 # How do I solve this, first I have to factor 2 x </mrow> x Kapalci 2022-06-24 Answered
How do I solve this, first I have to factor $\frac{2x}{x-1}+\frac{3x+1}{x-1}-\frac{1+9x+2{x}^{2}}{{x}^{2}-1}$
the solution is
$\frac{3x}{x+1}$
The only advance that I have done is factor ${x}^{2}-1$=$\left(x-1\right)$$\left(x+1\right)$
do not know how can I factor $1+9x+2{x}^{2}$ can someone please guide me in how to solve this exercise.
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The first two terms are already over a common denominator.
First simplify
$\frac{2x}{x-1}+\frac{3x+1}{x-1}-\frac{1+9x+2{x}^{2}}{{x}^{2}-1}=\frac{5x+1}{x-1}-\frac{1+9x+2{x}^{2}}{{x}^{2}-1}$
Then put the first fraction over the same denominator as the second
$\frac{\left(5x+1\right)\left(x+1\right)}{{x}^{2}-1}-\frac{1+9x+2{x}^{2}}{{x}^{2}-1}=\frac{\left(5x+1\right)\left(x+1\right)-\left(1+9x+2{x}^{2}\right)}{{x}^{2}-1}$
Now simplify the numerator

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Notice that
$\frac{A}{B}+\frac{{A}^{\prime }}{BC}=\frac{C}{C}\frac{A}{B}+\frac{{A}^{\prime }}{BC}=\frac{AC+{A}^{\prime }}{BC}$
Then
$\begin{array}{rl}\frac{\left(x+1\right)}{\left(x+1\right)}\frac{2x}{\left(x-1\right)}+& \frac{\left(x+1\right)}{\left(x+1\right)}\frac{3x+1}{\left(x-1\right)}+\frac{1+9x+2{x}^{2}}{\left(x-1\right)\left(x+1\right)}\\ & =\frac{2x\left(x+1\right)+\left(3x+1\right)\left(x+1\right)-\left(1+9x+2{x}^{2}\right)}{\left(x-1\right)\left(x+1\right)}\\ & =\frac{2{x}^{2}+2x+3{x}^{2}+3x+x+1-1-9x-2{x}^{2}}{\left(x-1\right)\left(x+1\right)}\\ & =\frac{3{x}^{2}-3x}{\left(x-1\right)\left(x+1\right)}\end{array}$

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