Question 241535
With -2 a zero of multiplicity 3, f(x) will be of the form {{{q(x - (-2))^3}}} or {{{q(x + 2)^3}}} where q is some constant. And we can use f(-1) to find q:
{{{4 = q((-1) + 2)^3}}}
{{{4 = q(1)^3}}}
{{{4 = q}}}
So {{{f(x) = 4(x+2)^3}}}. Since we are asked to find f(x) as a polynomial, we will need to multiply this out:
{{{f(x) = 4(x+2)^3}}}
{{{f(x) = 4(x+2)^2(x+2)}}}
Since {{{(x+2)^2 = x^2 + 4x + 4}}}:
{{{f(x) = 4(x^2 + 4x + 4)(x+2)}}}
{{{f(x) = 4(x^3 + 2x^2 +4x^2 +8x + 4x + 8)}}}
{{{f(x) = 4(x^3 +6x^2 + 12x + 8)}}}
{{{f(x) = 4x^3 + 24x^2 +48x + 32}}}