How do you solve #2x^3 + x^2 - 5x + 2 < 0#?

1 Answer
Oct 1, 2015

Answer:

#2x^3+x^2-5x+2 = (x+2)(2x-1)(x-1)#

Hence #2x^3+x^2-5x+2 < 0# when #x in (-oo, -2) uu (1/2, 1)#

Explanation:

Let #f(x) = 2x^3+x^2-5x+2#.

By the rational root theorem, any rational roots of #f(x) = 0# must be of the form #p/q# in lowest terms, where #p, q in ZZ#, #q != 0#, #p# a divisor of the constant term #2# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational roots are #+-1/2#, #+-1# and #+-2#.

We find #f(-2) = f(1/2) = f(1) = 0#, so #-2#, #1/2# and #1# are the three roots.

#f(x)# can potentially change sign at each of these roots, and will do since none of them is a repeated root.

#f(x) = (x+2)(2x-1)(x-1)#

So when #x < -2#, the signs of the three factors are #-#, #-# and #-#, so their product #f(x) < 0#.

When #-2 < x < 1/2#, the signs of the three factors are #+#, #-# and #-#, so #f(x) > 0#.

When #1/2 < x < 1#, the signs of the factors are #+#, #+# and #-#, so #f(x) < 0#.

When #1 < x#, the signs of the factors are #+#, #+# and #+#, so #f(x) > 0#.

So we find #f(x) < 0# when #x in (-oo, -2) uu (1/2, 1)#