Question

28v^3+16v^2-21v-12 factor by group please

Answer 1

The factor of
`28v^3+16v^2-21v-12` =
`(7v+4)(2v+√(3))(2v+√(3))` or
`(7v+4)(4v^2-3)`

Explanation:

The given equation:

`28v^3+16v^2-21v-12`

To find, the factors of
`28v^3+16v^2-21v-12` = ?

∴
`28v^3+16v^2-21v-12`

`=(4* 7)v^3+(4* 4)v^2-(3* 7)v-(3* 4)`

=
`4v^2(7v+4)-3(7v+4)`

=
`(7v+4)(4v^2-3)`

=
`(7v+4)[(2v)^2-√(3)^2]`

Using the algebraic identity,

`a^(2)-b^(2)=(a+b)(a-b)`

=
`(7v+4)(2v+√(3))(2v+√(3))`

∴ The factor of
`28v^3+16v^2-21v-12` =
`(7v+4)(2v+√(3))(2v+√(3))` or
`(7v+4)(4v^2-3)`