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

1 Answer
Jul 11, 2015

Answer:

Construct the coefficient of each power of #x# in descending order to find:

#(x-2)(2x+3)(3-x) = -2x^3+7x^2+3x-18#

Explanation:

For each power of #x# in descending order, pick out the different combinations of terms, using one from each binomial, such that when multiplied together they will give the target power of #x# and add them together. For brevity, omit the #x#'s as you are multiplying and adding the coefficients...

So:

#color(blue)(x^3)# : #(1*2*-1) = color(red)(-2)#

#color(blue)(x^2)# : #(1*2*3)+(1*3*-1)+(-2*2*-1)#

#= 6-3+4 = color(red)(7)#

#color(blue)(x)# : #(1*3*3)+(-2*2*3)+(-2*3*-1)#

#= 9-12+6 = color(red)(3)#

#color(blue)(1)# : #(-2*3*3) = color(red)(-18)#

Hence #(x-2)(2x+3)(3-x) = -2x^3+7x^2+3x-18#

Check: Try #x=1# ...

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

#= -1*5*2 = color(red)(-10)#

#-2x^3+7x^2+3x-18 = -2+7+3-18 = color(red)(-10)#