Answer:
-2, 3 & 4
Step-by-step explanation:
The integer roots of the polynomial function f(x)=x^3-5x^2-2x+24f(x)=x
3
−5x
2
−2x+24 can be only among the divisors of free term.
The divisors of free term are:
\pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 12, \pm 24.±1,±2,±3, ±4,±6,±8,±12,±24.
Check them:
1. f(1)=1^3-5\cdot 1^2-2\cdot 1+24=18\neq 0f(1)=1
3
−5⋅1
2
−2⋅1+24=18
=0 , 1 is not a root.
2. f(-1)=(-1)^3-5\cdot (-1)^2-2\cdot (-1)+24=20\neq 0f(−1)=(−1)
3
−5⋅(−1)
2
−2⋅(−1)+24=20
=0 , -1 is not a root.
3. f(2)=2^3-5\cdot 2^2-2\cdot 2+24=8\neq 0f(2)=2
3
−5⋅2
2
−2⋅2+24=8
=0 , 2 is not a root.
4. f(-2)=(-2)^3-5\cdot (-2)^2-2\cdot (-2)+24=0f(−2)=(−2)
3
−5⋅(−2)
2
−2⋅(−2)+24=0 , -2 is a root.
5. f(3)=3^3-5\cdot 3^2-2\cdot 3+24=0f(3)=3
3
−5⋅3
2
−2⋅3+24=0 , 3 is a root.
6. f(-3)=(-3)^3-5\cdot (-3)^2-2\cdot (-3)+24=-42\neq 0f(−3)=(−3)
3
−5⋅(−3)
2
−2⋅(−3)+24=−42
=0 , -3 is not a root.
7. f(4)=4^3-5\cdot 4^2-2\cdot 4+24=0f(4)=4
3
−5⋅4
2
−2⋅4+24=0 , 4 is a root.
The cubic function has at most 3 roots, then
Answer: roots of the function are -2, 3 and 4.
