How do you solve #2x ^ { 3} - x ^ { 2} - 6x < 0#?

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Answer 1 · 2017-09-28T16:31:08

#(-oo, -3/2) U (0,2)#

Given: Solve #2x^3 - x^2 - 6x < 0#

Pull out the greatest common factor (GCF): #x(2x^2 - x - 6) < 0#

Now factor the rest of the expression: #x(2x +3)(x-2) < 0#

The function #f(x)=2x^3 - x^2 - 6x =0# has three #x#-intercepts:

#x = 0, x = -3/2, x = 2#

To solve we need to set up intervals and test a value of #x# in the interval to see if it is < 0:

Intervals: #(-oo, -3/2), (-3/2, 0), (0, 2), (2, oo)#
test value: #x = -2, " " x = -1, " " x = 1, " " x = 3#

#f(-2) = 2(-2)^3 - (-2)^2 - 6(-2) = -8 < 0#

#f(-1) = 2(-1)^3 - (-1)^2 - 6(-1) = 3 > 0#

#f(1) = 2(1)^3 - (1)^2 - 6(1) = -5 < 0#

#f(3) = 2(3)^3 - (3)^2 - 6(3) = 27 > 0#

The intervals that are < 0 are:#(-oo, -3/2) U (0, 2)#