According to the rational root theorem, which answer is not a possible rational root of x³+6x²-x-10=0 ? 1.) 42.) 23.) 14.) 5

choice 1) 4



Explanation:

The equation x^3+6x^2-x-10 = 0 is the same as 1x^3+6x^2-x-10 = 0
The first term is 1x^3 and the last term is -10
Pull away the coefficient of the first term and it is 1

Focus on the first coefficient (1) and the last term (-10)

List out the factors of each:
Factors of 1: -1 and 1
Factors of -10: -10, -5, -2, -1, 1, 2, 5, 10

Divide the factors of the last term (-10) over the factors of the first coefficient (1) to get...
-10/(-1) = 10
-10/(1) = -10
-5/(-1) = 5
-5/(1) = -5
-2/(-1) = 2
-2/(1) = -2
-1/(-1) = 1
-1/(1) = -1

10/(-1) = -10
10/(1) = 10
5/(-1) = -5
5/(1) = 5
2/(-1) = -2
2/(1) = 2
1/(-1) = -1
1/(1) = 1

The unique results we get are: -1, 1, -2, 2, -5, 5, -10, 10

So those are the possible rational roots of x^3+6x^2-x-10 = 0
This rules out choice 2, choice 3, choice 4. The only thing left is choice 1. 
The value 4 is not a possible rational root of x^3+6x^2-x-10 = 0

So that's why choice 1 is the answer

40%=2/550=1/230=3/1056=14/25