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

Feb 14, 2015

Hi,

1) First, you use this famous formula :

${x}^{4} - {y}^{4} = \left(x - y\right) \left({x}^{3} + {x}^{2} y + x {y}^{2} + {y}^{3}\right)$

You can prove that if you expand the second member.

Take now $y = 4$. Because ${4}^{4} = 256$, you get :

${x}^{4} - 256 = \left(x - 4\right) \left({x}^{3} + 4 {x}^{2} + 16 x + 64\right)$.

If $x \setminus \ne 4$, you can divide by $x - 4$ and

$\setminus \frac{{x}^{4} - 256}{x - 4} = {x}^{3} + 4 {x}^{2} + 16 x + 64$.

2) If you know complex numbers, you can have a better factorization.

The equation ${x}^{4} = 256$ has 4 solutions in

$\setminus m a t h \boldsymbol{C} : 4 , 4 i , - 4 i , - 4$

Then, you can write, for all

$x \setminus \in \setminus m a t h \boldsymbol{C}$,

${x}^{4} - 256 = \left(x - 4\right) \left(x - 4 i\right) \left(x + 4 i\right) \left(x + 4\right)$

and $\setminus \frac{{x}^{4} - 256}{x - 4} = \left(x + 4\right) \left(x - 4 i\right) \left(x + 4 i\right)$.

If you want a real factorization, write

$\left(x - 4 i\right) \left(x + 4 i\right) = \left({x}^{2} + 16\right)$.

Conclusion $\setminus \frac{{x}^{4} - 256}{x - 4} = \left(x + 4\right) \left({x}^{2} + 16\right)$.

3) If you don't know complex numbers, no stress!

Remark that

${x}^{3} + 4 {x}^{2} + 16 x + 64$ has a easy root : $x = - 4$

then ${x}^{3} + 4 {x}^{2} + 16 x + 64 = \left(x + 4\right) \left(a {x}^{2} + b x + c\right)$

Develop and find that $a = 1$, $b = 0$ and $c = 16$.

You find again

$\setminus \frac{{x}^{4} - 256}{x - 4} = \left(x + 4\right) \left({x}^{2} + 16\right)$.