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Here is an example:

$\frac{x^2 - 3x + 2}{x + 1} -5 > 0$

My approach would be to factor, find the undefined areas and the zeros, and then pick some points in the intervals left to see what I find. I'm not really sure if that's the right way to go, and to find the zeros I would just throw an equals sign in instead of the inequality and find the roots. Is this the right approach, and are there any techniques to improve it?

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up vote 1 down vote accepted

This is not a general approach, but in this example $x=-1$ is 'special', so split the consideration into $I_- = (-\infty,-1)$ and $I_+ = (-1, +\infty)$.

On $I_+$, we have $x^2 - 3x + 2 - 5 (x + 1) = x^2-8x-3 > 0$. The factors are $4 \pm \sqrt{19}$, hence if $x \in I_+$, then $\frac{x^2 - 3x + 2}{x + 1} -5 > 0$ iff $x > 4 + \sqrt{19}$ or $x < 4 - \sqrt{19}$.

On $I_-$, we have $x^2 - 3x + 2 - 5 (x + 1) =x^2-8x-3 < 0$. From the previous reasoning, we have that if $x \in I_-$, then $\frac{x^2 - 3x + 2}{x + 1} -5 \leq 0$.

Hence $\frac{x^2 - 3x + 2}{x + 1} -5 > 0$ iff $ x \in (-1,4 - \sqrt{19}) \cup (4 + \sqrt{19}, +\infty)$.

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I would take the $5$ up into the rational function.

$$\frac{x^2-3x+2-5(x+1)}{x+1}>0$$ $$\frac{x^2-8x-3}{x+1}>0$$

Solving the quadratic on top gives


So we have


Now we have $3$ places where it can change sign: $-1$ and $4\pm \sqrt{19}$. For very large positive values this will clearly be positive, and it will flip signs (as each factor flips sign individually) as we move right to left along $\mathbb{R}$. This implies that it's positive on $(4+\sqrt{19},\infty)\cup (-1,4-\sqrt{19})$.

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Your approach will work, but the $-5$ makes some trouble. Factoring makes it easy to compare with $0$, but not so much with other numbers. My first try would be to incorporate the $5$ in the fraction, getting $$\frac {x^2-8x-3}{x+1} \gt 0$$ Unfortunately, the numerator doesn't factor easily, but I would just use the quadratic formula: $$\frac{(x-4-\sqrt{19})(x-4+\sqrt{19)}}{x+1}\gt 0$$ and the two roots and one asymptote are visible.

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