The ability to find k^{th} roots can also be used to factor some polynomials.
Example 1.5.4Factor the polynomial x^{3}− 27.
First find the cube roots of 27. By the above procedure using De Moivre’s theorem, these cube roots are
3,3
( √ -)
−-1+ i--3
2 2
, and 3
( √ -)
−-1− i--3
2 2
. Therefore, x^{3}− 27 =
( ( √- )) ( ( √-) )
− 1 -3- − 1 -3-
(x− 3) x − 3 2 + i 2 x − 3 2 − i2 .
Note also
( (−1 √3))
x − 3 2 + i2
( (−1 √3-))
x− 3 2 − i 2
= x^{2} + 3x + 9 and so
( )
x3 − 27 = (x − 3) x2 + 3x+ 9
where the quadratic polynomial x^{2} + 3x + 9 cannot be factored without using complex numbers.
Note that even though the polynomial x^{3}− 27 has all real coefficients, it has some complex zeros,
− 1
---
2
+ i
√3-
---
2
and
− 1
---
2
− i
√3
---
2
. These zeros are complex conjugates of each other. It is always
this way. You should show this is the case. To see how to do this, see Problems 17 and 18
below.
Another fact for your information is the fundamental theorem of algebra. This theorem says that any
polynomial of degree at least 1 having any complex coefficients always has a root in ℂ. This is sometimes
referred to by saying ℂ is algebraically complete. Gauss is usually credited with giving a proof of this
theorem in 1797 but many others worked on it and the first completely correct proof was due to Argand in
1806. For more on this theorem, you can google fundamental theorem of algebra and look at the
interesting Wikipedia article on it. Proofs of this theorem usually involve the use of techniques from
calculus even though it is really a result in algebra. A proof and plausibility explanation is given
later.