This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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You're on a roll — Step 1: Use the Factor Theorem to find a and b. The polynomial is f(x) = ax^4 + 7x^3 + x^2 + bx - 3. Given that (x-1) and (x+1) are factors of f(x), we know that f(1)=0 and f(-1)=0. For f(1)=0: a(1)^4 + 7(1)^3 + (1)^2 + b(1) - 3 = 0 a + 7 + 1 + b - 3 = 0 a + b + 5 = 0 a + b = -5 (*) For f(-1)=0: a(-1)^4 + 7(-1)^3 + (-1)^2 + b(-1) - 3 = 0 a(1) + 7(-1) + 1 - b - 3 = 0 a - 7 + 1 - b - 3 = 0 a - b - 9 = 0 a - b = 9 (**) Step 2: Solve the system of linear equations for a and b. We have: 1) a + b = -5 2) a - b = 9 Add equation (1) and equation (2): (a+b) + (a-b) = -5 + 9 2a = 4 a = 2 Substitute a=2 into equation (1): 2 + b = -5 b = -5 - 2 b = -7 The values of the constants are a=2 and b=-7. a=2, b=-7 Step 3: Find the set of real values of x for which f(x) > 0. Substitute a=2 and b=-7 into f(x): f(x) = 2x^4 + 7x^3 + x^2 - 7x - 3 Since (x-1) and (x+1) are factors, their product (x-1)(x+1) = x^2-1 is also a factor. Perform polynomial long division to find the other factor: (2x^4 + 7x^3 + x^2 - 7x - 3) ÷ (x^2 - 1) c|cc cc cc 2r2x^2 & +7x & +3 \\ 2-6 x^2-1 & 2x^4 & +7x^3 & +x^2 & -7x & -3 \\ 2r-(2x^4 & & -2x^2) \\ 2-4 2r0 & +7x^3 & +3x^2 & -7x \\ 2r & -(7x^3 & & -7x) \\ 3-5 2r & 0 & +3x^2 & 0 & -3 \\ 2r & & -(3x^2 & & -3) \\ 4-6 2r & & 0 & 0 & 0 \\ So, f(x) = (x^2-1)(2x^2+7x+3). Factor the quadratic term 2x^2+7x+3: 2x^2+7x+3 = 2x^2+x+6x+3 = x(2x+1)+3(2x+1) = (x+3)(2x+1) Thus, the fully factored form of f(x) is: f(x) = (x-1)(x+1)(x+3)(2x+1) Step 4: Determine the intervals where f(x) > 0. The roots of f(x) are the values of x where f(x)=0: x-1=0 x=1 x+1=0 x=-1 x+3=0 x=-3 2x+1=0 x = -(1)/(2) Order the roots on a number line: -3, -1, -(1)/(2), 1. These roots divide the number line into five intervals. We test the sign of f(x) in each interval: • For x < -3 (e.g., x=-4): f(-4) = (-5)(-3)(-1)(-7) = 105 > 0 • For -3 < x < -1 (e.g., x=-2): f(-2) = (-3)(-1)(1)(-3) = -9 < 0 • For -1 < x < -(1)/(2) (e.g., x=-0.75): f(-0.75) = (-1.75)(0.25)(2.25)(-0.5) ≈ 0.49 > 0 • For -(1)/(2) < x < 1 (e.g., x=0): f(0) = (-1)(1)(3)(1) = -3 < 0 • For x > 1 (e.g., x=2): f(2) = (1)(3)(5)(5) = 75 > 0 We are looking for f(x) > 0. This occurs in the intervals (-, -3), (-1, -(1)/(2)), and (1, ). The set of real values of x for which f(x) > 0 is: x (-, -3) (-1, -(1)/(2)) (1, ) Got more? Send 'em!