Question

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

\[
\boxed{-3x^3 - 6x^2 + 21x - 12}
\]