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AngelaB97
 one year ago
can someone please help me factor y^4y^3
AngelaB97
 one year ago
can someone please help me factor y^4y^3

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AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0in my book its says y(y31) but i don't understand how they got that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0between y\(^4\) and y\(^3\) which one is the LCM?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0As in which term has the smallest power between the two?

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0so wouldn't it be y^3(1y)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Although their way can also be another possible solution since between y\(^4\) and y\(^3\) there is a common y\(^1\), therefore, \[y^3(y1) \equiv y(y^3y^2)\] which i'm realizing is not what's written lol.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it was just the book's typo.

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0yeah i think so but what they did was use the difference of cubes so they furthered the equation to y(y^31) = y(y1)(y^2+y+1)

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.0can you explain how they did that please @Jhannybean

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well, \((y^31) \iff (a^3b^3) \implies (ab)(a^2+ab+b^2)\)
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