DeterminantHard
Question
Let ᐃ = 
, then
, then Options
A.1 - x3 is a factor of ᐃ
B.(1 - x3)2 is factor of ᐃ
C.ᐃ(x) = 0 has 4 real roots
D.ᐃ′(1) = 0
Solution
ᐃ = 
, ᐃ = (1 + x + x2) 
= (1 + x + x2) {1(1 - x3) - x(1 - x) + x2(x2 - 1)}
= (1 + x + x2) {1 - x3 - x + x2 + x4 - x2)
= (1 + x + x2) {x4 - x3 - x + 1)
ᐃ = (1 - x3)2
ᐃ′= 2(1 - x3) (-3 x2)
ᐃ′(1) = 0
, ᐃ = (1 + x + x2) = (1 + x + x2) {1(1 - x3) - x(1 - x) + x2(x2 - 1)}
= (1 + x + x2) {1 - x3 - x + x2 + x4 - x2)
= (1 + x + x2) {x4 - x3 - x + 1)
ᐃ = (1 - x3)2
ᐃ′= 2(1 - x3) (-3 x2)
ᐃ′(1) = 0
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