In the following problem is (x-1) a factor of the polynomial explain why or why not using the remainder theorem?

Step-by-step explanation:

Suppose the polynomial is of the form

P (x) = (x - 1) * Something. It does not matter unless it contains another (x - 1). All that matters is that something has x's in it.

So if you put 1 in for x, you get 1 - 1 = 0 and there cannot be any remainder because x - 1 is 0 when 1 is put in P (x).

Suppose when you put 1 into whatever something is and you get 14. That is not a remainder because 0 * 14 is still zero.

If x - 1 is not a factor of P (x) and it is not in something, you will get a remainder. x-1 is then not a factor of the polynomial and cannot be divided by x - 1 evenly.