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How do you solve $2 x^{3} - 5 x^{2} < 2 x - 5$ $2x^3-5x^2<2x-5$ ?

1 Answer

A. S. Adikesavan

Aug 12, 2016

$x \in (- 1, 1)$ $x in (-1, 1)$ .

Explanation:

$x^{2} (2 x - 5) < (2 x - 5)$ $x^2(2x-5)<(2x-5)$

To preserve the inequality sign, 2x-5 is not 0. Canceling this

factor, $x^{2} < 1$ $x^2<1$ . So, $| x | < 1 .$ $|x|<1.$ .

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