How do you factor completely x^4-81x481?

2 Answers
Aug 14, 2016

(x^4 - 81) = (x^2+9)(x^2-9)(x481)=(x2+9)(x29)

(x^2+9)(x^2-9) = (x^2+9)(x+3)(x-3)(x2+9)(x29)=(x2+9)(x+3)(x3)

Aug 14, 2016

(x-3)(x+3)(x^2+9)(x3)(x+3)(x2+9)

Explanation:

This is a color(blue)"difference of squares"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)