Question 78159
Factor completely:
{{{m^4-9n^4}}} Do you see this binomial as the difference of two squares? {{{(m^2)^2-(3n^2)^2}}}
The difference of two squares can be factored thus:
{{{A^2-B^2 = (A+B)(A-B)}}} Applying this to the given binomial, you get:
{{{m^4-9n^4 = (m^2+3n^2)(m^2-3n^2)}}} Now you'll notice that the second parentheses contains the difference of two squares {{{(m)^2-(sqrt(3)n)^2}}}which can be factored, as explained before:
{{{m^2-3n^2 = (m+sqrt(3)n)(m-sqrt(3)n)}}}
While the sum of two squares {{{m^2+3n^2}}} can be factored, the factors do entail the use of complex numbers.
{{{A^2+B^2 = (A+Bi)(A-Bi)}}}
So, if you want to factor "completely", you would get:
{{{m^4-9n^4 = (m+sqrt(3)ni)(m-sqrt(3)ni)(m+sqrt(3)n)(m-sqrt(3)n)}}} where:{{{i = sqrt(-1)}}}