This expression should be extremely easy to simplify, but for some reason I can't do it.
$$\frac{x^4-1}{x-1}$$
I know it simplifies down to this, but I don't know how to get there
$$x^3+x^2+x+1$$
This is a very basic question on my calculus worksheet, I would appreciate if anyone could explain how the first expression simplifies down to the second.
On
you have $(x^2-1)(x^2+1)/(x-1)$,so $(x^2-1)=(x-1)(x+1)$,if we divide it by $(x-1)$,we will have $(x+1)(x^2+1)=x^3+x+x^2+1$
On
To clarify Mark Dominus's comment, suppose that you don't know the expression on the right, but you want to simplify $\displaystyle{\frac{x^4-1}{x-1}}$ as much as possible by simplifying the numerator as much as possible. Well, you can't dive $x^4$ directly by $x-1$, but you know that $x^3\cdot(x-1)=x^4-x^3$, so add and subtract $x^3$: write the numerator as $x^4-1=(x^4-x^3)+(x^3-1)$ and your fraction as $\displaystyle{\frac{x^4-1}{x-1} = \frac{(x^4-x^3)+(x^3-1)}{x-1}=\frac{x^4-x^3}{x-1}+\frac{x^3-1}{x-1} = x^3+\frac{x^3-1}{x-1}}$. Now can you see how to keep going from here?
Write it as the difference of two squares $$\frac{(x^2-1)(x^2+1)}{(x-1)}$$ $$\frac{(x+1)(x-1)(x^2+1)}{(x-1)}$$ $$(x+1)(x^2+1)=x^3+x^2+x+1$$