Question 628683
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:<ul><li>The other roots are a pair of complex conjugates</li><li>One or more of the three given roots are multiple roots.</li></ul>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.<br>
I am going to assume that explanation for the "missing" roots is the second one.<br>
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.<br>
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