Q. ( t = frac{(2 – x)(3 – x)}{3x^2 – 7x + 6} ).

We are given the rational expression

\(t=\dfrac{(2-x)(3-x)}{3x^{2}-7x+6}\).

1. Rewrite the numerator in a standard factor form.

   Observe that each factor of the numerator can be written by factoring out a negative sign:

   \(2-x = -(x-2)\) and \(3-x = -(x-3)\).

   Multiplying these two equalities gives

   \((2-x)(3-x) = \bigl(-(x-2)\bigr)\bigl(-(x-3)\bigr) = (+1)\,(x-2)(x-3) = (x-2)(x-3).\)

   Thus the numerator simplifies to \((x-2)(x-3)\).

2. Attempt to factor the denominator and check for common factors.

   The denominator is the quadratic \(3x^{2}-7x+6\). To see whether it factors over the real numbers, compute its discriminant:

   \(\Delta = b^{2}-4ac = (-7)^{2}-4\cdot 3\cdot 6 = 49-72 = -23.\)

   Because \(\Delta = -23 < 0\), the quadratic has no real roots and therefore does not factor over the real numbers into linear real factors. In particular, it does not share any linear factor with the numerator \((x-2)\) or \((x-3)\).

3. Write the simplified form and state the domain.

   Using the numerator rewrite and the fact that no cancellation with the denominator is possible, the expression simplifies to

   \(t = \dfrac{(x-2)(x-3)}{3x^{2}-7x+6}.\)

   Because the denominator \(3x^{2}-7x+6\) has no real zeros (discriminant negative) and the leading coefficient 3 is positive, the denominator is never zero for any real x. Therefore the expression is defined for every real number x; the domain is all real numbers, \(\mathbb{R}\).

4. Optional: expanded numerator form (equivalent expression).

   Expanding the numerator gives \( (x-2)(x-3) = x^{2}-5x+6\), so an equivalent form is

   \(t = \dfrac{x^{2}-5x+6}{3x^{2}-7x+6}.\)

   No further simplification by cancellation is possible over the real numbers.

Final result: \(t = \dfrac{(x-2)(x-3)}{3x^{2}-7x+6}\), domain: all real numbers.