To construct a polynomial \( f(x) \) that satisfies the given conditions, we start with the zeros and their respective multiplicities:
- Zero at \( x = -4 \) with multiplicity 1.
- Zero at \( x = 1 \) with multiplicity 2.
The general form of the polynomial can be expressed based on its zeros:
\[ f(x) = k (x + 4)(x - 1)^2 \]
where \( k \) is a constant that we will determine using the condition \( f(0) = -12 \).
First, let's expand \( f(x) \):
\[ f(x) = k (x + 4)(x^2 - 2x + 1) \]
Now we further expand \( (x + 4)(x^2 - 2x + 1) \):
\[ f(x) = k [(x + 4)(x^2 - 2x + 1)] \] \[ = k [x^3 - 2x^2 + x + 4x^2 - 8x + 4] \] \[ = k [x^3 + 2x^2 - 7x + 4] \]
Thus, we have:
\[ f(x) = k (x^3 + 2x^2 - 7x + 4) \]
Next, we apply the condition \( f(0) = -12 \):
\[ f(0) = k (0^3 + 2(0)^2 - 7(0) + 4) = k \cdot 4 \]
Setting this equal to -12 gives:
\[ k \cdot 4 = -12 \]
Solving for \( k \):
\[ k = \frac{-12}{4} = -3 \]
Now substituting \( k \) back into our polynomial:
\[ f(x) = -3 (x^3 + 2x^2 - 7x + 4) \]
Expanding this gives:
\[ f(x) = -3x^3 - 6x^2 + 21x - 12 \]
Therefore, the polynomial of lowest degree that satisfies the given conditions is:
\[ \boxed{-3x^3 - 6x^2 + 21x - 12} \]