someone please show or explain the easiest way to understand this 
Find the polynomial with leading coefficient 1 and degree 3 that 
has -1, 1, and 3 as roots. 
Possible answers: 
x³ - 3x² - x + 3
x³ - 3x² + x - 3
x³ + 3x² + x + 3
x³ + 3x² + x + 3 
Thanks to anyone who can help me

Two methods.  The first method works only for multiple choice 
tests, but not on a test if you have to show your work.

The first way is to plug them all three, -1, 1, and 3, in and 
see which one gives you 0 for all three numbers 

Plug x = -1 in the first answer:

   x³ - 3x² - x + 3
(-1)³ - 3(-1)² - (-1) + 3
    -1 - 3 + 1 + 3
           0

Plug x = 1 in the first answer:

   x³ - 3x² - x + 3
(1)³ - 3(1)² - (1) + 3
   1 - 3 - 1 + 3
         0
 
Plug x = 3 in the first answer:

   x³ - 3x² - x + 3
(3)³ - 3(3)² - (3) + 3
  27 - 27 - 1 + 3
         0

Wow, they all three gave 0 for the answer, so that's the 
correct choice!

However, on another test it might not have been the first 
one.

If you plug -1, 1, and 3 in the 2nd choice you get 
-8,-4, and 0. They aren't all 0, so that's not the
answer.

If you plug -1, 1, and 3 in the 3rd choice you get 
4, 8, and 60.  They aren't all 0, so that's not the 
answer.

Notice that the 4th choice is identical to the 3rd choice.

Now for method 2:
This is the real way to work the problem. 
Solve the problem in reverse.

We begin with what we would have ended up with if we 
had solved it:

x = -1, x = 1, and x = 3

Now ask: What would we have set = 0 to get those, if we had
solved it?

x + 1 = 0, x - 1 = 0, and x - 3 = 0

Next ask: Then what must the factored form of the polynomial 
equation have been to have given us that?

(x + 1)(x - 1)(x - 3) = 0

Next ask: What must the original polynomial equation have been?

Let's multiply the left side out and see

[(x + 1)(x - 1)](x - 3) = 0

[x² - x + x - 1)(x - 3) = 0

        (x² - 1)(x - 3) = 0

       x² - 3x² - x + 3 = 0

That's why it's the first choice.

Edwin