Asked by patti
how do I start to solve this
3y^2+17y+20=
3y^2+17y+20=
Answers
Answered by
Reiny
is your equation equal to zero?
if so then
3y^2+17y+20=0 factors to
(y + 4)(3y + 5) = 0
y = -4 or y = -5/3
if so then
3y^2+17y+20=0 factors to
(y + 4)(3y + 5) = 0
y = -4 or y = -5/3
Answered by
patti
It just says to factor
Answered by
Reiny
then just eliminate my last step
You must have learned a method to do this
You must have learned a method to do this
Answered by
patti
so if I have this problem to factor 2x^2+7xy-30y I would do this to factor it x^2+xy+7+2-30y^2
Answered by
Reiny
you probably meant to type
2x^2+7xy-30y^2 or else it won't work
the method most students appear to use these days is the method of decomposition of the middle term.
In this method you multiply the first and last number coefficients ... 2*(-30) = -60
now look for two factors of -60 which have a sum of 7
after about 3 tries I got 12 and -5
so I will break up the middle term of 7xy as -5xy+12xy
so we have
2x^2 - 5xy + 12xy - 30y^2
= x(2x - 5y) + 6y(2x - 5y)
= (2x-5y)(x+6y)
Although this seems to be the popular method taught these days, us old-timers don't use this method.
I have my own quick way where I can do about 5 questions in the time it takes somebody to do one using the above method.
(I wouldn't be at all surprised if other math tutors on here like bobpurley and drwls use the same method I do)
2x^2+7xy-30y^2 or else it won't work
the method most students appear to use these days is the method of decomposition of the middle term.
In this method you multiply the first and last number coefficients ... 2*(-30) = -60
now look for two factors of -60 which have a sum of 7
after about 3 tries I got 12 and -5
so I will break up the middle term of 7xy as -5xy+12xy
so we have
2x^2 - 5xy + 12xy - 30y^2
= x(2x - 5y) + 6y(2x - 5y)
= (2x-5y)(x+6y)
Although this seems to be the popular method taught these days, us old-timers don't use this method.
I have my own quick way where I can do about 5 questions in the time it takes somebody to do one using the above method.
(I wouldn't be at all surprised if other math tutors on here like bobpurley and drwls use the same method I do)
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