How do you multiply #(x-1)(x-2)(x-3)#?

1 Answer
Jun 11, 2018

Answer:

#(x-1)(x-2)(x-3) = x^3-6x^2+11x-6#

Explanation:

It is helpful to know that:

#(x-alpha)(x-beta)(x-gamma)#

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

In this identity the expressions forming the coefficients of #x^k# are (apart from the alternating signs), the elementary symmetric polynomials in #alpha, beta, gamma#, namely:

  • #alpha+beta+gamma#

  • #alphabeta+betagamma+gammaalpha#

  • #alphabetagamma#

With #alpha=1#, #beta=2# and #gamma=3#, we find:

#alpha+beta+gamma=1+2+3=6#

#alphabeta+betagamma+gammaalpha=(1 * 2) + (2 * 3) + (3 * 1) = 2+6+3=11#

#alphabetagamma = 1 * 2 * 3 = 6#

So:

#(x-1)(x-2)(x-3) = x^3-6x^2+11x-6#