How to simplify a fraction like this one?

Question

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

Is there a rule for factorizing polynomials in the numerator?

Answer 1

"Simplification" may depend on each one, but

$$\frac{x^2-3x+1}{x-3}=\frac{x(x-3)+1}{x-3}=x+\frac1{x-3}$$

We did "not" simplify by factoring since the factoring of the numerator isn't nice at all in regards with this problem, as the numerator's roots are $\;\frac12\left(3\pm\sqrt5\right)\;$, yet for some use the rightmost expression could probably called "simpler" than the original one. For example, to differentiate the expression or to find the limit when $\;x\to\infty\;$ , etc.

Answer 2

alright, i will complete the squares and use the difference of squares to factor. there is no way you can solve the inequality without factoring.

$$x^2 - 3x + 1 = x^2 - 3x + \frac 94 - \frac 94 + 1=\left(x-\frac32\right)^2 -\left(\frac {\sqrt 5} 2\right)^2 = \left(x -\frac{3-\sqrt 5}2\right)\left(x -\frac{3+\sqrt 5}2\right) $$ therefore you want solve
$$\frac{ \left(x -\frac{3-\sqrt 5}2\right)\left(x -\frac{3+\sqrt 5}2\right)}{(x-3)} > 0 $$ has the three critical numbers $x = {(3-\sqrt5)}/2, 3, {(3-\sqrt5)}/2$ and switches across each. and you can verify by taking one test point in each of the region, if you must, that the solution is $${(3 -\sqrt 5)}/2 < x < 3, {(3 +\sqrt 5)}/2 < x < \infty. $$