Find a polynomial of degree 3 with real coefficients and zeros of minus​3,minus​1, and​ 4, for which ​f(minus​2)equals12.

To find a polynomial \( f(x) \) of degree 3 with real coefficients that has zeros at \( x = -3 \), \( x = -1 \), and \( x = 4 \), and satisfies \( f(-2) = 12 \), we can start by constructing the polynomial from its roots.

### Step 1: Set up the polynomial
Since the polynomial has zeros at \( x = -3 \), \( x = -1 \), and \( x = 4 \), we can write it in factored form as:
\[
f(x) = a(x + 3)(x + 1)(x - 4)
\]
where \( a \) is a constant to be determined.

### Step 2: Use the condition \( f(-2) = 12 \) to find \( a \)
To find \( a \), substitute \( x = -2 \) and set \( f(-2) = 12 \):
\[
f(-2) = a(-2 + 3)(-2 + 1)(-2 - 4)
\]

Now calculate each term:
1. \( -2 + 3 = 1 \)
2. \( -2 + 1 = -1 \)
3. \( -2 - 4 = -6 \)

Substitute these values into the expression:
\[
f(-2) = a \cdot 1 \cdot (-1) \cdot (-6)
\]
\[
f(-2) = a \cdot 6
\]

Since \( f(-2) = 12 \), we can set up the equation:
\[
6a = 12
\]
\[
a = 2
\]

### Step 3: Write the final polynomial
Now that we know \( a = 2 \), substitute back into the polynomial:
\[
f(x) = 2(x + 3)(x + 1)(x - 4)
\]

### Step 4: Expand the polynomial (optional)
Expanding \( f(x) = 2(x + 3)(x + 1)(x - 4) \) gives:
\[
f(x) = 2(x^3 - 8x + 3x^2 -12x -4x -12)
\

Find a polynomial of degree 3 with real coefficients and zeros of -3, -1, and 4, for which f(-2) = 12.

Discover how to find a polynomial of degree 3 with real coefficients, given the roots -3, -1, and 4, and the condition f(-2) = 12. This problem explores polynomial factorization, solving for constants, and key algebraic concepts suitable for high school students.

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