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

1 Answer
Aug 18, 2016

Answer:

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

Explanation:

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

By the rational roots theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p, 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#

Trying each in turn, we find:

#f(-2) = -8-4(4)+11(2)+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)#

Factor the remaining quadratic by completing the square and using the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

as follows:

#x^2-6x+1#

#=x^2-6x+9-8#

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

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

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

Putting it all together:

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