How do you create a polynomial p which has zeros $c = 1, c = 3$ $c=1, c=3$ , $c = - 3$ $c=-3$ is a zero of multiplicity 2, the leading term is $- 5 x^{3}$ $-5x^3$ ?

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Answer 1 · 2017-08-06T07:21:02

$- 5 x^{3} + 5 x^{2} + 45 x - 45$ $-5x^3+5x^2+45x-45$

Note that $x$ $x$ is a zero of a polynomial if and only if $(x - c)$ $(x-c)$ is a factor of that polynomial.

So in order to have zeros $1$ $1$ , $3$ $3$ and $- 3$ $-3$ , our polynomial must be a multiple of:

$(x - 1) (x - 3) (x + 3) = (x - 1) (x^{2} - 9) = x^{3} - x^{2} - 9 x + 9$ $(x-1)(x-3)(x+3) = (x-1)(x^2-9) = x^3-x^2-9x+9$

In order that its leading term be $- 5 x^{3}$ $-5x^3$ , we just need to multiply it by $- 5$ $-5$ to get:

$- 5 (x^{3} - x^{2} - 9 x + 9) = - 5 x^{3} + 5 x^{2} + 45 x - 45$ $-5(x^3-x^2-9x+9) = -5x^3+5x^2+45x-45$

How do you create a polynomial p which has zeros c=1, c=3c=1,c=3c=1, c=3, c=-3c=−3c=-3 is a zero of multiplicity 2, the leading term is -5x^3−5x3-5x^3?