How do you find all of the real zeros of #f(x)=x^3-2x^2-6x+12# and identify each zero as rational or irrational?

1 Answer
Nov 1, 2016

Answer:

The zeros of #f(x)# are:

#x = +-sqrt(6)" "# (irrational)

#x = 2" "# (rational)

Explanation:

The difference of squares identity can be written:

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

We use this with #a=x# and #b=sqrt(6)# later.

Given:

#f(x) = x^3-2x^2-6x+12#

Note that the ratio between the first and second terms is the same as that between the third and fourth terms. So this cubic will factor by grouping:

#x^3-2x^2-6x+12 = (x^3-2x^2)-(6x-12)#

#color(white)(x^3-2x^2-6x+12) = x^2(x-2)-6(x-2)#

#color(white)(x^3-2x^2-6x+12) = (x^2-6)(x-2)#

#color(white)(x^3-2x^2-6x+12) = (x^2-(sqrt(6))^2)(x-2)#

#color(white)(x^3-2x^2-6x+12) = (x-sqrt(6))(x+sqrt(6))(x-2)#

Hence the zeros of #f(x)# are:

#x = +-sqrt(6)" "# (irrational)

#x = 2" "# (rational)