How do you factor #2x^3 + 7x^2 - 3x -18#?

1 Answer
George C.
May 10, 2015

#2x^3+7x^2-3x-18=(x+3)(x+2)(2x-3)#

The #2x^3# term quickly becomes dominant as #|x|# increases, so any solutions of #2x^3+7x^2-3x-18 = 0# will be quite small.

Integer solutions will be factors of 18, so try #x = -3, -2, -1, 0, 1, 2 or 3#. #x = -3# is a solution, so #(x + 3)# is a factor. #x = -2# is also a solution, so #(x+2)# is also a factor. If the remaining factor is of the form #(ax+b)# then looking at the coefficient of #x^3# (#2#) and the constant term (#18#) we must have #a=2# and #b=-3#.

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