How do you simplify #(x^4-256)/(x-4)#?

Question

Eva Abramson · Accepted Answer

Hi, I propose another answer. 1) First, you use this famous formula : #x^4-y^4 = (x-y)(x^3+x^2y + xy^2+y^3)# You can prove that if you expand the second member. Take now #y=4#. Because #4^4=256#, you get : #x^4-256 = (x-4)(x^3+4x^2 + 16x+64)#. If #x
e 4#, you can divide by #x-4# and  #\frac{x^4-256}{x-4} = x^3+4x^2+16x+64#. 2) If you know complex numbers, you can have a better factorization. The equation #x^4=256# has 4 solutions in  #\mathbb{C} : 4,4i,-4i,-4# Then, you can write, for all  #x\in \mathbb{C}#, #x^4-256 = (x-4)(x-4i)(x+4i)(x+4)# and #\frac{x^4-256}{x-4} =(x+4)(x-4i)(x+4i)#. If you want a real factorization, write  #(x-4i)(x+4i) = (x^2+16)#. Conclusion #\frac{x^4-256}{x-4} =(x+4)(x^2+16)#. 3) If you don't know complex numbers, no stress!  Remark that  #x^3+4x^2+16x+64# has a easy root : #x=-4# then #x^3+4x^2+16x+64 = (x+4)(ax^2+bx+c)# Develop and find that #a=1#, #b=0# and #c=16#.  You find again  #\frac{x^4-256}{x-4} =(x+4)(x^2+16)#.

Ethan Adkins · Answer

undefined To simplify the expression (x^4-256)/(x-4), we can use the difference of squares formula. This formula states that a^2 - b^2 can be factored as (a + b)(a - b). In this case, we have x^4 - 256, which can be written as (x^2 + 16)(x^2 - 16). The second factor, x^2 - 16, is also a difference of squares and can be further factored as (x + 4)(x - 4). Therefore, the simplified expression is (x^2 + 16)(x + 4).