How do you write a polynomial whose zeros are -9, multiplicity of 1, 4 multiplicity of 2 and degree 3?

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How do you write a polynomial whose zeros are -9, multiplicity of 1, 4 multiplicity of 2 and degree 3?

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Answer 1 · 2018-04-17T22:49:58

$f (x) = (x + 9) {(x - 4)}^{2}$ $f(x)=(x+9)(x-4)^2$

Explanation:

A zero of $- 9$ $-9$ with a multiplicity of $1$ $1$ tells us that $(x + 9)$ $(x+9)$ is a factor of our polynomial.

The multiplicity of a zero simply tells us how many times the factor involving that zero shows up. A zero of $4$ $4$ with a multiplicity of $2$ $2$ tells us that $(x - 4) (x - 4) = {(x - 4)}^{2}$ $(x-4)(x-4)=(x-4)^2$ is another factor of our polynomial. Since the multiplicity is $2$ $2$ , it must show up twice.

So, we obtain

$f (x) = (x + 9) {(x - 4)}^{2}$ $f(x)=(x+9)(x-4)^2$

This is indeed of degree $3$ $3$ -- if we multiplied everything out, the leading term would be $x^{3}$ $x^3$ , the term of the highest exponent in the polynomial, making it $3 r d$ $3rd$ degree.

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How do you write a polynomial whose zeros are -9, multiplicity of 1, 4 multiplicity of 2 and degree 3?

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