Question

Please help me this question is very overdue and I need to submit it now. Will mark brainliest.

Answer 1

Write an equation that each term can be divisible by 2y. 
6y³ + 10y² + 4y

Break down each term into prime factors. 
(3 · 2 · y · y · y) + (5 · 2 · y · y) + (2 · 2 · y) 

Now, each term has a GCF of 2y.

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Factor out term 2y: 
6y³ + 10y² + 4y / 2y =
2y(3y² + 5y + 2) 

Factor 3y² + 5y + 2: 
(3y² + 2y) + (3y + 2) <-------- I split up the 5y between the 2
Factor out the y from (3y² + 2y) =
y(3y + 2) + (3y + 2)
Factor out common term 3y + 2 = 
(3y + 2) + (y + 1) = 
2y(3y + 2)(y + 1) 

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Two equivalent expressions are: 
2y(3y + 2)(y + 1)
6y³ + 10y² + 4y

Answer 2

Example,
1).   2y(y³+4y²-9y)= I took 2y because it is GCF , just wrote something like (y³+4y²-9y)
to make second one i opened parentheses using distributive property
2y*y³+2y*4y²-2y*9y and then i got 2)
2).   =2y⁴+8y³-18y²=   to make 3) I took y² and made back distributive property
3).  =y²(2y²+8y-18)=
4).  =  2y²(y²+4y-9)=
5) =2y(y³+4y²)-18y²

all these equations are really equals to each other and to 2y⁴+8y³-18y², they are all the same, or equivalent