Mathematics

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)


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Lyndsey863

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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. 

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