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How do you write a cubic polynomial function with zeros -3, 2, and 1?

1 Answer

Nov 24, 2016

$f (x) = (x + 3) (x - 2) (x - 1) = x^{3} - 7 x + 6$ $f(x) = (x+3)(x-2)(x-1) = x^3-7x+6$

Explanation:

Multiply together linear factors with each of these zeros:

$f (x) = (x + 3) (x - 2) (x - 1) = x^{3} - 7 x + 6$ $f(x) = (x+3)(x-2)(x-1) = x^3-7x+6$

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

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