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 + 10y<span>² + 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 (3y<span>² + 2y) = </span></span>y(3y <span>+ 2) + (3y + 2) Factor out common term 3y + 2 = </span>(3y + 2) + (y + 1) = 2y(3y + 2)(y + 1)
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
