Question 661682
One answer is
{{{(x+2)^2(x-4)=x^3-12x-16}}}
You could multiply the whole thing by any non-zero number to get another solution.
 
To have {{{z}}} for a zero with multiplicity {{{m}}},
a polynomial must have {{{(x-z)}}} as a factor {{{m}}} times.
An example would be {{{(x-z)^m}}}, but we can multiply that by any factor,
other than zero, and {{{z}}} would still be a zero with multiplicity {{{m}}}.
 
To get {{{-2}}} as a zero, you need the factor{{{(x-(-2))=(x+2)}}}
To get {{{-2}}} as a zero with multiplicicty {{{2}}}, you need to include as a factor
{{{(x+2)(x+2)=(x+2)^2}}}
 
To get {{{4}}} as a zero, you need the factor{{{(x-4)}}}
 
To have both,
{{{-2}}} as a zero with multiplicicty {{{2}}},
and {{{4}}} as a zero with multiplicicty {{{1}}},
you need to include {{{(x+2)^2(x-4)}}} as a factor.
 
Since {{{(x+2)^2(x-4)=x^3-12x-16}}}
is a polynomial of degree 3, you cannot include any other factors with the variable {{{x}}},
but you can multiply that polynomial by any real number other than zero,
and it will still be a polynomial of degree 3, with the required zeros.