Question

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

Answer 1

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`

Answer 2

`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`