Explain by completing the square that $x^2 - 2x + 3$ is always positive for all real values of $x$. Hence, find the range of values of $p$ if the inequality $\frac{3x^2 + px + 3}{x^2 - 2x + 3} > 0$ is satisfied for all real values of $x$.

1. Complete the square for the quadratic expression

$x^2 - 2x + 3 = (x - 1)^2 - (-1)^2 + 3$
$= (x - 1)^2 - 1 + 3$
$= (x - 1)^2 + 2$

2. Explain why the expression is always positive

For all real values of $x$, $(x - 1)^2 \ge 0$.
Therefore, $(x - 1)^2 + 2 \ge 2$.
Since $2 > 0$, the expression $x^2 - 2x + 3$ is always positive.

1. Deduce the condition for the numerator based on the denominator's sign.

From part (a), we know that $x^2 - 2x + 3 > 0$ for all real values of $x$.
For the fraction $\frac{3x^2 + px + 3}{x^2 - 2x + 3}$ to be strictly positive, the numerator must also be strictly positive for all real $x$:
$3x^2 + px + 3 > 0$

2. Apply the discriminant condition for a quadratic expression to be always positive.

For the quadratic expression $3x^2 + px + 3$ to be always positive, since the coefficient of $x^2$ is $3 > 0$, its discriminant must be strictly less than zero:
$b^2 - 4ac < 0$
$p^2 - 4(3)(3) < 0$

3. Solve the quadratic inequality for $p$.

$p^2 - 36 < 0$
$(p + 6)(p - 6) < 0$
$-6 < p < 6$