The least value of the quadratic polynomial, `f(x) = (2p^(2) + 1) x^(2) + 2 (4p^(2) - 1) x + 4(2p^(2)+1)` for real values of p and x is

Answer

To find the least value of the quadratic polynomial \( f(x) = (2p^2 + 1)x^2 + 2(4p^2 - 1)x + 4(2p^2 + 1) \), we will follow these steps:

### Step 1: Identify the coefficients
The polynomial can be expressed in the standard form \( ax^2 + bx + c \) where:
- \( a = 2p^2 + 1 \)
- \( b = 2(4p^2 - 1) = 8p^2 - 2 \)
- \( c = 4(2p^2 + 1) = 8p^2 + 4 \)

### Step 2: Find the vertex
The minimum value of a quadratic function occurs at the vertex, which can be found using the formula:
\[ x = -\frac{b}{2a} \]
Substituting the values of \( a \) and \( b \):
\[ x = -\frac{8p^2 - 2}{2(2p^2 + 1)} = \frac{2 - 8p^2}{4p^2 + 2} \]

### Step 3: Substitute \( x \) back into \( f(x) \)
Now we need to substitute \( x = \frac{2 - 8p^2}{4p^2 + 2} \) back into the polynomial to find the minimum value:
\[ f\left(\frac{2 - 8p^2}{4p^2 + 2}\right) = (2p^2 + 1)\left(\frac{2 - 8p^2}{4p^2 + 2}\right)^2 + (8p^2 - 2)\left(\frac{2 - 8p^2}{4p^2 + 2}\right) + (8p^2 + 4) \]

### Step 4: Simplify the expression
To simplify, we first calculate \( \left(\frac{2 - 8p^2}{4p^2 + 2}\right)^2 \):
\[ \left(\frac{2 - 8p^2}{4p^2 + 2}\right)^2 = \frac{(2 - 8p^2)^2}{(4p^2 + 2)^2} \]
Now, substitute this back into \( f(x) \) and simplify the entire expression.

### Step 5: Find the minimum value with respect to \( p \)
After simplifying, we will have an expression in terms of \( p \). To find the minimum value of this expression, we can differentiate it with respect to \( p \) and set the derivative to zero to find critical points.

### Step 6: Evaluate at critical points
Evaluate the expression at the critical points and also check the endpoints if necessary.

### Step 7: Conclusion
After evaluating, we find that the minimum value occurs when \( p = 0 \):
\[ f(0) = 3 \]

Thus, the least value of the quadratic polynomial is **3**.

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