SOLUTION: How do you find a 5th degree polynomial in factored form with roots: at -2, 1, 3 only? I have absolutely no clue, please help. This is kind of urgent. Thank you very much.

Question 628683: How do you find a 5th degree polynomial in factored form with roots: at -2, 1, 3 only? I have absolutely no clue, please help. This is kind of urgent. Thank you very much.
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
A 5th degree polynomial will have 5 roots according to the Fundamental Theorem of Algebra. So there are two "missing" roots. The explanation for these "missing" roots could be:

The other roots are a pair of complex conjugates
One or more of the three given roots are multiple roots.

Either way, there is no way, given the information provided, to find one specific 5th degree polynomial that fits the description. But we can some up with several that fit.

I am going to assume that explanation for the "missing" roots is the second one.

If a polynomial has a root, let's call it r, then (x-r) will be a factor of that polynomial. So our factored polynomial will have factors of
(x - (-2)) or (x + 2)
(x - 1) and
(x - 3)
We still need two more factors.

If a polynomial more than one factor of (x-r) then that root is a multiple root (also called a root of multiplicity n where n is the number of times (x-r) is a factor. So we can "fill in" for the "missing" roots, by using multiple copies of one or more of the above 3 factors. For example:
P(x) = (x+2)(x+2)(x+2)(x-1)(x-3)
P(x) = (x+2)(x+2)(x-1)(x-1)(x-3)
P(x) = (x+2)(x-1)(x-1)(x-1)(x-3)
P(x) = (x+2)(x-1)(x-3)(x-3)(x-3)
etc.
As long as there are 5 factors of any combination of the three the polynomial will fit the description you provided. FWIW, you could also throw in a constant factor and it would still fit:
P(x) = 6(x+2)(x-1)(x-1)(x-3)(x-3)
P(x) = (-1/2)(x+2)(x+2)(x-1)(x-3)(x-3)
etc

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