For real values of x, the value of the expression `(11x^(2)+12x+6)/(x^(2)+4x+2)` lies between -5 and 3.True or False.

Answer

To determine whether the expression \(\frac{11x^2 + 12x + 6}{x^2 + 4x + 2}\) lies between -5 and 3 for real values of \(x\), we can follow these steps:

### Step 1: Set up the inequality
We want to check if:
\[ -5 < \frac{11x^2 + 12x + 6}{x^2 + 4x + 2} < 3 \]
This can be split into two inequalities:
1. \(\frac{11x^2 + 12x + 6}{x^2 + 4x + 2} > -5\)
2. \(\frac{11x^2 + 12x + 6}{x^2 + 4x + 2} < 3\)

### Step 2: Solve the first inequality
Starting with:
\[ \frac{11x^2 + 12x + 6}{x^2 + 4x + 2} > -5 \]
Multiply both sides by \(x^2 + 4x + 2\) (assuming \(x^2 + 4x + 2 > 0\)):
\[ 11x^2 + 12x + 6 > -5(x^2 + 4x + 2) \]
This simplifies to:
\[ 11x^2 + 12x + 6 > -5x^2 - 20x - 10 \]
Combine like terms:
\[ 16x^2 + 32x + 16 > 0 \]
Factoring out 16:
\[ 16(x^2 + 2x + 1) > 0 \]
This simplifies to:
\[ 16(x + 1)^2 > 0 \]
Since \((x + 1)^2 \geq 0\) for all \(x\) and is equal to 0 when \(x = -1\), this inequality holds for all \(x \neq -1\).

### Step 3: Solve the second inequality
Now, consider:
\[ \frac{11x^2 + 12x + 6}{x^2 + 4x + 2} < 3 \]
Again, multiply both sides by \(x^2 + 4x + 2\) (assuming \(x^2 + 4x + 2 > 0\)):
\[ 11x^2 + 12x + 6 < 3(x^2 + 4x + 2) \]
This simplifies to:
\[ 11x^2 + 12x + 6 < 3x^2 + 12x + 6 \]
Combine like terms:
\[ 11x^2 < 3x^2 \]
This leads to:
\[ 8x^2 < 0 \]
This inequality is never true for real \(x\) since \(8x^2 \geq 0\) for all \(x\).

### Step 4: Conclusion
Since the second inequality \(8x^2 < 0\) is never satisfied for any real \(x\), the original statement that the expression lies between -5 and 3 is **False**.