How do you find a polynomial function that has zeros $x=-5, 1, 2$ and degree n=4?

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Answer 1 · 2018-08-10T21:23:08

$f(x) = (x+5)^2(x-1)(x-2) = x^4+7x^3-3x^2-55x+50$

If a polynomial function $f(x)$ has zeros $x = -5$ , $x=1$ and $x=2$ , then it has factors $(x+5)$ , $(x-1)$ and $(x-2)$ .

If these were its only factors, then it would be a cubic.

It is not clear from the question whether $-5$ , $1$ and $2$ are supposed to be the only zeros. If so then one of them must be of multiplicity $2$ .

In any case, a suitable quartic would be:

$f(x) = (x+5)^2(x-1)(x-2) = x^4+7x^3-3x^2-55x+50$

How do you find a polynomial function that has zeros x=-5, 1, 2x=-5, 1, 2 and degree n=4?