Question 812405
A general formula for a 3rd degree polynomial in factored form would be:
{{{f(x)=a(x-z[1])(x-z[2])(x-z[3])}}}
where the z's are zeros of the polynomial and the "a" is a non-zero constant. Using the given zeros we get:
{{{f(x)=a(x-(-3))(x-(-2))(x-2)}}}
which simplifies to:
{{{f(x)=a(x+3)(x+2)(x-2)}}}
Now we must multiply this out and find the right value for "a", either one first. First I will find the "a". For this we will use the fact that f(1) = -12:
{{{-12=a((1)+3)((1)+2)((1)-2) = 0}}}
Simplifying...
{{{-12=a(4)(3)(-1)}}}
{{{-12=-12a}}}
Dividing by -12:
{{{1 = a}}}
So:
{{{f(x)=1(x+3)(x+2)(x-2)}}}
Now we multiply this out. I'll start by using the {{{(a+b)(a-b)=a^2-b^2}}} pattern to multiply the last two factors:
{{{f(x)=(x+3)((x)^2-(2)^2)}}}
which simplifies to:
{{{f(x)=(x+3)(x^2-4)}}}
Using FOIL on what's left:
{{{f(x)=x^3-4x+3x^2-12}}}
Reordering into standard form:
{{{f(x)=x^3+3x^2-4x-12}}}