3-x^2-4x+4\).
SOLUTION
Step 1
Try \(x=1\): \(1-1-4+4=0\) ✓ So \((x-1)\) is factor.
Divide
\(\div(x-1)\) gives \(x^2-4=(x-2)(x+2)\)
Answer
\(\boldsymbol{(x-1)(x-2)(x+2)}\)
Medium 1
[4] Find the value of \(a\) if \((x-3)\) is a factor of \(x^3+ax^2-7x+3\).
SOLUTION
Factor Thm
Sub \(x=3\): \(27+9a-21+3=0 \Rightarrow 9a=-9 \Rightarrow \boldsymbol{a=-1}\)
Medium 2
[4] Given \(f(x)=2x^3+3x^2+px+q\) has remainder 6 when divided by \((x-1)\) and is divisible by \((x+2)\), find \(p\) and \(q\).
SOLUTION
f(1)=6
\(2+3+p+q=6 \Rightarrow p+q=1\) ...[1]
f(-2)=0
\(-16+12-2p+q=0 \Rightarrow -2p+q=4\) ...[2]
Solve
[1]-[2]: \(3p=-3 \Rightarrow p=-1,\;q=2\)
Answer
\(\boldsymbol{p=-1,\;q=2}\)
Medium 3
[5] Solve \(2x^3-3x^2-11x+6=0\).
SOLUTION
Trial
\(x=3\): \(54-27-33+6=0\) ✓ So \((x-3)\) is a factor.
Divide
\(2x^3-3x^2-11x+6 = (x-3)(2x^2+3x-2)=(x-3)(2x-1)(x+2)\)
Answer
\(\boldsymbol{x=3,\;\tfrac{1}{2},\;-2}\)
Hard 1
[5] The polynomial \(f(x)=x^3+ax^2+bx-8\) has factor \((x-2)\) and \(f(1)=-6\). Find \(a,b\) and fully factorise.
SOLUTION
f(2)=0
\(8+4a+2b-8=0 \Rightarrow 4a+2b=0 \Rightarrow 2a+b=0\) ...[1]
f(1)=-6
\(1+a+b-8=-6 \Rightarrow a+b=1\) ...[2]
Solve
[1]-[2]: \(a=-1,\;b=2\)
Factorise
\(f(x)=x^3-x^2+2x-8=(x-2)(x^2+x+4)\) \((x^2+x+4\) has no real roots)
Hard 2
[5] Given \(x^4-5x^2+4=(x^2-1)(x^2-4)\), solve \(x^4-5x^2+4=0\) and hence \(e^{4t}-5e^{2t}+4=0\).
SOLUTION
Factorise
\((x-1)(x+1)(x-2)(x+2)=0 \Rightarrow x=\pm1,\pm2\)
Substitution
Let \(u=e^t\): \(u=1,2\) (reject \(-1,-2\) since \(e^t>0\))
\(e^t=1\Rightarrow t=0\); \(e^t=2\Rightarrow t=\ln2\)
\(e^t=1\Rightarrow t=0\); \(e^t=2\Rightarrow t=\ln2\)
Hard 3
[6] Find the range of values of \(k\) for which \(x^3-kx+k=0\) has exactly one real root by using the substitution \(u=x-\frac{k}{3x}\).
SOLUTION
Approach
Exactly one real root means discriminant of auxiliary quadratic \(<0\). For cubic \(x^3+px+q=0\): \(\Delta=4p^3+27q^2<0\).
Here \(p=-k,\;q=k\): \(-4k^3+27k^2<0 \Rightarrow k^2(27-4k)<0\)
\(k\neq0\) and \(27-4k<0 \Rightarrow \boldsymbol{k>\tfrac{27}{4},\;k\neq0}\)
Here \(p=-k,\;q=k\): \(-4k^3+27k^2<0 \Rightarrow k^2(27-4k)<0\)
\(k\neq0\) and \(27-4k<0 \Rightarrow \boldsymbol{k>\tfrac{27}{4},\;k\neq0}\)
Past Year Paper Questions
Cambridge 0606 — Factor & Remainder Theorem
0606
[8] (a) Given \(f(x)=x^3+3x^2-4\), show that \((x-1)\) is a factor. [1]
(b) Factorise \(f(x)\) completely. [3]
(c) Solve \(f(x-1)=0\). [2]
(b) Factorise \(f(x)\) completely. [3]
(c) Solve \(f(x-1)=0\). [2]
SOLUTION
(a)
\(f(1)=1+3-4=0\) ✓
(b)
\((x-1)(x^2+4x+4)=(x-1)(x+2)^2\)
(c)
Replace \(x\) with \(x-1\): \((x-2)(x+1)^2=0 \Rightarrow \boldsymbol{x=2\text{ or }x=-1}\)
0606 Style
Factor Theorem Application
[8]
(a) Show that \((x-3)\) is a factor of \(p(x)=2x^3-5x^2-4x+3\) and fully factorise \(p(x)\). [4]
(b) Solve \(p(x) = 0\). [2]
(c) Find the remainder when \(p(x+1)\) is divided by \(x\). [2]
(b) Solve \(p(x) = 0\). [2]
(c) Find the remainder when \(p(x+1)\) is divided by \(x\). [2]
FULL WORKED SOLUTION
(a)
\(p(3)=54-45-12+3=0\) ✓. Divide to get \(2x^2+x-1=(2x-1)(x+1)\).
\(\boldsymbol{p(x)=(x-3)(2x-1)(x+1)}\)
\(\boldsymbol{p(x)=(x-3)(2x-1)(x+1)}\)
(b)
\(\boldsymbol{x=3,\; x=\tfrac{1}{2},\; x=-1}\)
(c)
Remainder = \(p(0+1)=p(1)=2-5-4+3=\boldsymbol{-4}\)
0606 Style
Cubic Sketch & Roots
[7]
\(f(x) = (2x+1)(x-2)^2\)
(a) State the roots of \(f(x) = 0\) and the y-intercept. [3]
(b) Describe the behaviour at each root (crossing or touching). [2]
(c) Sketch \(y = f(x)\). [2]
(a) State the roots of \(f(x) = 0\) and the y-intercept. [3]
(b) Describe the behaviour at each root (crossing or touching). [2]
(c) Sketch \(y = f(x)\). [2]
FULL WORKED SOLUTION
(a)
Roots: \(x=-\frac{1}{2}\) and \(x=2\). y-intercept: \(f(0)=(1)(4)=\boldsymbol{4}\)
(b)
\(x=-\frac{1}{2}\): simple root → crosses x-axis. \(x=2\): repeated root → touches (bounces off) x-axis.
(c)
Positive leading coeff (\(2x^3\)) → starts low-left, ends high-right. Crosses at \(x=-\frac{1}{2}\), bounces at \(x=2\), through \((0,4)\).