Answer:
(x + 2) (x − 1) (x − 3)
Explanation:
The leading coefficient is 1, and the constant is 6. Using the rational root theorem, the possible real roots are:
±1, ±2, ±3, ±6
Trying them all:
(-1)³ − 2(-1)² − 5(-1) + 6 = 8
(1)³ − 2(1)² − 5(1) + 6 = 0
(-2)³ − 2(-2)² − 5(-2) + 6 = 0
(2)³ − 2(2)² − 5(2) + 6 = -4
(-3)³ − 2(-3)² − 5(-3) + 6 = -24
(3)³ − 2(3)² − 5(3) + 6 = 0
(-6)³ − 2(-6)² − 5(-6) + 6 = -252
(6)³ − 2(6)² − 5(6) + 6 = 120
So -2, 1, and 3 are all roots. Since the equation is cubic, there can only be three roots, so there are no irrational roots to find.
Therefore:
x³ − 2x² − 5x + 6 = (x + 2) (x − 1) (x − 3)