How do you factor #y= 6x^3+13x^2-14x+3# ?

1 Answer
Alan P. · Aviv S.
Mar 21, 2018

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

Explanation:

We can optimistically hope that #y=color(lime)6x^3+13x^2-14x+color(magenta)3#
has at least one rational root.

By the Rational Root Theorem we know that any such root must be of the form #color(blue)(p/q)#
where #color(blue)p# is an integer factor of #color(magenta)3# (for this expression)
#color(white)("XXX")color(blue)p in {+-1,+-3}#
and
#color(blue)q# is an integer factor of #color(lime)6# (again, for this expression)
#color(white)("XXX")color(blue)q in {+-1,+-2,+-3,+-6}#

There turn out to only be 10 possible rational roots using this information:
#color(white)("XXX")color(blue)(p/q) in {+-3,+-1,+-1/2,+-1/3,+-1/6}#

You could evaluate these manually, but I used a spreadsheet:
enter image source here

In this case, we have been extremely lucky, since we have all 3 possible roots identified #{0.5=1/2, 0.333bar3=1/3, -3}#

Which implies the factors:
#color(white)("XXX")color(lime)6 * (x-1/2) * (x-1/3) * (x+3)#
or (after using the #color(lime)6# to clear the fractions)
#color(white)("XXX")(2x-1)(3x-1)(x+3)#

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