# How do you solve (x-1)(x-2)(x-3)(x-4) = 24?

Apr 19, 2016

$x = \left\{\begin{matrix}0 \\ 5 \\ \frac{5}{2} \pm \frac{\sqrt{15}}{2} i\end{matrix}\right.$

#### Explanation:

Let $f \left(x\right) = \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right)$

Then:

f(0) = (-1)(-2)(-3)(-4) = 4! = 24

f(5) = (5-1)(5-2)(5-3)(5-4) = 4*3*2*1 = 4! = 24

So both $x = 0$ and $x = 5$ are roots and $x$ and $\left(x - 5\right)$ are factors.

$f \left(x\right) - 24$

$= \left(x - 1\right) \left(x - 2\right) \left(x - 3\right) \left(x - 4\right) - 24$

$= {x}^{4} - 10 {x}^{3} + 35 {x}^{2} - 50 x$

$= x \left(x - 5\right) \left({x}^{2} - 5 x + 10\right)$

The remaining quadratic factor is in the form $a {x}^{2} + b x + c$, with $a = 1$, $b = - 5$ and $c = 10$.

This has zeros given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{5 \pm \sqrt{{5}^{2} - \left(4 \cdot 1 \cdot 10\right)}}{2}$

$= \frac{5 \pm \sqrt{- 15}}{2}$

$= \frac{5}{2} \pm \frac{\sqrt{15}}{2} i$