# Factor by grouping 28y^3+21y^2+36y+27?

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We'll group the first two terms of the sum and the last two term and we'll get:

`(28y^3 + 21y^2) + (36y + 27)`

We'll factor `7y^2` within the first pair of brackets and we'll factor 9 within the 2nd pair of brackets:

`7y^2(4y + 3) + 9(4y + 3)`

We notice the common factor (4y + 3):

`(4y + 3)(7y^2 + 9)`

**Therefore, the factorised result is `(4y+3)(7y^2 + 9)` .**

`28y^3+21y^2+36y+27 `

`(28y^3+21y^2) (+36y+27)`

`7y^2(4y+3) 9(4y+3)`

`(4y+3) (7y^2+9)`