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

Question

Genesis Allen · Accepted 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# 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)#

Nathan Axel · Answer

undefined To multiply $(x-2)(2x+3)(3-x)$, you can use the distributive property and the associative property of multiplication.

First, multiply the first two binomials $(x-2)(2x+3)$ using the distributive property:

$(x-2)(2x+3) = x(2x) + x(3) - 2(2x) - 2(3) = 2x^2 + 3x - 4x - 6 = 2x^2 - x - 6$

Now, multiply the result by the third binomial $3-x$:

$(2x^2 - x - 6)(3-x) = (2x^2)(3) - (2x^2)(x) - (x)(3) + (x)(x) - (6)(3) + (6)(x)$

$= 6x^2 - 2x^3 - 3x + x^2 - 18 + 6x$

$= -2x^3 + 7x^2 + 3x - 18 + 6x$

$= -2x^3 + 7x^2 + 9x - 18$

So, $(x-2)(2x+3)(3-x) = -2x^3 + 7x^2 + 9x - 18$