How do I solve this, first I have to factor $ 2x\over x-1$ + $ 3x +1\over x-1$ - $ 1 + 9x + 2x^2\over x^2-1$?

Question

I am doing calculus exercises but I'm in trouble with this

$$\frac{ 2x}{x-1} + \frac{3x +1}{ x-1} - \frac{1 + 9x + 2x^2}{x^2-1}$$

the solution is

$$ 3x\over x+1$$

The only advance that I have done is factor $ x^2-1$ = $( x-1)$ $ (x+1)$.

I do not know how can I factor $1 + 9x + 2x^2$, can someone please guide me in how to solve this exercise.

Answer 1

Notice that $$\frac{A}{B} + \frac{A'}{BC} = \frac{C}{C}\frac{A}{B} + \frac{A'}{BC} = \frac{AC + A'}{BC}$$

Then

$$\begin{align}\frac{(x+1)}{(x+1)}\frac{2x}{(x-1)}+&\frac{(x+1)}{(x+1)}\frac{3x+1}{(x-1)}+\frac{1 + 9x + 2x^2}{(x-1)(x+1)}\\&=\frac{2x(x+1) + (3x+1)(x+1) - (1+9x + 2x^2)}{(x-1)(x+1)} \\&= \frac{2x^2 + 2x + 3x^2+3x + x + 1 - 1-9x - 2x^2}{(x-1)(x+1)}\\&=\frac{3x^2 - 3x}{(x-1)(x+1)}\end{align}$$

Answer 2

Multiply the first two terms by $x+1$ in nominator and denominator. Then add all three terms and you obtain an expression $$\frac{f(x)}{(x-1)(x+1)}. $$ Now see how to factor $f(x)=3x^2-3x$.

Answer 3

The first two terms are already over a common denominator.

First simplify $$\frac {2x}{x-1}+\frac {3x+1}{x-1}-\frac {1+9x+2x^2}{x^2-1}=\frac {5x+1}{x-1}-\frac {1+9x+2x^2}{x^2-1}$$

Then put the first fraction over the same denominator as the second $$\frac {(5x+1)(x+1)}{x^2-1}-\frac {1+9x+2x^2}{x^2-1}=\frac {(5x+1)(x+1)-(1+9x+2x^2)}{x^2-1}$$

Now simplify the numerator