Question

\frac{x^4-3x^3+2x^2}{30-x^2-x}<0

Answer 1

$\dfrac{x^4-3x^3+2x^2}{30-x^2-x}<0\ \ \ \to \ \ \ \ \dfrac{x^2\, (x^2-3x+2)}{-(x^2+x-30)}<0\ ,\\\\\\x^2-3x+2=0\ \ \to \ \ x_1=1\ ,\ x_2=2\ \ (teorema\ Vieta)\\\\x^2+x-30=0\ \ \to \ \ \ x_1=-6\ ,\ x_2=5\ \ (teorema\ Vieta)\\\\\\\dfrac{x^2\, (x-1)(x-2)}{-(x+6)(x-5)}<0\ \ ,\ \ \dfrac{x^2(x-1)(x-2)}{(x+6)(x-5)}>0\\\\\\znaki:\ \ \ +++(-6)---(0)---(1)+++(2)---(5)+++\\\\x\in (-\infty ;-6\, )\cup (\, 1\, ;\, 2\, )\cup (\, 5\, ;+\infty )$