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W(x)= x⁵+(k³+3k²)x³-2(k²+2k)x-k
W(1) ≤ -2
(1)⁵+(k³+3k²)*(1)³-2(k²+2k)*(1)-k ≤ -2
1 + k³+3k² - 2k² - 4k - k ≤-2
k³ + k² - 5k + 3 ≤ 0
(k - 1) * (k² + 2k - 3) ≤ 0
(k - 1) * (k + 3) * (k - 1) ≤ 0
(k - 1)² * (k + 3) ≤ 0
Zatem k ≤ - 3 lub k = 1.
R(x) = W(x)
x = 1, bo mamy dany dwumian (x-1)
Zatem:
W(1) = 1⁵+(k³+3k²)1³-2(k²+2k)1-k = 1+k³+3k²-2k²-4k-k = k³+k²-5k+1
W(1) ≤ -2
=> k³+k²-5k+3 ≤ 0
(k-1)(k-1)(k+3) ≤ 0
=> k ∈ (-₀₀, -3>