How do you simplify #(x^4-256)/(x-4)#?
1 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
#x^4-256 = (x-4)(x^3+4x^2 + 16x+64)# .
If
#\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
#\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
You find again
#\frac{x^4-256}{x-4} =(x+4)(x^2+16)# .
