University:
High School
Course:
High School Algebra
Academic year:

2023
Views:

494

Pages:

8
Author:

demdainssi

The Fundamental Theorem of Algebra An nth degree polynomial has n solutions. Complex Conjugates Theorem: If a + bi is a zero, then a - bi is also a zero. Complex Solutions come in pairs! B. EXAMPLES Determine how many solutions. a. x³ + 3x² + 16x + 48 = 0 degree: 3 ⇒ 3 solutions b. x⁴ - 2x³ + 5x + 12 = 0 degree: 4 ⇒ 4 solutions How many zeros? a. f(x) = x⁴ + 6x³ + 12x² + 8x degree: 4 ⇒ 4 zeros b. f(x) = x³ - 6x + 2 degree: 3 ⇒ 3 zeros remember possible rational roots: factors of constant factors of leading coefficient 3. Find All Zeros a. f(x) = x³ + 7x² + 16x + 12 possible rational roots: ±1, 2, 3, 4, 6, 12 1 | 1 7 16 12 | 1 8 24 36 | 1 8 24 36 2 | 1 7 16 12 | 2 18 68 | 1 9 34 80 -1 | 1 7 16 12 | -1 -6 -10 | 1 6 10 2 -2 | 1 7 16 12 | -2 -10 -12 | 1 5 6 0 ✓ ⇒ -2 is a root (x + 2) is a factor x³ + 7x² + 16x + 12 = (x + 2)(x² + 5x + 6) = (x + 2)(x + 3)(x + 2) By Zero Product Property x = -2, -2, -3 b. f(x) = x⁵ + x³ - 2x² - 12x - 8 There are 5 zeros (Solutions). possible roots: ± 1, 2, 4, 8 Don't forget space holder! x⁴ term 1 | 1 0 1 -2 -12 -8 | 1 1 2 0 -12 | 1 1 2 0 -12 -20 -1 | 1 0 1 -2 -12 -8 | -1 1 -2 4 8 | 1 -1 2 -4 -4 0 ✓ x = -1 is a Solution when you find a Solution use the new coefficients to continue finding other solutions 2 | 1 -1 2 -4 -8 | 2 2 8 8 | 1 1 4 4 0 ✓ x = 2 is a Solution x³ + x² + 4x + 4 = 0 Factor by grouping x²(x + 1) + 4(x + 1) = 0 (x + 1)(x² + 4) = 0 x + 1 = 0 x² + 4 = 0 -1 -1 -4 -4 x = -1 x² = -4 √x² = √-4 x = ±2i x = -1, -1, 2, 2i, -2i C. f(x) = x⁵ + 3x⁴ + 9x³ + 23x² - 36 There are 5 zeros possible roots: ±1, 2, 3, 4, 6, 9, 12, 18, 36 Don't forget space holder! No X term 1 | 1 3 9 23 0 -36 | 1 4 13 36 36 | 1 4 13 36 36 0 ✓ x = 1 2 | 1 3 9 23 0 -36 | 2 10 38 122 244 | 1 5 19 61 122 208 -2 | 1 3 9 23 0 -36 | -2 -2 -14 -18 36 | 1 1 7 9 -18 0 ✓ x = -2 x³ + 2x² + 9x + 18 = 0 Factoring by grouping x²(x + 2) + 9(x + 2) = 0 (x + 2)(x² + 9) = 0 x + 2 = 0