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j-dog100000

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#1 j-dog100000

Member since 2005 • 680 Posts

Im having trouble factoring down this trinomial, can someone help me out. The question is

3y^+y-30

(^=squared)

17 years ago

Vax45

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#2 Vax45

Member since 2005 • 4834 Posts

I don't quite understand that problem, can you put it in word form?

17 years ago

andyxm

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#3 andyxm

Member since 2006 • 6194 Posts

what is this formula equilalent to? it has to equal something....

17 years ago

j-dog100000

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#4 j-dog100000

Member since 2005 • 680 Posts

No, this is factoring. So you have to factor the trinomial down to 2 binomials.

17 years ago

j-dog100000

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#5 j-dog100000

Member since 2005 • 680 Posts

I don't quite understand that problem, can you put it in word form?Vax45

What do you mean by word form.

17 years ago

simer190

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#6 simer190

Member since 2003 • 38 Posts

y=3, 1. 3 works 1 doesn't.

17 years ago

bobwill1

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#7 bobwill1

Member since 2003 • 2487 Posts

You just need to find the factors, and then use trial and error to find the right combination. (3y + 2) (y - 15) 3y^2 + 2y - 45y - 30 (3y + 15) (y - 2) 3y^2 + 15y - 6y - 30 (3y - 15) (y + 2) 3y^2 - 15y + 5y - 30 (3y - 2) (y + 15) 3y^2 + 45y - 2 y - 30 (3y + 10) (y - 3) 3y^2 - 9y + 10y - 30 3y^2 +y -30

17 years ago

math trouble...

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