How do you factor completely #x^4-81#?

2 Answers
Aug 14, 2016

#(x^4 - 81) = (x^2+9)(x^2-9)#

#(x^2+9)(x^2-9) = (x^2+9)(x+3)(x-3)#

Aug 14, 2016

Answer:

#(x-3)(x+3)(x^2+9)#

Explanation:

This is a #color(blue)"difference of squares"# and, in general, factorises as follows.

#color(red)(|bar(ul(color(white)(a/a)color(black)(a^2-b^2=(a-b)(a+b))color(white)(a/a)|)))........ (A)#

here #(x^2)^2=x^4" and " (9)^2=81#

#rArra=x^2" and " b=9#

substituting into (A)

#rArrx^4-81=(x^2-9)(x^2+9)........ (B)#

Now, the factor #x^2-9 " is also a "color(blue)"difference of squares"#

#rArrx^2-9=(x-3)(x+3)#

substituting into (B) to complete the factorising.

#rArrx^4-81=(x-3)(x+3)(x^2+9)#