Write a polynomial f(x) that satisfies the given conditions.

Question

Polynomial of lowest degree with zeros of -4 (multiplicity 1), 1 (multiplicity 2), and with f (0) = -12.

Answer 1

To construct a polynomial $f(x)$ that satisfies the given conditions, we start with the zeros and their respective multiplicities:

1. Zero at $x=-4$ with multiplicity 1.
2. 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:

$$\boxed{{\displaystyle -3x^3-6x^2+21x-12}}$$