How do you find all the zeros of f(x)=x3−1?

1 Answer
George C.
Aug 8, 2016

f(x) has zeros 1, ω=−12+√32i and ¯¯ω=−12−√32i

Explanation:

The difference of squares identity can be written:

a2−b2=(a−b)(a+b)

We use this below with a=(x+12) and b=√32i.

f(x)=x3−1

Note that f(1)=1−1=0 so x=1 is a zero and (x−1) a factor...

x3−1

=(x−1)(x2+x+1)

=(x−1)((x+12)2−14+1)

=(x−1)((x+12)2+34)

=(x−1)⎛⎝(x+12)2−(√32i)2⎞⎠

=(x−1)(x+12−√32i)(x+12+√32i)

Hence the other two zeros are:

x=−12±√32i

One of these is often denoted by the Greek letter ω

ω=−12+√32i

This is called the primitive Complex cube root of 1.

The other is ¯¯ω=ω2=−12−√32i

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