How do you find all of the real zeros of f(x)=x3−2x2−6x+12 and identify each zero as rational or irrational?
1 Answer
Nov 1, 2016
The zeros of
x=±√6 (irrational)
x=2 (rational)
Explanation:
The difference of squares identity can be written:
a2−b2=(a−b)(a+b)
We use this with
Given:
f(x)=x3−2x2−6x+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:
x3−2x2−6x+12=(x3−2x2)−(6x−12)
x3−2x2−6x+12=x2(x−2)−6(x−2)
x3−2x2−6x+12=(x2−6)(x−2)
x3−2x2−6x+12=(x2−(√6)2)(x−2)
x3−2x2−6x+12=(x−√6)(x+√6)(x−2)
Hence the zeros of
x=±√6 (irrational)
x=2 (rational)
