List of possible integral roots = 1, -1, 2, -2, 3, -3, 6, -6
List of corresponding remainders = 0, -16, -4, 0, 0, 96, 600, 1764
Check out the table below for a more organized way to represent the answer. The x values are the possible roots while the P(x) values are the corresponding remainders.
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Explanation:
We'll use the rational root theorem. This says that the factors of the last term divide over the factors of the first coefficient to get the list of all possible rational roots.
We'll be dividing factors of 6 over factors of 1. We'll do the plus and minus version of each. Since we're dividing over +1 or -1, this means that we're basically just looking at the plus minus of the factors of 6.
Those factors are: 1, -1, 2, -2, 3, -3, 6, -6
This is the list of possible integral roots.
Basically we list 1,2,3,6 with the negative versions of each value thrown in as well.
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From there, you plug each value into the P(x) function
If we plugged in x = 1, then,
P(x) = x^4 - 3x^3 - 3x^2 + 11x - 6
P(1) = (1)^4 - 3(1)^3 - 3(1)^2 + 11(1) - 6
P(1) = 1 - 3 - 3 + 11 - 6
P(1) = 0
This shows that x = 1 is a root, since we get a remainder 0. Do the same for the other possible rational roots listed above. You should find (through trial and error) that x = -2 and x = 3 are the other two roots.
