P(x)=6x3+5x2−2x−1
=6x3+6x2−x2−x−x−1
=6x2(x+1)−x(x+1)−(x+1)
=(x+1)(6x2−x−1) Let 6x2−x−1=0.
D=(−1)2−4(6)(−1)=25
x=2(6)1±25=121±5
x1=121−5=−31,x2=121+5=21 Then
6x2−x−1=6(x+31)(x−21)
=(3x+1)(2x−2)
P(x)=6x3+5x2−2x−1
=(x+1)(3x+1)(2x−2) Zeros: −1,−31,21
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