Solve the equation: $\;$ $\left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) = 15$
Given equation: $\;$ $\left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) = 15$
i.e. $\;$ $\left(x^2 - 5x + 4\right) \left(x^2 - 5x + 6\right) = 15$ $\;\;\; \cdots \; (1)$
Let $\;$ $x^2 - 5x = p$ $\;\;\; \cdots \; (2)$
Then $(1)$ becomes
$\left(p +4\right) \left(p + 6\right) = 15$
i.e. $\;$ $p^2 + 10p + 24 = 15$
i.e. $\;$ $p^2 + 10p + 9 = 0$
i.e. $\;$ $\left(p + 9\right) \left(p + 1\right) = 0$
i.e. $\;$ $p = -9$ $\;$ or $\;$ $p = -1$
When $\;$ $p = -9$, $(2)$ becomes
$x^2 - 5x = -9$
i.e. $\;$ $x^2 - 5x + 9 = 0$ $\;\;\; \cdots \; (3)$
Discriminant of $(3)$ is
$\Delta = \left(-5\right)^2 - 4 \times 1 \times 9 = = 25 - 36 = - 9 < 0$
$\implies$ $(3)$ does not have any real roots.
When $\;$ $p = -1$, $\;$ $(2)$ becomes
$x^2 - 5x = -1$
i.e. $\;$ $x^2 - 5x + 1 = 0$ $\;\;\; \cdots \; (4)$
Solving $(4)$ for $x$ gives
$x = \dfrac{5 + \sqrt{21}}{2}$ $\;$ or $\;$ $x = \dfrac{5 - \sqrt{21}}{2}$
$\therefore \;$ Solution of the given equation is
$x = \left\{\dfrac{5 + \sqrt{21}}{2}, \; \dfrac{5 - \sqrt{21}}{2} \right\}$