# 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)?

Apr 21, 2016

$P \left(x\right) = {x}^{5} + {x}^{4} - 5 {x}^{3} + 3 {x}^{2}$

#### Explanation:

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

$P \left(x\right) = {x}^{2} {\left(x - 1\right)}^{2} \left(x + 3\right)$

$= {x}^{2} \left({x}^{2} - 2 x + 1\right) \left(x + 3\right)$

$= {x}^{5} + {x}^{4} - 5 {x}^{3} + 3 {x}^{2}$

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

Footnote

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