Question

The polynomial of degree 5, P(x) has leading coefficient 1, has roots of multiplicity 2 at x=1 and x=0, and a root of multiplicity 1 at x=-3, how do you find a possible formula for P(x)?

Answer 1

`P(x) = x^5+x^4-5x^3+3x^2`

Explanation:

Each root corresponds to a linear factor, so we can write:

`P(x) = x^2(x-1)^2(x+3)`

`=x^2(x^2-2x+1)(x+3)`

`= x^5+x^4-5x^3+3x^2`

Any polynomial with these zeros and at least these multiplicities will be a multiple (scalar or polynomial) of this `P(x)`

Footnote

Strictly speaking, a value of `x` that results in `P(x) = 0` is called a root of `P(x) = 0` or a zero of `P(x)`. So the question should really have spoken about the zeros of `P(x)` or about the roots of `P(x) = 0`.