How do you find a polynomial function of degree 6 with -1 as a zero of multiplicity 3, 0 as a zero of multiplicity 2, and 1 as a zero of multiplicity 1?

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How do you find a polynomial function of degree 6 with -1 as a zero of multiplicity 3, 0 as a zero of multiplicity 2, and 1 as a zero of multiplicity 1?

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Answer 1 · 2016-11-05T08:46:57

$f (x) = x^{6} + 2 x^{5} - 2 x^{3} - x^{2}$ $f(x) = x^6+2x^5-2x^3-x^2$

Explanation:

Each zero (e.g. $a$ $a$ ) corresponds to a linear factor (e.g. $(x - a)$ $(x-a)$ ).

Multiplicity corresponds to a repetition of that factor.

So in our example, the following polynomial fits the criteria:

$f (x) = {(x - (- 1))}^{3} {(x - 0)}^{2} (x - 1)$ $f(x) = (x-(-1))^3(x-0)^2(x-1)$

$f (x) = {(x + 1)}^{3} x^{2} (x - 1)$ $color(white)(f(x)) = (x+1)^3x^2(x-1)$

$f (x) = x^{2} {(x + 1)}^{2} (x - 1) (x + 1)$ $color(white)(f(x)) = x^2(x+1)^2(x-1)(x+1)$

$f (x) = x^{2} (x^{2} + 2 x + 1) (x^{2} - 1)$ $color(white)(f(x)) = x^2(x^2+2x+1)(x^2-1)$

$f (x) = x^{2} (x^{4} + 2 x^{3} - 2 x - 1)$ $color(white)(f(x)) = x^2(x^4+2x^3-2x-1)$

$f (x) = x^{6} + 2 x^{5} - 2 x^{3} - x^{2}$ $color(white)(f(x)) = x^6+2x^5-2x^3-x^2$

Any polynomial in $x$ $x$ with these zeros in these multiplicities will be a multiple (scalar or polynomial) of this $f (x)$ $f(x)$ .

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How do you find a polynomial function of degree 6 with -1 as a zero of multiplicity 3, 0 as a zero of multiplicity 2, and 1 as a zero of multiplicity 1?

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