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

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Answer 1 · 2015-02-14T12:36:54

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\ne 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)#.