# What is the factorisation of 2y^{2} - 38y - 228

2
by jonas

2014-10-03T17:57:35+05:30

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2Y² - 38Y -228
=2(Y²-19Y-114)
GENERALLY YOU NEED TO FIND TWO NUMBERS SO THAT
1.THEIR SUM IS -19
2.PRODUCT IS -114

BUT IN THIS CASE, IT IS NOT POSSIBLE.

SO FIRST FIND THE SOLUTION BY
Y =  AND
OR Y = (19 + √817) AND (19 - √817)

NOW ONCE YOU GET THE SOLUTIONS, YOU CAN DIRECTLY WRITE THE FACTORIZATION AS

2(Y - (19 + √817))(Y - (19 - √817))

If p and q are roots of an equation, x^2 + ax + b =0, it can be factorized as (x-p)(x-q)
ok ....ty :)
2014-10-03T17:59:49+05:30

### This Is a Certified Answer

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2 (y² - 19 y - 114 )

114 = 3 * 2 * 19   - factorization does not seem to be possible directly with these factors.

roots are  = [ - b +- sqrt(b² - 4a c) ] / 2
= [19 +- sqrt(361 + 456) ] / 2 = (19+ √817 )/ 2  and (19-√817)/2

So three factors are : 2 ,  [ y - (19 + √817) /2 ] and  [ y - (19 - √817)/2 ]

thanx n u r welco