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#