In my opinion, I find that the best war to divide a cubic or a quadratic by a linear equation is by long division. The picture below shows how to do it for an example:
Basically, what you do is divide x³ by x to get x², and multiply the divisor by that to get x³+x², and take that away from x³+2x² to get x², and then move the -5x down and then repeat the process to get the final answer of x²+x-6 remainder 8.
Say there is a polynomial, f(x) and it is divided by the linear equation x-c. We get the linear equation (x-c) multiplied by the new equation (g(x)) plus a remainder r(x). But as the remainder is a constant, we can express is as 'r', giving us the equation f(x) = ((x-c)*g(x))+r. Now, if we set x equal to c we get f(c) = (c-c)*g(c)+r => f(c) = 0*g(c)+r => f(c) = r. Therefore we get the Remainder Theorem, that states that when a polynomial f(x) is divided by (x-c) the remainder is equal to c. For example, if the polynomial we used above, x³+2x²-5x+2 is divided by x+1 then the remainder equals f(-1) = (-1)³+2(-1)²-5(-1)+2 = -1+2+5+2 = 8, as shown above. This is easily the quicker way to find the remainder of a division without doing long division.
Following on from the remainder theorem, the factor theorem states that if when a polynomial f(x) is divided by x+c and f(c) is 0, then x+c must be a factor of f(x). The reverse is also true, if it is a facotr then f(c) must equal 0. This can be used in some questions, for example you may be asked to verify if a linear equation is a factor of a cubic, then replace x with the value of the constant with its sign changed and see if it equals 0. Then, you can divide by it to get a quadratic which can be factorised