Question

select the set of all possible solutions to the polynomial.
$x^3 + 3x + 2$

a) $left\\{-1, -\frac{1}{2}, \frac{1}{2}, 1\
ight\\}$
d) ${1, 3}$

b) $left\\{-\frac{1}{3}, -1, 1, \frac{1}{3}\
ight\\}$
e) ${-2, -1, 1, 2}$

c) ${1, 2}$
f) $left\\{\frac{1}{2}, 1\
ight\\}$

Explanation:

Step1: Apply Rational Root Theorem

Possible roots: $\pm1,\pm2$

Step2: Check root validity

$(-1)^3 + 3(-1) + 2 = -1 -3 +2 = -2
eq 0$; $(-2)^3 +3(-2)+2=-8-6+2=-12
eq0$; $1^3+3(1)+2=6
eq0$; $2^3+3(2)+2=16
eq0$
But the question asks for all possible solutions (rational candidates per theorem), so the set is $\{-2,-1,1,2\}$.

Answer:

e) $\{-2,-1,1,2\}$