QUICK ALGEBRA II HELP PLEASE! 1). (look on first picture) 2). (Question on second pic) A) 3 B) -4 C) -3 3). (Look on third pic) 4).(look on last 2 pics)
1. f(x) = x³ + 5x² + 7x + 3 f(x) = x³ + 3x² + 2x² + 6x + x + 3 f(x) = x²(x) + x²(3) + 2x(x) + 2x(3) + 1(x) + 1(3) f(x) = x²(x + 3) + 2x(x + 3) + 1(x + 3) f(x) = (x² + 2x + 1)(x + 3) f(x) = (x² + x + x + 1)(x + 3) f(x) = (x(x) + x(1) + 1(x) + 1(3))(x + 3) f(x) = (x(x + 1) + 1(x + 1))(x + 3) f(x) = (x + 1)(x + 1)(x + 3) f(x) = (x + 1)²(x + 3) The root -1 has a multiplicity of 2, making the answer equal to C. 2. root of f(x) = 0, multiplicity of 2 root of y = 0, multiplicity of 2 ⇵ (x - 3)² The answer is A. 3. x⁴ - x³ - 9x² + 7x + 14 = 0 x⁴ - 2x³ + x³ - 2x² - 7x² + 14x - 7x + 14 = 0 x³(x) - x³(2) + x²(x) - x²(2) - 7x(x) + 7x(2) - 7(x) + 7(2) = 0 x³(x - 2) + x²(x - 2) - 7x(x - 2) - 7(x - 2) = 0 (x³ + x² - 7x - 7)(x - 2) = 0 (x²(x) + x²(1) - 7(x) - 7(1))(x - 2) = 0 (x²(x + 1) - 7(x + 1))(x - 2) = 0 (x² - 7)(x + 1)(x - 2) = 0 x² - 7 = 0 or x + 1 = 0 or x - 2 = 0 + 7 + 7 - 1 - 1 + 2 + 2 x² = 7 or x = -1 or x = 2 There are three roots, making the answer equal to C. 4. f(x) = x³ + 4x² + 7x + 6 f(x) = x³ + 2x² + 2x² + 3x + 4x + 6 f(x) = x³ + 2x² + 3x + 2x² + 4x + 6 f(x) = x(x²) + x(2x) + x(3) + 2(x²) + 2(2x) + 2(3) f(x) = x(x² + 2x + 3) + 2(x² + 2x + 3) f(x) = (x + 2)(x² + 2x + 3) f(x) = (x + 2)(x² + 2x + 1 + 2) f(x) = (x + 2)(x² + 2x + 1 - (-2)) f(x) = (x + 2)(x² + x + x + 1 - (-1)(2) + xi√(2) - xi√(2) + i√(2) - i√(2)) f(x) = (x + 2)(x² + x + xi√(2) + x + 1 + i√(2) - xi√(2) - i√(2) - i²√(4)) f(x) = (x + 2)(x(x) + x(1) + x(i√(2)) + 1(x) + 1(1) + 1(i√(2)) - i√(2)(x) - i√(2)(1) - i√(2)(i√(2)) f(x) = (x + 2)(x(x + 1 + i√(2)) + 1(x + 1 + i√(2)) - i√(2)(x + 1 + i√(2)) f(x) = (x + 2)(x + 1 - i√(2)))(x + 1 + i√(2)) f(x) = (x + 2)(x - (1 + i√(2)))(x - (-1 - i√(2))) The answer is D.