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

Question

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

Savannah Anderson · Accepted 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)# 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)#

Abigail Allen · Answer

undefined To solve the inequality $2x^3 + x^2 - 5x + 2 < 0$, follow these steps:

1. Factor the polynomial if possible.
2. Identify the critical points by setting each factor equal to zero and solving for $x$.
3. Use test points in each interval determined by the critical points to determine the sign of the polynomial in each interval.
4. Determine the intervals where the polynomial is less than zero.

After solving, you'll get the intervals where $2x^3 + x^2 - 5x + 2 < 0$.

How do you solve #2x^3 + x^2 - 5x + 2 &lt; 0#?

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