How do you solve #5/(x^3-x^2-4X+4) < 0#?

1 Answer
George C.
Oct 9, 2015

This reduces to finding out when #x^3-x^2-4x+4 < 0#, which happens when #x in (-oo, -2) uu (1, 2)#

Explanation:

#5/(x^3-x^2-4x+4) < 0# when #x^3-x^2-4x+4 < 0#,

so let's find where that happens...

Let #f(x) = x^3-x^2-4x+4#

We can factor #f(x)# by grouping...

#f(x) = x^3-x^2-4x+4#

#= (x^3-x^2)-(4x-4)#

#= x^2(x-1) - 4(x-1)#

#= (x^2-4)(x-1)#

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

The graph of #f(x)# will cross the #x# axis when #x = -2#, #x = 1# and #x = 2#. I say cross rather than intersect, since none of these roots of #f(x) = 0# are repeated.

Since the coefficient of #x^3# is positive, this means that #f(x) < 0# when #x in (-oo, -2) uu (1, 2)#

#x^3-x^2-4x+4# ...

graph{x^3-x^2-4x+4 [-9.92, 10.08, -3.12, 6.88]}

#5/(x^3-x^2-4x+4)# ...

graph{5/(x^3-x^2-4x+4) [-18.6, 17.45, -10.69, 7.33]}

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