How do you solve #-2x^3-8x^2+6x+36#?

1 Answer
Jun 13, 2015

Answer:

#-2x^3-8x^2+6x+36# can be factored or its zeros found using a combination of the rational roots theorem and by recognising a perfect square trinomial factor.

Explanation:

#-2x^3-8x^2+6x+36#

#=-2(x^3+4x^2-3x-18)#

Let #f(x) = x^3+4x^2-3x-18#

By the rational roots theorem, any rational root of #f(x) = 0# of the form #p/q# in lowest terms must be such that #p# is a divisor of #18# and #q# is a divisor of #1#. That is:

#x = +-1#, #+-2#, #+-3#, #+-6#, #+-9# or #+-18#

Trying the first few:

#f(1) = 1+4-3-18 = -16#
#f(-1) = -1+4+3-18 = -12#
#f(2) = 8+16-6-18 = 0#

So #x=2# is a root of #f(x) = 0# and #(x-2)# is a factor of #f(x)#

#x^3+4x^2-3x-18 = (x-2)(x^2+6x+9)#

#x^2+6x+9# is recognisable as a perfect square trinomial:
It is of the form #a^2+2ab+b^2 = (a+b)^2# with #a=x# and #b=3#, so

#(x^2+6x+9) = (x+3)^2#

Putting this all together we get:

#-2x^3-8x^2+6x+36 = -2(x-2)(x+3)(x+3)#

So #-2x^3-8x^2+6x+36=0# has one root #x=2# and one repeated root #x=-3# with multiplicity #2#.