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Mathematics 21 Online
OpenStudy (anonymous):

Can someone help me factor 5y^3+10y^2-4y-8 using the grouping method?

OpenStudy (anonymous):

Please I really need help

OpenStudy (anonymous):

Let's try ;) \[5y^3+10y^2-4y-8\]\[5y^2*(y+2)-4*(y+2)\]the answer is:\[(y+2)*(5y^2-4)\]

OpenStudy (anonymous):

Why y+2?? I'm confused

OpenStudy (anonymous):

\[5y^2*(y+2)=5y^3+10y^2\]and\[-4*(y+2)=-4y-8\]You have to use the signal rule

OpenStudy (anonymous):

Ohhh okay factoring them both gives you y+2, then what?

OpenStudy (anonymous):

you can factoring again because (y+2) is common in both members

OpenStudy (anonymous):

okay so 5y^2(y+2)+-4(y+2)? Or how is it written out exactly?

OpenStudy (anonymous):

@D3xt3R ?

OpenStudy (anonymous):

\[5y^2*(y+2)-4*(y+2)\]\[(5y^2-4)*(y+2)\]

OpenStudy (anonymous):

Okay, so with the commutative property of multiplication we can rewrite it as (y+2)(5y^2-4) yes?

OpenStudy (anonymous):

Exactly

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