How do you find a polynomial function that has zeros x=-3, 0, 1 and degree n=3?

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How do you find a polynomial function that has zeros x=-3, 0, 1 and degree n=3?

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Answer 1 · 2017-06-25T18:26:12

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

Explanation:

$given zeros x = - 3, x = 0, x = 1$ $"given zeros " x=-3,x=0,x=1$

$then factors are (x + 3), (x - 0), (x - 1)$ $"then factors are " (x+3),(x-0),(x-1)$

$\Rightarrow f (x) = x (x + 3) (x - 1)$ $rArrf(x)=x(x+3)(x-1)$

$\Rightarrow f (x) = x (x^{2} + 2 x - 3)$ $color(white)(rArrf(x))=x(x^2+2x-3)$

$\Rightarrow f (x) = x^{3} + 2 x^{2} - 3 x \leftarrow is a possible function$ $color(white)(rArrf(x))=x^3+2x^2-3xlarr" is a possible function"$
graph{x^3+2x^2-3x [-10, 10, -5, 5]}

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How do you find a polynomial function that has zeros x=-3, 0, 1 and degree n=3?

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