The minimum value of the polynimial `x(x+1)(x+2)(x+3)` is

Options

A. a) `0`
B. b) `(9)/(16)`
C. c) `-1`
D. d) `-(3)/(2)`

Answer

To find the minimum value of the polynomial \( p(x) = x(x+1)(x+2)(x+3) \), we can follow these steps:

### Step 1: Expand the Polynomial
First, we will expand the polynomial \( p(x) \).

\[ p(x) = x(x+1)(x+2)(x+3) \]

We can group the terms:

\[ p(x) = (x(x+3))((x+1)(x+2)) \]

Calculating \( x(x+3) \):

\[ x(x+3) = x^2 + 3x \]

Now, calculating \( (x+1)(x+2) \):

\[ (x+1)(x+2) = x^2 + 3x + 2 \]

Now we can write:

\[ p(x) = (x^2 + 3x)(x^2 + 3x + 2) \]

Let \( y = x^2 + 3x \). Then we can rewrite \( p(x) \) as:

\[ p(x) = y(y + 2) = y^2 + 2y \]

### Step 2: Find the Derivative
Next, we need to find the derivative of \( p(y) \) to locate the critical points.

\[ p'(y) = 2y + 2 \]

### Step 3: Set the Derivative to Zero
To find the critical points, set the derivative equal to zero:

\[ 2y + 2 = 0 \]

Solving for \( y \):

\[ 2y = -2 \implies y = -1 \]

### Step 4: Determine if it is a Minimum or Maximum
To determine if this critical point is a minimum or maximum, we need to find the second derivative.

\[ p''(y) = 2 \]

Since \( p''(y) = 2 > 0 \), this indicates that \( p(y) \) has a minimum at \( y = -1 \).

### Step 5: Calculate the Minimum Value
Now we can find the minimum value of \( p(y) \) at \( y = -1 \):

\[ p(-1) = (-1)^2 + 2(-1) = 1 - 2 = -1 \]

### Conclusion
Thus, the minimum value of the polynomial \( p(x) = x(x+1)(x+2)(x+3) \) is:

\[ \boxed{-1} \]