To simplify the expression \((12r^2 + 6r + 3) / (1 + 2r + 4r^2)\), we can start by factoring both the numerator and the denominator, if possible.
Step 1: Factor the numerator.
The numerator is \(12r^2 + 6r + 3\). We can factor out the greatest common factor, which is \(3\):
\[ 12r^2 + 6r + 3 = 3(4r^2 + 2r + 1) \]
Next, we can check if \(4r^2 + 2r + 1\) is factorable. To do this, we can use the quadratic formula, \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
- Here, \(a = 4\), \(b = 2\), and \(c = 1\).
- The discriminant \(b^2 - 4ac = 2^2 - 4 \cdot 4 \cdot 1 = 4 - 16 = -12\).
Since the discriminant is negative, \(4r^2 + 2r + 1\) has no real roots and cannot be factored further over the reals.
Step 2: Factor the denominator.
The denominator is \(1 + 2r + 4r^2\). We can rewrite it as \(4r^2 + 2r + 1\). Using the same approach, we check if it can be factored or if it has real roots:
Using the quadratic formula again with \(a = 4\), \(b = 2\), and \(c = 1\), we find:
- The discriminant \(b^2 - 4ac = 2^2 - 4 \cdot 4 \cdot 1 = 4 - 16 = -12\).
As with the numerator, the discriminant is negative, indicating \(4r^2 + 2r + 1\) cannot be factored.
Step 3: Rewrite the expression.
Since both numerator and denominator yield \(3(4r^2 + 2r + 1)\) in the numerator and \(4r^2 + 2r + 1\) in the denominator, we can simplify the expression:
\[ \frac{3(4r^2 + 2r + 1)}{4r^2 + 2r + 1} \]
Provided \(4r^2 + 2r + 1 \neq 0\) (which it cannot be for real \(r\)), we can cancel these terms:
\[ = 3 \quad (\text{for } 4r^2 + 2r + 1 \neq 0) \]
Thus, the simplified expression is:
\[ \boxed{3} \]