Hi Peter,

Suppose the problem was to factor x² - 729. This I can do since 729 = 27² and thus x² - 729 = x² - 27² is a difference of squares and thus

x² - 729 = x² - 27² = (x - 27)(x + 27)

In particular this means that if x is an integer then the integer x² - 729 factors. For example if x = 2 then

x² - 729 = 2² - 729 = -725 = -25 × 29.

Now what about x² + 729? In this case if x = 2 then x² + 729 = 733 which is a prime and hence doesn't factor except as 1 × 733. Thus the only possible factorization of x² + 729 is 1 × (x² + 729) and hence we say that x² + 729 doesn't factor.

Harley

Peter wrote back

Question from peter, a student:

I,am having trouble factorising a polynomial x^6+729 into polynomial factors,
with real coefficients, where the factors are either linear or quadratic.
you have explained the process when x is x^2 what happens when x is x^3 or x^4
etc.

thanks Pete

Since 729 = 9³, if you write z = x² then

x⁶ + 729 = z³ + 9³

which is a sum of cubes. Sums of cubes do factor nicely

z³ + a³ = (z + a)(z² - az + a²)

Harley