(12r^2+6r+3) / (1+2r+4r^2)

Answer 1

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 $4r^2+2r+1\ne 0$ (which it cannot be for real $r$), we can cancel these terms:

$$=3(\text{for }4r^2+2r+1\ne 0)$$

Thus, the simplified expression is:

$$\boxed{{\displaystyle 3}}$$