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)?

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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)?

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Answer 1 · 2016-04-21T20:53:35

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

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 $P(x) = 0$ .

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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)?

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