Key Concepts in Polynomials
- Definition of a Polynomial: A polynomial is an algebraic expression consisting of variables, coefficients, and exponents, where the exponents are whole numbers. Examples:
- Polynomial: 2×3−3×2+5x−72x^3 – 3x^2 + 5x – 7
- Not a Polynomial: 2x−1+x2x^{-1} + \sqrt{x} (exponent is not a whole number).
- Degrees of a Polynomial:
- The degree is the highest power of the variable in the polynomial.
- Examples:
- 5×3−2×2+45x^3 – 2x^2 + 4: Degree = 3.
- 7y5+3y4+y7y^5 + 3y^4 + y: Degree = 5.
- Types of Polynomials Based on Degree:
- Linear Polynomial: Degree = 1 (e.g., 3x+23x + 2).
- Quadratic Polynomial: Degree = 2 (e.g., x2−4x+3x^2 – 4x + 3).
- Cubic Polynomial: Degree = 3 (e.g., 2×3−x2+3x−52x^3 – x^2 + 3x – 5).
- Types of Polynomials Based on Terms:
- Monomial: One term (e.g., 3x3x).
- Binomial: Two terms (e.g., x2+2xx^2 + 2x).
- Trinomial: Three terms (e.g., x3+x2+2x^3 + x^2 + 2).
Key Operations and Concepts
1. Addition, Subtraction, and Multiplication
- Combine like terms while performing operations.
Example: Simplify (2×2+3x+4)+(x2−5x+6)(2x^2 + 3x + 4) + (x^2 – 5x + 6): =(2×2+x2)+(3x−5x)+(4+6).= (2x^2 + x^2) + (3x – 5x) + (4 + 6). =3×2−2x+10.= 3x^2 – 2x + 10.
2. Division of Polynomials
Division is performed using long division.
Example: Divide 2×3+3×2−x+52x^3 + 3x^2 – x + 5 by x+1x + 1.
- Divide 2x32x^3 by xx: Quotient = 2x22x^2.
- Multiply 2x22x^2 by x+1x + 1: 2×3+2x22x^3 + 2x^2.
- Subtract: (2×3+3×2−x+5)−(2×3+2×2)=x2−x+5(2x^3 + 3x^2 – x + 5) – (2x^3 + 2x^2) = x^2 – x + 5.
- Repeat until the degree of the remainder is less than the degree of the divisor.
The Remainder Theorem
If p(x)p(x) is divided by x−ax – a, the remainder is p(a)p(a).
Example: Find the remainder when p(x)=x3−3×2+4x−2p(x) = x^3 – 3x^2 + 4x – 2 is divided by x−2x – 2.
- Substitute x=2x = 2 into p(x)p(x): p(2)=23−3(22)+4(2)−2.p(2) = 2^3 – 3(2^2) + 4(2) – 2. =8−12+8−2=2.= 8 – 12 + 8 – 2 = 2.
- The remainder is 2.
The Factor Theorem
A polynomial p(x)p(x) has x−ax – a as a factor if p(a)=0p(a) = 0.
Example: Check if x−1x – 1 is a factor of p(x)=x3−4×2+3xp(x) = x^3 – 4x^2 + 3x.
- Substitute x=1x = 1 into p(x)p(x): p(1)=13−4(12)+3(1)=1−4+3=0.p(1) = 1^3 – 4(1^2) + 3(1) = 1 – 4 + 3 = 0.
- Since p(1)=0p(1) = 0, x−1x – 1 is a factor.
Zeros of a Polynomial
The zeros of a polynomial are the values of xx for which p(x)=0p(x) = 0.
Key Results:
- A polynomial of degree nn has at most nn zeros.
- Zeros can be real or complex numbers.
Example: Find the zeros of p(x)=x2−5x+6p(x) = x^2 – 5x + 6.
- Factorize: p(x)=(x−2)(x−3).p(x) = (x – 2)(x – 3).
- Zeros: x=2,x=3x = 2, x = 3.
Graphical Representation
The graph of a polynomial depends on its degree and coefficients:
- Linear Polynomials: Straight line.
- Quadratic Polynomials: Parabola.
- Cubic Polynomials: S-shaped curve.
Application Problems
Problem 1: Sum and Product of Zeros
If p(x)=ax2+bx+cp(x) = ax^2 + bx + c, the sum and product of zeros are: Sum of zeros=−ba,Product of zeros=ca.\text{Sum of zeros} = -\frac{b}{a}, \quad \text{Product of zeros} = \frac{c}{a}.
Example: For p(x)=2×2−5x+3p(x) = 2x^2 – 5x + 3:
- Sum of zeros: =−−52=52.= -\frac{-5}{2} = \frac{5}{2}.
- Product of zeros: =32.= \frac{3}{2}.
Problem 2: Verify Factor Theorem
Verify that x−2x – 2 is a factor of p(x)=x3−3×2+4x−8p(x) = x^3 – 3x^2 + 4x – 8.
- Substitute x=2x = 2 into p(x)p(x): p(2)=23−3(22)+4(2)−8=8−12+8−8=0.p(2) = 2^3 – 3(2^2) + 4(2) – 8 = 8 – 12 + 8 – 8 = 0.
- Since p(2)=0p(2) = 0, x−2x – 2 is a factor.
Critical Observations
- Importance of Remainder and Factor Theorem:
- Simplifies finding roots and divisors of polynomials.
- Basis for synthetic division.
- Role of Degree:
- The degree dictates the behavior of polynomials (number of zeros, shape of graph).
- Graphical Interpretation:
- Understanding graphs builds intuition about polynomial behavior.
- Applications:
- Used in physics (trajectory of objects), economics (profit functions), and data fitting.
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