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3 votes
Factor by grouping
(4y^3 + 28y^2) + (y + 7)

User Ravena
by
5.2k points

1 Answer

4 votes

We have the following expression given:


(4y^3+28y^2)+(y+7)

We can start selecting as common factor 4y^2 and we got:


4y^2(y+7)+(y+7)

Now we can select y+7 as common factor and we got:


(y+7)\left\lbrack 4y^2+1\right\rbrack

So then our final answer would be:


(y+7)(4y^2+1)

User Plieb
by
5.7k points