What’s is the fully factored form of #3x^3-2x^2-7x-2# ?

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Answer 1 · 2018-03-09T18:54:33

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

#3x^3-2x^2-7x-2#

=#3x^3+3x^2-5x^2-5x-2x-2#

=#3x^2*(x+1)-5x*(x+1)-2*(x+1)#

=#(x+1)*(3x^2-5x-2)#

=#(x+1)(3x^2-6x+x-2)#

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

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

Answer 2 · 2018-03-09T19:03:59

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

It is obvious that -1 is a root of #3x^3-2x^2-7x-2#:

#3(-1)^3-2(-1)^2-7(1)-2= 0#

Therefore, #(x+1)# is a factor.

Either synthetic or long division of #(3x^3-2x^2-7x-2)/(x+1)# gives us the quadratic:

#(3x^2-5x-2)#

2 is obviously a root of the quadratic, therefore, #(x-2)# must be a factor.

#(3x^2-5x-2)= (x-2)(?x" ?")#

The only other factor must have 3 for the coefficient of x and +1 the other term:

#(3x+1)#

The factorization is:

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