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

Question

Isaiah Anthony · Accepted Answer

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

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#

Avery Alexander · Answer

undefined To multiply (x-1)(x-2)(x-3), you can use the distributive property and multiply the expressions two at a time, then simplify the result. Here's the step-by-step process:

1. Multiply (x-1) by (x-2):
   (x-1)(x-2) = x(x) + x(-2) - 1(x) - 1(-2)
              = x^2 - 2x - x + 2
              = x^2 - 3x + 2

2. Now, multiply the result from step 1 by (x-3):
   (x^2 - 3x + 2)(x-3) = (x^2)(x) + (x^2)(-3) - (3x)(x) - (3x)(-3) + 2(x) + 2(-3)
                        = x^3 - 3x^2 - 3x^2 + 9x - 3x + 9 + 2x - 6
                        = x^3 - 6x^2 + 8x - 6

So, (x-1)(x-2)(x-3) = x^3 - 6x^2 + 8x - 6.