Find the remainder when $6x^3 - 13x^2 + 17x - 6$ is divided by $2x - 1$. Show that there is only one real root of the equation
$6x^3 - 13x^2 + 17x - 6 = 0.$

1. Apply the Remainder Theorem by substituting $x = \frac{1}{2}$ into the polynomial

Let $f(x) = 6x^3 - 13x^2 + 17x - 6$.
By the Remainder Theorem, the remainder when $f(x)$ is divided by $2x - 1$ is $f\left(\frac{1}{2}\right)$.
$f\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^3 - 13\left(\frac{1}{2}\right)^2 + 17\left(\frac{1}{2}\right) - 6$

2. Evaluate the expression to find the remainder

$f\left(\frac{1}{2}\right) = 6\left(\frac{1}{8}\right) - 13\left(\frac{1}{4}\right) + \frac{17}{2} - 6$
$= \frac{3}{4} - \frac{13}{4} + \frac{34}{4} - \frac{24}{4}$
$= \frac{3 - 13 + 34 - 24}{4}$
$= 0$

1. Use the result from part (a) to factorise the cubic polynomial

From part (a), $P\left(\frac{1}{2}\right) = 6\left(\frac{1}{8}\right) - 13\left(\frac{1}{4}\right) + 17\left(\frac{1}{2}\right) - 6 = 0$, so $2x - 1$ is a factor.
By long division or comparing coefficients:
$6x^3 - 13x^2 + 17x - 6 = (2x - 1)(3x^2 - 5x + 6)$

2. Find the real root from the linear factor

For $(2x - 1)(3x^2 - 5x + 6) = 0$,
$2x - 1 = 0 \implies x = \frac{1}{2}$

3. Show that the quadratic factor has no real roots using the discriminant

For the quadratic factor $3x^2 - 5x + 6 = 0$,
Discriminant $\Delta = b^2 - 4ac = (-5)^2 - 4(3)(6)$
$\Delta = 25 - 72 = -47$
Since $\Delta < 0$, the quadratic equation $3x^2 - 5x + 6 = 0$ has no real roots.
Therefore, $x = \frac{1}{2}$ is the only real root.