How do you evaluate #(x - 1) ( x - 4) ( x - 2)#?

1 Answer
George C.
Sep 2, 2017

#(x-1)(x-4)(x-2) = x^3-7x^2+14x-8#

Explanation:

Note that in general:

#(x-alpha)(x-beta)(x-gamma) = x^3-(alpha+beta+gamma)x^2+(alphabeta+betagamma+gammaalpha)x-alphabetagamma#

So with #alpha=1#, #beta=4# and #gamma=2# we find:

#(x-1)(x-4)(x-2)#

#= x^3-(1+4+2)x^2+((1)(4)+(4)(2)+(2)(1))x-(1)(4)(2)#

#= x^3-7x^2+14x-8#

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