How to factor $(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$ to get $\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$?

Question

How can I factor this: $$\left(1+\frac{1}{x}\right)\left(-\frac{6}{x^2}\right)+\left(\frac{6}{x^3}\right)\left(1+\frac{1}{x}\right)^2$$

in order to get this result: $$\frac{6}{x^3}\left(1+\frac{1}{x}\right)\left(1+\frac{2}{x}\right)$$

I have tried to factor out the $(1+\frac{1}{x})$ use the distributive property, and failed multiple times. Every online calculator I tried gives me different results.

Answer 1

My first thought was to factor out the $\frac 6{x^2}$ but yours was to factor out the $1 + \frac 1x$. So we'll do it your way:

Factor out the $1 + \frac 1x$ and we get

$(1 + \frac 1x) (\frac {-6}{x^2} + \frac 6{x^3}(1 + \frac 1x))$.

Now factor out the $\frac 6{x^2}$ to get:

$\frac 6{x^2}(1 + \frac 1x) (-1 + \frac 1x(1 + \frac 1x))$

$\frac 6{x^2}(1 + \frac 1x)(-1 + \frac 1x + \frac 1{x^2})$

now, I can struggle with trying to factor $(-1+\frac 1x + \frac 1{x^2})$[1] but I must point out that if you claimed factorization and my demonstrated factorization were both true we'd have $-1 + \frac 1x + \frac 1{x^2} = \frac 1x(1+\frac 2x)$. WHich it doesn't.

Are you sure the question wasn't to solve for the $x$ where it does factor to that? In which case we'd solve for $-1 + \frac 1x + \frac 1{x^2} = \frac 1x(1+\frac 2x)$ or $-1 = \frac 1{x^2}$ which is $x =\pm i$ if we are allowed to use complex numbers... So... I guess that wasn't the question.

[1] Long story short; by the rational roots theorem we can't.

Answer 2

Hint

$$-\frac{6}{x^2}\left(1+\frac{1}{x}\right)+\frac{6}{x^3}\left(1+\frac{1}{x}\right)^2=\frac{6}{x^2}\left(1+\frac 1x\right)\left(\frac 1x+\frac 1{x^2}-1\right).$$

Answer 3

Eqn 1 $$(1+\frac{1}{x})(-\frac{6}{x^2})+(\frac{6}{x^3})(1+\frac{1}{x})^2$$ $$(-\frac{6}{x^2})+(-\frac{6}{x^3})+(\frac{6}{x^3})(1+\frac{1}{x^2}+\frac{2}{x})$$ $$(-\frac{6}{x^2})+ (-\frac{6}{x^3}) + \frac{6}{x^3}+\frac{6}{x^5}+\frac{12}{x^4}$$

$$\frac6{x^2}[\frac{1}{x^3}+\frac{2}{x^2}-1]$$

Eqn 2 $$\frac{6}{x^3}(1+\frac{1}{x})(1+\frac{2}{x})$$ $$(\frac{6}{x^3}+\frac{6}{x^4})(1+\frac2{x})$$ $$\frac6{x^3}[1+\frac3{x} +\frac2{x^2}]$$

On comparision we know that both the equations are different and there is is no way to convert Eqn 1 to Eqn 2.