Question #b89e4

1 Answer
Cem Sentin
Feb 21, 2018

#x_1=-1#, #x_2=1/2# and #x_3=2#

Explanation:

#2x^3-3x^2-3x+2#

=#2x^3+2-(3x^2+3x)#

=#2(x^3+1)-3x(x+1)#

=#2(x^2-x+1)(x+1)-3x(x+1)#

=#(x+1)*[2(x^2-x+1)-3x]#

=#(x+1)*(2x^2-5x+2)#

=#(x+1)(2x-1)(x-2)#

Hence, zeros of #f(x)# are #x_1=-1#, #x_2=1/2# and #x_3=2#

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