How do you solve #24x^3+18x^2-168x = 0#?

1 Answer
Truong-Son N.
Jun 11, 2015

First off, notice that you can divide out #x#. That means one of the solutions is #x = 0#.

#0 = 24x^2 + 18x - 168#

Now you can divide by 3.

#0 = 8x^2 + 6x - 56#

You can't really factor it into integers though. The factors of 56 include 1, 2, 4, 7, 8, 14, 28, and 56. The ones that could work with #8x# and #x# to give #6# are #7# with #8# or #4# with #14#; the rest are too far apart. Regardless, neither of those give #6x# in the end. So:

#x = [-b pm sqrt(b^2 - 4ac)]/(2a)#

#= [-6 pm sqrt(36 - 4*8*-56)]/(16)#

#= [-6 pm sqrt(1828)]/(16)#

#= [-6 pm 2sqrt(457)]/(16)#

Overall:

#x = 0#
#x ~~ 2.3125#
#x ~~-3.0625#

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