How do you solve #x^3 - 4x^2 - 11x + 2 = 0#?

1 Answer
George C.
May 13, 2016

#x = -2# or #x = 3+-2sqrt(2)#

Explanation:

#f(x) = x^3-4x^2-11x+2#

By the rational root theorem, any rational zeros of #f(x)# will be expressible in the form #p/q# for integers #p# and #q# with #p# a divisor of the constant term #2# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-2#

We find:

#f(-2) = -8-16+22+2 = 0#

So #x=-2# is a zero and #(x+2)# a factor:

#x^3-4x^2-11x+2 = (x+2)(x^2-6x+1)#

We can factor #x^2-6x+1# by completing the square:

#x^2-6x+1#

#=(x-3)^2-9+1#

#=(x-3)^2-8#

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

#=((x-3)-2sqrt(2))((x-3)+2sqrt(2))#

#=(x-3-2sqrt(2))(x-3+2sqrt(2))#

Hence:

#x = 3+-2sqrt(2)#

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