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anonymous
 5 years ago
where do i go from,
15(5x^21)(x2)
anonymous
 5 years ago
where do i go from, 15(5x^21)(x2)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are you trying to do?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0factor the polynomial.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0egh. I'd start by getting it into the standard polynomial form

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the original is, 75x^3150x^215x+30

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so factor a 15 out of everything.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well i do factor by grouping

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'd start by getting smaller numbers.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well thats why i broke it up into 75x^3150x^2 abd 15x+30

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[75x^3150x^215x+30\] \[= 15(5x^3  10x^2  x + 2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then regrouping and pulling a 5x from the first two terms you should have something nice.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that way you would pull out a 75 and get 75x^2(x2)15(x2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so then its (75x^215) and (x2) thenn 15(5x^2+1)(x2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[15(5x^3  10x^2  x + 2) \] \[= 15[5x(x2) (x2)]\] \[= 15(5x1)(x2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so i basically had the right answer right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Mine should be \[15[5x^2(x−2)−(x−2)]\] \[=15(5x^21)(x2)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright i will be back in a few with some more
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