University:
York University
Course:
MATH 1013 | Applied Calculus
Academic year:

2023
Views:

153

Pages:

4
Author:

Niklas

I. COMPLEX NUMBERS Complex numbers are not real Ex. 2+i, 3-4i i is an imaginary number Definition: i^2 = -1 Add, subtract, multiply, and divide complex numbers (3-i) + (5+2i) (4+2i) * (7-i) (3+i)(-3-2i) (11+2i) / (2-i) (2i)(-5-4i) (5+7i) / (4+i) II. POLYNOMIAL BASICS OPERATIONS (x^3+4x^2-7) + (2x^2-7x+4) (2x^3+4x^2-7x) - (4x^2+5x-9) (x-3)^2 (x^2-7x+4)(x-3) (3x^3+4x^2-11x-6) / (x+3) (x^4+x^2-3x+7) / (2x-1) Factor Theorem If the remainder of a quotient is 0, then the divisor is a factor of the dividend 7. In questions 5 and 6, which one has a factor been found? 5 III. POLYNOMIAL EQUATIONS AND INEQUALITIES Fundamental Theorem of Algebra: Every equation of degree n has n roots in the complex numbers Linear Factorization Theorem: A polynomial of degree n will have n factors of the form (x-c), where c ∈ ℂ x-intercepts: Solutions to P(x) = 0 Zeroes: Solutions to P(x) = 0 Factor: A divisor of P(x) 8. The zeroes of a polynomial are x = 3, x = 1, and x = 2 + 3i. x-int: [1, 3] Factors: (x-3)(x-1)(x-2+3i)(x-2-3i) IV. SOLVING POLYNOMIAL EQUATIONS FACTOR FIRST Descartes' Rule of Signs: Number of positive roots is the number of sign changes in P(x) Number of negative roots is the number of sign changes in P(-x) Number of complex roots come in pairs and are determined by the amount of roots left over Rational Root Theorem: Possible rational roots of P(x) are ±p/q, where p is the set of factors of the constant term, and q is the set of factors of the leading coefficient Example: P(x) = 2x^3 + 2x^2 - 3x + 6 P(-x) = -2x^3 + 2x^2 + 3x + 6 ±p/q = {±1, ±2, ±3, ±6, ±1/2} 9. P(x) = 9x^3 - 7x - 2 a. Maximum number of zeroes: 3 b. Descartes' Rule of Signs: 2 positive, 1 negative c. Rational Root Theorem: ±1, ±2, ±1/3, ±2/3, ±1/9, ±2/9 10. P(x) = x^4 + 2x^3 - 4x^2 + 5x a. Maximum number of zeroes: 4 b. Descartes' Rule of Signs: 2 or 0 positive, 2 or 0 negative c. Rational Root Theorem: ±1, ±5 I. Solve each polynomial equation: 11. (3x-1)(x+2)^2(x-1) = 0 12. x^3 + 3x^2 - x - 3 = 0 13. x^3 - 2x^2 + 3x - 6 = 0 II. Solve each polynomial inequality: 14. x^3 + x^2 - 5x + 3 ≤ 0 15. x^4 - 16 > 5x^3 - 20x V. IRRATIONAL ROOTS: Must estimate using the Location Principle Let P(x) be a polynomial function and P(a) > 0 and P(b) < 0. P(x) = 0 on (a, b) 16. Consider x^4 - 6x^3 + 10x^2 - 14x - 15 = 0 has a real root between 4 and 5. Using the Location Principle, approximate the real root to two decimal places. VI. REVIEW Solve x^3 - x^2 - 10x - 8 = 0 Solve x^4 + 5x^3 - 5x - 5 > 0 Solve x^3 + 2x^2 - 3x - 2 = 0. Approximate to one decimal place.