Many ways to do this. I'll be going with "by inspection"
let f(x) = 3x^3 + 2x^2 - 19x + 6
If 3x - 1 is a factor, then it can be assumed that:
(3x - 1)(x^2 + Ax - 6) = 3x^3 + 2x^2 - 19x + 6 = f(x)
If you're wondering how I got that, multiply 3x by x^2 to get the 3x^3 term. Multiply -1 by - 6 to get the +6 term. In the MIDDLE, the Ax term represents an unknown because we don't yet know what value of A will give us the two middle terms, 2x^2 and -19x.
HOWEVER, we now expand out:
let q(x) = f(x) = (3x - 1)(x^2 + Ax - 6) = 3x^3 + 3Ax^2 - 18x - x^2 - Ax + 6
= 3x^3 + (3A - 1)x^2 - (A + 18)x + 6
BUT we know that q(x) = f(x)
Therefore, 3x^3 + (3A - 1)x^2 - (A + 18)x + 6 = 3x^3 + 2x^2 - 19x + 6
equation co-efficients: 3 = 3, 3A - 1 = 2, -(A + 18) = -19, and 6 = 6
See? We obtain the correctly found co-efficients from before.
To solve the UNKNOWN 'A' co-efficient, set 3A - 1 = 2 => 3A = 3 and therefore, A = 1. You can check that A = 1 by solving equation 2: -(A + 18) = -19, therefore A + 18 = 19 => A = 1
so if A = 1 you get f(x) = q(x) = (3x-1)(x^2 + x - 6)
Using cross method: x 3
x -2
therefore f(x) = (3x - 1)(x + 3)(x - 2) = 3x^3 + 2x^2 - 19x + 6
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