Complete notes with examples, important theorems, and formula sheet
Polynomials are algebraic expressions that consist of variables and coefficients, involving operations of addition, subtraction, multiplication, and non-negative integer exponents. They are fundamental in algebra and appear everywhere in mathematics.
A polynomial is an expression of the form: p(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + ... + a₂x² + a₁x + a₀
Where aₙ, aₙ₋₁, ..., a₂, a₁, a₀ are constants (called coefficients), and n is a non-negative integer.
Examples: 3x² + 2x + 1, 5x³ - 4x + 7, 2x - 3
Q1: Find the degree of p(x) = 5x⁴ - 3x² + 7x - 2
Solution: The highest power of x is 4, so degree = 4
Q2: Find the degree of p(x) = 7x - 5
Solution: The highest power of x is 1, so degree = 1
Q3: Find the degree of p(x) = 9
Solution: This is a constant polynomial, so degree = 0
| Type | Degree | General Form | Example |
|---|---|---|---|
| Constant Polynomial | 0 | p(x) = a | 5, -3, 7 |
| Linear Polynomial | 1 | p(x) = ax + b | 2x + 3, 5x - 7 |
| Quadratic Polynomial | 2 | p(x) = ax² + bx + c | x² - 5x + 6, 3x² + 2 |
| Cubic Polynomial | 3 | p(x) = ax³ + bx² + cx + d | 2x³ - 4x² + x - 1 |
| Biquadratic Polynomial | 4 | p(x) = ax⁴ + bx³ + cx² + dx + e | x⁴ - 3x² + 2 |
A polynomial is in standard form when its terms are arranged in descending order of their degrees.
Example: 5x³ + 2x² - 7x + 4 (degree 3 → 2 → 1 → 0)
A zero of a polynomial is a value of the variable that makes the polynomial equal to zero.
If p(a) = 0, then 'a' is called a zero (or root) of the polynomial p(x).
Geometrically, zeros are the x-coordinates where the graph of the polynomial touches or crosses the x-axis.
Q1: Find the zero of p(x) = 2x - 6
Solution:
Set p(x) = 0
2x - 6 = 0
2x = 6
x = 3
Therefore, zero = 3
Q2: Find the zeros of p(x) = x² - 5x + 6
Solution:
Set p(x) = 0
x² - 5x + 6 = 0
(x - 2)(x - 3) = 0
x - 2 = 0 or x - 3 = 0
Therefore, zeros = 2 and 3
The Remainder Theorem is a powerful tool that helps us find the remainder when a polynomial is divided by a linear expression.
If a polynomial p(x) is divided by (x - a), then the remainder is p(a).
In other words: Simply substitute x = a in the polynomial to get the remainder!
Q1: Find the remainder when p(x) = x³ - 3x² + 2x - 5 is divided by (x - 2)
Solution:
Here, divisor is (x - 2), so a = 2
By Remainder Theorem, remainder = p(2)
p(2) = (2)³ - 3(2)² + 2(2) - 5
p(2) = 8 - 12 + 4 - 5
p(2) = -5
Therefore, remainder = -5
Q2: Find the remainder when p(x) = 2x² + 3x - 1 is divided by (x + 1)
Solution:
Divisor is (x + 1) = (x - (-1)), so a = -1
Remainder = p(-1)
p(-1) = 2(-1)² + 3(-1) - 1
p(-1) = 2 - 3 - 1 = -2
Therefore, remainder = -2
The Factor Theorem is a special case of the Remainder Theorem. It helps us determine whether a linear expression is a factor of a polynomial.
(x - a) is a factor of polynomial p(x) if and only if p(a) = 0.
In simple words: If substituting x = a makes the polynomial zero, then (x - a) is a factor!
Q1: Check if (x - 2) is a factor of p(x) = x³ - 4x² + 5x - 2
Solution:
By Factor Theorem, (x - 2) is a factor if p(2) = 0
p(2) = (2)³ - 4(2)² + 5(2) - 2
p(2) = 8 - 16 + 10 - 2
p(2) = 0
Yes! (x - 2) is a factor since p(2) = 0
Q2: Check if (x + 3) is a factor of p(x) = x² + 4x + 3
Solution:
(x + 3) = (x - (-3)), so check if p(-3) = 0
p(-3) = (-3)² + 4(-3) + 3
p(-3) = 9 - 12 + 3 = 0
Yes! (x + 3) is a factor
Remainder Theorem: Find remainder when dividing. Answer: p(a)
Factor Theorem: Check if it's a factor. Answer: Yes (if p(a)=0) or No
Factorization means writing a polynomial as a product of two or more polynomials. It's like breaking down a number into prime factors.
Q1: Factorize x² - 16
Solution: Using identity a² - b² = (a+b)(a-b)
x² - 16 = x² - 4²
= (x + 4)(x - 4)
Q2: Factorize x² + 7x + 12
Solution: Split middle term (7x = 3x + 4x)
x² + 7x + 12 = x² + 3x + 4x + 12
= x(x + 3) + 4(x + 3)
= (x + 3)(x + 4)
Q3: Factorize x³ - 8
Solution: Using identity a³ - b³ = (a-b)(a² + ab + b²)
x³ - 8 = x³ - 2³
= (x - 2)(x² + 2x + 4)
Algebraic identities are equations that are true for all values of the variables. They are extremely useful in factorization and simplification.
Q1: Expand (2x + 3)³
Solution: Using (a + b)³ = a³ + 3a²b + 3ab² + b³
Here a = 2x, b = 3
(2x + 3)³ = (2x)³ + 3(2x)²(3) + 3(2x)(3)² + (3)³
= 8x³ + 36x² + 54x + 27
Q2: Factorize x³ + 27
Solution: Using a³ + b³ = (a + b)(a² - ab + b²)
x³ + 27 = x³ + 3³
= (x + 3)(x² - 3x + 9)
Q3: If x + y + z = 0, find x³ + y³ + z³
Solution: Using: If a + b + c = 0, then a³ + b³ + c³ = 3abc
x³ + y³ + z³ = 3xyz
Polynomial: p(x) = aₙxⁿ + ... + a₁x + a₀
Degree: Highest power of variable
Zero: Value where p(a) = 0
Constant term: a₀ (no variable)
Maximum zeros = degree of polynomial
Degree 0: Constant (a)
Degree 1: Linear (ax + b)
Degree 2: Quadratic (ax² + bx + c)
Degree 3: Cubic (ax³ + bx² + cx + d)
Degree 4: Biquadratic
Remainder Theorem:
If p(x) ÷ (x-a), remainder = p(a)
Factor Theorem:
(x-a) is factor ⟺ p(a) = 0
Zero and factor are related!
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
a² - b² = (a+b)(a-b)
(x+a)(x+b) = x² + (a+b)x + ab
(a+b)³ = a³ + 3a²b + 3ab² + b³
(a-b)³ = a³ - 3a²b + 3ab² - b³
a³ + b³ = (a+b)(a² - ab + b²)
a³ - b³ = (a-b)(a² + ab + b²)
(a+b+c)² = a² + b² + c² + 2ab + 2bc + 2ca
a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)
Special Case:
If a+b+c = 0, then a³+b³+c³ = 3abc
1. Take out common factors first
2. Use identities when possible
3. Split middle term for quadratics
4. Use Factor Theorem for higher degrees
Always check by multiplying back!
Linear zero: x = -b/a (from ax+b=0)
Quadratic zeros: 2 values max
Perfect square: b² = 4ac
Sum of zeros (quadratic): -b/a
Product of zeros: c/a
• Memorize all the algebraic identities - they're used everywhere!
• Practice identifying which identity to use for different problems.
• Always verify your factorization by multiplying the factors back.
• For Factor Theorem, try simple values like ±1, ±2, ±3 first.
• Remainder Theorem is faster than long division - use it when possible!
• When factorizing, look for patterns: common factors first, then identities.
• Remember: Degree tells you maximum number of zeros possible.