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

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Answer 1 · 2016-04-19T20:00:57

$x={ (0), (5), (5/2+-sqrt(15)/2i) :}$

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

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 $(x-5)$ are factors.

$f(x)-24$

$= (x-1)(x-2)(x-3)(x-4)-24$

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

$=x(x-5)(x^2-5x+10)$

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

This has zeros given by the quadratic formula:

$x = (-b+-sqrt(b^2-4ac))/(2a)$

$=(5+-sqrt(5^2-(4*1*10)))/2$

$=(5+-sqrt(-15))/2$

$=5/2+-sqrt(15)/2i$

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