📊 Types of Polynomials
General Form
p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ (aₙ ≠ 0)
n = degree (non-negative integer) · aₙ = leading coefficient · a₀ = constant term
Degree 0
Constant Polynomial
p(x) = k (k ≠ 0)
A non-zero constant. Graph is a horizontal line that never crosses the x-axis.
e.g. p(x) = 7, p(x) = −3
Zeroes: None
Degree 1
Linear Polynomial
p(x) = ax + b (a ≠ 0)
Graph is a straight line. Always crosses x-axis at exactly one point.
e.g. 2x + 3, 5x − 10
Zero: x = −b/a (exactly 1)
Degree 2
Quadratic Polynomial
p(x) = ax² + bx + c (a ≠ 0)
Graph is a parabola. Opens up if a > 0, down if a < 0.
e.g. x² − 5x + 6, 2x² + x − 1
Zeroes: 0, 1 or 2
Degree 3
Cubic Polynomial
p(x) = ax³+bx²+cx+d (a ≠ 0)
S-shaped curve. Always has at least one real zero (cubic always crosses x-axis).
e.g. x³ − 6x² + 11x − 6
Zeroes: 1, 2 or 3
💡 Degree Rules to Remember
- The degree is the highest power of x with a non-zero coefficient.
- A polynomial of degree n has at most n zeroes.
- The zero polynomial (p(x) = 0) has no defined degree.
- Every real number is a zero of the zero polynomial.
- Binomial = 2 terms · Trinomial = 3 terms · Monomial = 1 term
💡 Quick check: Is 1/(x+1) a polynomial? NO — polynomials cannot have variables in the denominator. Is √x a polynomial? NO — powers must be whole numbers.
Value of Polynomial
Evaluating p(x) at x = k
Substitute x = k into p(x) to find p(k).
Example: p(x) = x² − 3x + 2
p(0) = 0 − 0 + 2 = 2
p(1) = 1 − 3 + 2 = 0 ← zero!
p(2) = 4 − 6 + 2 = 0 ← zero!
p(3) = 9 − 9 + 2 = 2
Identifying Degree
Find degree of composite expressions
5x³ − 2x + 1 → degree 3
3x²y + 2xy² → not a polynomial in one variable
(x+1)(x²−x+1) → expand = x³+1, degree 3
7 → degree 0 (constant)
📈 Graphs & Zeroes of Polynomials
Core Principle
Zero of p(x) = x-intercept of the graph y = p(x)
The number of x-intercepts tells you the number of zeroes. The shape of the graph reveals the type of polynomial.
🎪 Interactive Graph Explorer
RULE 01
Linear polynomial → exactly 1 zero
The graph of y = ax + b is a straight line. It always crosses the x-axis exactly once at x = −b/a. So a linear polynomial always has exactly one zero.
RULE 02
Quadratic polynomial → 0, 1 or 2 zeroes
The graph is a parabola. If it cuts the x-axis at 2 points → 2 distinct zeroes. If it just touches → 1 repeated zero. If it doesn't touch → no real zeroes. Use discriminant b²−4ac to determine which case.
RULE 03
Cubic polynomial → always at least 1 zero
The S-shaped cubic graph MUST cross the x-axis at least once. So a cubic polynomial always has at least 1 real zero. It may have 1, 2 or 3 zeroes depending on its shape.
RULE 04
Max zeroes = degree of polynomial
A polynomial of degree n can have AT MOST n real zeroes. This is the fundamental theorem behind all the rules above. It also means a degree 4 polynomial can have 0, 1, 2, 3 or 4 zeroes.
🎯 Exam tip: If asked "how many zeroes can a polynomial of degree n have?" → Answer: "at most n zeroes". If asked "how many zeroes does a linear polynomial have?" → Always exactly 1.
⍺ Quadratic — Zeroes & Coefficient Relations
For p(x) = ax² + bx + c with zeroes α and β
α + β = −b/a αβ = c/a
These hold for ANY quadratic, even without finding α and β explicitly. Derived from Vieta's formulas.
Sum of Zeroes
α + β = −b/a
= −(coefficient of x) ÷ (coefficient of x²)
The NEGATIVE sign is critical! Many students write +b/a by mistake.
Product of Zeroes
αβ = c/a
= (constant term) ÷ (coefficient of x²)
No negative sign here — just c/a directly.
Forming Polynomial
x² − (S)x + (P)
Given sum S and product P of zeroes, the quadratic is x² − Sx + P (or any non-zero scalar multiple).
Derivation of the Formulas
If α and β are the zeroes of ax² + bx + c:
⇒ ax² + bx + c = a(x − α)(x − β)
= a[x² − (α+β)x + αβ]
= ax² − a(α+β)x + a·αβ
Comparing coefficients of x: b = −a(α+β) ⇒ α+β = −b/a
Comparing constant terms: c = a·αβ ⇒ αβ = c/a
⇒ ax² + bx + c = a(x − α)(x − β)
= a[x² − (α+β)x + αβ]
= ax² − a(α+β)x + a·αβ
Comparing coefficients of x: b = −a(α+β) ⇒ α+β = −b/a
Comparing constant terms: c = a·αβ ⇒ αβ = c/a
Q.E.D. ■
📝 Worked Example 1 — Find sum and product of zeroes
p(x) = 6x² − 7x − 3 (a = 6, b = −7, c = −3)
α + β = −b/a = −(−7)/6 = 7/6
αβ = c/a = −3/6 = −1/2
Verification: zeroes of 6x²−7x−3 = 0 are x = 3/2 and x = −1/3
Sum = 3/2 + (−1/3) = 9/6 − 2/6 = 7/6 ✓
Product = 3/2 × (−1/3) = −3/6 = −1/2 ✓
α + β = −b/a = −(−7)/6 = 7/6
αβ = c/a = −3/6 = −1/2
Verification: zeroes of 6x²−7x−3 = 0 are x = 3/2 and x = −1/3
Sum = 3/2 + (−1/3) = 9/6 − 2/6 = 7/6 ✓
Product = 3/2 × (−1/3) = −3/6 = −1/2 ✓
Both formulas verified successfully.
📝 Worked Example 2 — Form a quadratic from given zeroes
Given zeroes: α = 2 + √3 and β = 2 − √3
Sum = (2+√3) + (2−√3) = 4
Product = (2+√3)(2−√3) = 4 − 3 = 1
Required polynomial: x² − (sum)x + (product)
= x² − 4x + 1
Sum = (2+√3) + (2−√3) = 4
Product = (2+√3)(2−√3) = 4 − 3 = 1
Required polynomial: x² − (sum)x + (product)
= x² − 4x + 1
Quadratic: x² − 4x + 1
Note: (2+√3)(2−√3) uses the identity (a+b)(a−b) = a²−b²
📝 Worked Example 3 — Find k using zero condition
One zero of p(x) = 3x² − kx + 4 is 2. Find k and the other zero.
Since 2 is a zero: p(2) = 0
3(4) − k(2) + 4 = 0 ⇒ 12 − 2k + 4 = 0 ⇒ 2k = 16 ⇒ k = 8
So p(x) = 3x² − 8x + 4. Other zero β:
α + β = 8/3 ⇒ 2 + β = 8/3 ⇒ β = 8/3 − 2 = 2/3
Since 2 is a zero: p(2) = 0
3(4) − k(2) + 4 = 0 ⇒ 12 − 2k + 4 = 0 ⇒ 2k = 16 ⇒ k = 8
So p(x) = 3x² − 8x + 4. Other zero β:
α + β = 8/3 ⇒ 2 + β = 8/3 ⇒ β = 8/3 − 2 = 2/3
k = 8, zeroes are 2 and 2/3
| Identity | Formula in terms of α+β and αβ | Use case |
|---|---|---|
| α² + β² | (α+β)² − 2αβ | Very common in board exams |
| (α−β)² | (α+β)² − 4αβ | Finding individual zeroes |
| α³ + β³ | (α+β)³ − 3αβ(α+β) | Higher order questions |
| 1/α + 1/β | (α+β) / αβ | Common in 2-mark questions |
| α/β + β/α | (α²+β²) / αβ = [(α+β)²−2αβ] / αβ | Requires both steps above |
💡 Sign trick: "Sum = Minus b over a, Product = Plus c over a". Always check: if both zeroes are positive, product and sum are both positive.
🎯 Cubic — Zeroes & Coefficient Relations
For p(x) = ax³ + bx² + cx + d with zeroes α, β, γ
α+β+γ = −b/a αβ+βγ+γα = c/a αβγ = −d/a
Note the sign pattern: −b/a, +c/a, −d/a (alternates). This mirrors the expansion of (x−α)(x−β)(x−γ).
Sum (1 at a time)
α+β+γ = −b/a
Sum of all three zeroes. The sign is NEGATIVE (−b/a), same pattern as the quadratic sum formula.
Sum of Pairs (2 at a time)
αβ+βγ+γα = c/a
Take all possible pairs: αβ, βγ, γα and add them. Sign is POSITIVE (+c/a).
Product (3 at a time)
αβγ = −d/a
Product of ALL three zeroes. Sign is NEGATIVE (−d/a). Most commonly forgotten!
Derivation for Cubic
ax³ + bx² + cx + d = a(x−α)(x−β)(x−γ)
Expanding (x−α)(x−β)(x−γ):
= x³ − (α+β+γ)x² + (αβ+βγ+γα)x − αβγ
So: ax³ + bx² + cx + d = a·x³ − a(α+β+γ)x² + a(αβ+βγ+γα)x − a·αβγ
Comparing x² terms: b = −a(α+β+γ) ⇒ α+β+γ = −b/a
Comparing x terms: c = a(αβ+βγ+γα) ⇒ αβ+βγ+γα = c/a
Comparing constants: d = −a·αβγ ⇒ αβγ = −d/a
Expanding (x−α)(x−β)(x−γ):
= x³ − (α+β+γ)x² + (αβ+βγ+γα)x − αβγ
So: ax³ + bx² + cx + d = a·x³ − a(α+β+γ)x² + a(αβ+βγ+γα)x − a·αβγ
Comparing x² terms: b = −a(α+β+γ) ⇒ α+β+γ = −b/a
Comparing x terms: c = a(αβ+βγ+γα) ⇒ αβ+βγ+γα = c/a
Comparing constants: d = −a·αβγ ⇒ αβγ = −d/a
Q.E.D. ■
📝 Worked Example 1 — Verify relations for a cubic
p(x) = x³ − 6x² + 11x − 6 (a=1, b=−6, c=11, d=−6)
Zeroes: α=1, β=2, γ=3
Sum: 1+2+3 = 6 −b/a = −(−6)/1 = 6 ✓
Pairs: 1×2+2×3+3×1 = 2+6+3 = 11 c/a = 11/1 = 11 ✓
Product: 1×2×3 = 6 −d/a = −(−6)/1 = 6 ✓
Zeroes: α=1, β=2, γ=3
Sum: 1+2+3 = 6 −b/a = −(−6)/1 = 6 ✓
Pairs: 1×2+2×3+3×1 = 2+6+3 = 11 c/a = 11/1 = 11 ✓
Product: 1×2×3 = 6 −d/a = −(−6)/1 = 6 ✓
All three relations verified.
📝 Worked Example 2 — Find all zeroes given one zero
p(x) = 2x³ + x² − 5x + 2. One zero is 1/2. Find all zeroes.
Divide p(x) by (x − 1/2), or equivalently by (2x − 1):
2x³ + x² − 5x + 2 = (2x − 1)(x² + x − 2)
Factor the quotient: x² + x − 2 = (x+2)(x−1)
So zeroes: 1/2, −2, 1
Verify with relations (a=2, b=1, c=−5, d=2):
Sum: 1/2+(−2)+1 = −1/2 = −b/a = −1/2 ✓
Product: (1/2)(−2)(1) = −1 = −d/a = −2/2 = −1 ✓
Divide p(x) by (x − 1/2), or equivalently by (2x − 1):
2x³ + x² − 5x + 2 = (2x − 1)(x² + x − 2)
Factor the quotient: x² + x − 2 = (x+2)(x−1)
So zeroes: 1/2, −2, 1
Verify with relations (a=2, b=1, c=−5, d=2):
Sum: 1/2+(−2)+1 = −1/2 = −b/a = −1/2 ✓
Product: (1/2)(−2)(1) = −1 = −d/a = −2/2 = −1 ✓
All zeroes: 1/2, −2, and 1
💡 Memory aid for the sign pattern:
"Sum: Minus, Pairs: Plus, Product: Minus" → −b/a, +c/a, −d/a
"Sum: Minus, Pairs: Plus, Product: Minus" → −b/a, +c/a, −d/a
➗ Division Algorithm for Polynomials
Division Algorithm (Theorem)
p(x) = g(x) · q(x) + r(x)
For any polynomials p(x) [dividend] and g(x) [divisor, g(x) ≠ 0], there exist unique polynomials q(x) [quotient] and r(x) [remainder] such that r(x) = 0 OR deg(r) < deg(g).
✎ Interactive Polynomial Division
Divide ax² + bx + c by (x + d). Enter the coefficients below.
STEP 01
Arrange in descending order of degree
Write both dividend and divisor with powers of x in decreasing order. Insert missing terms with coefficient 0 as placeholders. Example: x³ + 2 should be written as x³ + 0·x² + 0·x + 2.
STEP 02
Divide leading term of dividend by leading term of divisor
This gives the first term of the quotient. Multiply the whole divisor by this term. Subtract from the dividend. This is exactly like integer long division.
STEP 03
Bring down the next term and repeat
After subtraction, bring down the next term of the dividend. Repeat step 2. Continue until the degree of the remainder is less than the degree of the divisor.
STEP 04
Verify: p(x) = g(x)·q(x) + r(x)
Multiply divisor × quotient, then add remainder. The result must equal the original dividend. Also check: degree(remainder) < degree(divisor).
📝 Worked Example — Finding remaining zeroes
p(x) = x³ − 3x² − x + 3. Given: one zero is x = 3. Find all zeroes.
Since 3 is a zero, (x − 3) is a factor. Divide p(x) by (x − 3):
x³ − 3x² − x + 3 ÷ (x − 3):
→ x³ ÷ x = x² ... multiply: x²(x−3) = x³−3x² ... subtract: −x+3
→ −x ÷ x = −1 ... multiply: −1(x−3) = −x+3 ... subtract: 0
Quotient = x² − 1 = (x−1)(x+1)
All zeroes: x = 3, x = 1, x = −1
Since 3 is a zero, (x − 3) is a factor. Divide p(x) by (x − 3):
x³ − 3x² − x + 3 ÷ (x − 3):
→ x³ ÷ x = x² ... multiply: x²(x−3) = x³−3x² ... subtract: −x+3
→ −x ÷ x = −1 ... multiply: −1(x−3) = −x+3 ... subtract: 0
Quotient = x² − 1 = (x−1)(x+1)
All zeroes: x = 3, x = 1, x = −1
The three zeroes are 3, 1 and −1.
Verification: (−b/a) = −(−3)/1 = 3 = 3+1+(−1) ✓ | −d/a = −3/1... wait d=3, so −d/a = −3 = 3×1×(−1) ✓
Application
When to use the Division Algorithm
- Finding all zeroes when one or two zeroes are given
- Checking if g(x) is a factor of p(x) (remainder = 0)
- Expressing a polynomial in factored form
- Finding the quotient and remainder in exam problems
Quick Check
Factor Theorem
x − a is a factor of p(x) ⟺ p(a) = 0
Equivalently: a is a zero of p(x) iff (x−a) divides p(x) with zero remainder.
This makes it easy to check whether a given value is a zero: just substitute!
💡 CBSE Exam tip: In 3-mark division problems, always write the full verification step p(x) = g(x)·q(x) + r(x) — examiners award 1 mark specifically for this check.