University:
University of North Georgia
Course:
MATH 2460 | Calculus II
Academic year:

2023
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168

Pages:

2
Author:

SpaceCadet101

Algebra 2 Honors Polynomial Notes (8.6/8.7) Name: Key Warm Up Find all the zeros of the following polynomial functions. a. f(x) = x³ + 7x² + 5x + 9 Not Factorable Rational Root Theorem Suppose that a polynomial equation with integral coefficients has the root p/q, where p and q are relatively prime integers. Then p must be a factor of the constant term of the polynomial and q must be a factor of the coefficient of the highest-degree term. EXAMPLE 1 - Rational Root Theorem List the possible rational roots of each equation. a. 9x³ + x² - 4x + 4 = 0 p: ±1, ±2, ±4 q: ±1, ±3, ±9 p/q: ±1, ±1/3, ±2/3, ±4/3, ±2, ±4, ±4/9 b. 6k⁴ - 7k³ + 8k² - 3k + 2 = 0 p: ±1, ±2 q: ±1, ±2, ±3, ±6 p/q: ±1, ±1/2, ±1/3, ±1/6, ±2, ±2/3 EXAMPLE 2 - Rational Root Theorem Tell how you know that the following equations have no rational roots. a. x³ + x² - x + 1 = 0 1: (1)³ + (1)² - (1) + 1 = 2 -1: (-1)³ + (-1)² - (-1) + 1 = 2 No b. x³ + 3x - 3 = 0 1: (1)³ + 3(1) - 3 = 1 -1: (-1)³ + 3(-1) - 3 = -7 3: (3)³ + 3(3) - 3 = 33 -3: (-3)³ + 3(-3) - 3 = -57 No Descartes' Rule of Signs Let P(x) be a simplified polynomial with real coefficients and terms arranged in decreasing degree of x. 1. The number of positive real roots of P(x) = 0 equals the number of variations of sign of P(x) or is fewer than this number by an even integer. 2. The number of negative real roots of P(x) = 0 equals the number of variations of sign of P(-x) or is fewer than this number by an even integer. EXAMPLE 3 - Descartes' Rule of Signs List the possibilities for the nature of the roots of each equation. a. f(x) = 2x⁶ - 3x² - x + 1 f(x) = + - - + 1 = 3 or 1 f(-x) = + - - + 1 = 3 or 1 b. h(x) = x⁶ - 2x⁵ + 3x⁴ - 10x³ - 6x² - 8x - 8 h(x) = + - + - - - - = 3 or 1 h(-x) = + + + - - - - = 3 or 1 Theorems Roots Theorem: Every polynomial equation with complex coefficients and positive degree n has exactly n roots. Conjugate Root Theorem: If a polynomial equation with real coefficients has a + bi as a root (a and b real, b ≠ 0), then a - bi is also a root. EXAMPLE 4 - Roots Find a polynomial equation with integral coefficients that has the given numbers as roots. a. -1, 3, -i, and i f(x) = (x+1)(x-3)(x+i)(x-i) = (x²+x-3x-3)(x²+1) = x⁴+x²-2x³-2x-3x²-3 f(x) = x⁴-2x³-2x²-2x-3 b. 5, 2+i Sum: 2+2+2-1 = 4/1 Product: (2+i)(2-i) = 4+1 = 5/1 f(x) = (x²-4x+5)(x-5) = x³-5x²-4x²+20x+5x-25 f(x) = x³-9x²+25x-25 EXAMPLE 5 - Synthetic Process A root of the equation is given. Solve the equation. a. 2x³-x²+10x-5=0; i√5 -> -i√5 i√5 | 2 -1 10 -5 | 2i√5 -10 -i√5 5 ----------------------- 2 -1+2i√5 0 0 (x-i√5)(x+i√5)(2x-1) = 0 x = i√5, -i√5, 1/2 b. x⁴-5x²-10x-6=0; -1+i -1+i | 1 0 -5 -10 -6 | -1+i 2+2i 3+3i 6 ----------------------- 1 -1+i -3-i 0 (x-(-1+i))(x-(-1-i))(x²-2x-3) = 0 (x+1-i)(x+1+i)(x-3)(x+1) = 0 x = -1±i, 3, -1

Math Polynomials Notes 6 Key

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