Question 716099
A polynomial function with zeros {{{z[1]}}}, {{{z[2]}}}, {{{z[3]}}}, etc. will have a factored form of:
{{{f(x) = a(x - z[1])(x - z[2])(x - z[3])}}}...
where "a" is any non-zero number.<br>
For a polynomial with the given zeros, its factored form will be:
{{{f(x) = a(x-1)(x-(-1))(x-(-2))}}}
Since we can pick any non-zero number for "a", there will be a different f(x) for every possible "a". So there is not just one function with the given zeros. To find one such function we will pick a value for "a". To make things easy, we'll make "a" be 1:
{{{f(x) = (x-1)(x-(-1))(x-(-2))}}}
All we have left to do is put the equation into the requested standard form. So we simplify:
{{{f(x) = (x-1)(x+1)(x+2)}}}
We can use the {{{(a+b)(a-b) = a^2-b^2}}} pattern to quickly multiply the first two factors:
{{{f(x) = (x^2-1)(x+2)}}}
Now we use FOIL to multiply the remaining factors:
{{{f(x) = x^2*x+x^2*2+(-1)*x+(-1)*2}}}
Continuing the simplifying...
{{{f(x) = x^3+2x^2+(-x)+(-2)}}}
This is a polynomial function, in standard form, with the given zeros.