How do you factor #m^4 - n ^4#?

2 Answers
Jun 24, 2018

Answer:

#(m^2-n^2)(m^2+n^2)#

Explanation:

Remember the #a^2-b^2=(a-b)(a+b)#

Note that #m^4# is the same as #(m^2)^2# and the same for #n#. So we may write this as:
#color(white)("d")a^2color(white)("dd")-color(white)("d")b^2color(white)("dd") ->color(white)("dd")(a-b)(a+b)#

#(m^2)^2-(n^2)^2 ->(m^2-n^2)(m^2+n^2)#

Jun 24, 2018

Answer:

#(m^2+n^2)(m+n)(m-n)#

Explanation:

What we have is a difference of squares. Recall that if we have an expression

#color(steelblue)a^2-color(steelblue)b^2#

This has an expansion of

#color(purple)((a+b))color(steelblue)((a-b))#

Let's rewrite our expression as

#color(steelblue)((m^2))^2-color(steelblue)((n^2))^2#

Since we have a difference of squares, we can rewrite this as

#color(purple)((m^2+n^2))color(steelblue)((m^2-n^2))#

What we have in purple is a sum of squares, which can't be factored with real numbers, but we have another difference of squares in blue.

Factoring this, we get

#bar( ul( | color(white)(2/2) color(purple)((m^2+n^2))color(steelblue)((m+n)(m-n)) color(white)(2/2)| ))#

Hope this helps!