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=-1 Find a possible formula for P(x)?

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Answer 1 · 2018-02-08T23:04:49

$P(x) = x^2(x-1)^2(x+1)$

Given that we have a root of multiplicity $2$ $at x=1$ , we know that $P(x)$ has a factor $(x-1)^2$

Given that we have a root of multiplicity $2$ at $x=0$ , we know that $P(x)$ has a factor $x^2$

Given that we have a root of multiplicity $1$ at $x=-1$ , we know that $P(x)$ has a factor $x+1$

We are given that $P(x)$ is a polynomial of degree $5$ , and we have therefore identified all five roots, and factors, so we can write

$P(x) = 0 => x^2(x-1)^2(x+1) = 0$

And we can therefore write

$P(x) = Ax^2(x-1)^2(x+1)$

We also know that the leading coefficient is $1 => A=1$

Hence,

$P(x) = x^2(x-1)^2(x+1)$