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(x+3)^(3)(x-1)^(2)(x-4) >= 0

(x+3)3(x−1)2(x−4)≥0 (x+3)^{3}(x-1)^{2}(x-4) \geq 0

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Full solution

Q. (x+3)3(x−1)2(x−4)≥0 (x+3)^{3}(x-1)^{2}(x-4) \geq 0
  1. Identify Zeros: Identify the zeros of the function (x+3)3(x−1)2(x−4)(x+3)^3(x-1)^2(x-4). Set each factor to zero: (x+3)=0(x+3) = 0 gives x=−3x = -3. (x−1)=0(x-1) = 0 gives x=1x = 1. (x−4)=0(x-4) = 0 gives x=4x = 4. Critical points: −3-3, 11, 44.
  2. Determine Intervals: Determine the intervals to test based on the critical points.\newlineThe critical points divide the number line into four intervals: (−∞,−3)(-\infty, -3), (−3,1)(-3, 1), (1,4)(1, 4), (4,∞)(4, \infty).
  3. Test Sign: Test the sign of (x+3)3(x−1)2(x−4)(x+3)^3(x-1)^2(x-4) in each interval.\newlineFor x < -3, choose x=−4x = -4: (−4+3)3(−4−1)2(−4−4)=(−1)3(−5)2(−8)=−1⋅25⋅(−8)=200(-4+3)^3(-4-1)^2(-4-4) = (-1)^3(-5)^2(-8) = -1\cdot25\cdot(-8) = 200 (positive).\newlineFor -3 < x < 1, choose x=0x = 0: (0+3)3(0−1)2(0−4)=33(−1)2(−4)=27⋅1⋅(−4)=−108(0+3)^3(0-1)^2(0-4) = 3^3(-1)^2(-4) = 27\cdot1\cdot(-4) = -108 (negative).\newlineFor 1 < x < 4, choose x=2x = 2: (2+3)3(2−1)2(2−4)=53(1)2(−2)=125⋅1⋅(−2)=−250(2+3)^3(2-1)^2(2-4) = 5^3(1)^2(-2) = 125\cdot1\cdot(-2) = -250 (negative).\newlineFor x < -300, choose x < -311: x < -322 (positive).
  4. Combine Results: Combine the results to form the solution.\newline(x+3)3(x−1)2(x−4)≥0(x+3)^3(x-1)^2(x-4) \geq 0 is satisfied when the expression is positive or zero.\newlineThe intervals where this occurs are (−∞,−3](-\infty, -3], [1,4][1, 4], and [4,∞)[4, \infty).

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