# 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)?

Feb 8, 2018

$P \left(x\right) = {x}^{2} {\left(x - 1\right)}^{2} \left(x + 1\right)$

#### Explanation:

Given that we have a root of multiplicity $2$ $a t x = 1$, we know that $P \left(x\right)$ has a factor ${\left(x - 1\right)}^{2}$

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

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

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

$P \left(x\right) = 0 \implies {x}^{2} {\left(x - 1\right)}^{2} \left(x + 1\right) = 0$

And we can therefore write

$P \left(x\right) = A {x}^{2} {\left(x - 1\right)}^{2} \left(x + 1\right)$

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

Hence,

$P \left(x\right) = {x}^{2} {\left(x - 1\right)}^{2} \left(x + 1\right)$