How do you simplify #(2x + 1)(3x + 1)(x + 4)#?

2 Answers
Jun 30, 2017

See a solution process below:

Explanation:

First, multiply the two terms on the left. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

#(color(red)(2x) + color(red)(1))(color(blue)(3x) + color(blue)(1))(x + 4)# becomes:

#((color(red)(2x) xx color(blue)(3x)) + (color(red)(2x) xx color(blue)(1)) + (color(red)(1) xx color(blue)(3x)) + (color(red)(1) xx color(blue)(1)))(x + 4)#

#(6x^2 + 2x + 3x + 1)(x + 4)#

We can now combine like terms:

#(6x^2 + (2 + 3)x + 1)(x + 4)#

#(6x^2 + 5x + 1)(x + 4)#

We now use the same process for the two remaining terms:

#(color(red)(6x^2) + color(red)(5x) + color(red)(1))(color(blue)(x) + color(blue)(4))# becomes:

#(color(red)(6x^2) xx color(blue)(x)) + (color(red)(6x^2) xx color(blue)(4)) + (color(red)(5x) xx color(blue)(x)) + (color(red)(5x) xx color(blue)(4)) + (color(red)(1) xx color(blue)(x)) + (color(red)(1) xx color(blue)(4))#

#6x^3 + 24x^2 + 5x^2 + 20x + 1x + 4#

We can now combine like terms:

#6x^3 + (24 + 5)x^2 + (20 + 1)x + 4#

#6x^3 + 29x^2 + 21x + 4#

Jun 30, 2017

#color(green)(6x^3+29x^2+21x+4#

Explanation:

#(2x+1)(3x+1)(x+4)#

#color(white)(aaaaaaaaaaaaa)##2x+1#
#color(white)(aaaaaaaaaa)## xx underline(3x+1)#
#color(white)(aaaaaaaaaaaaa)##6x^2+3x#
#color(white)(aaaaaaaaaaaaaaaaa)##2x+1#
#color(white)(aaaaaaaaaaaaa)##overline(6x^2+5x+1)#
#color(white)(aaaaaaaaaaa)## xx x+4#
#color(white)(aaaaaaaaaaaaa)##overline(6x^3+5x^2+x)#
#color(white)(aaaaaaaaaaaaaaaaa)##24x^2+20x+4#
#color(white)(aaaaaaaaaaaaa)##overline(6x^3+29x^2+21x+4)#

#color(white)(aaaaaaaaaaaaa)##color(green)(6x^3+29x^2+21x+4#