How do you factor #3y^2+21y-36#?

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Answer 1 · 2015-05-16T12:43:18

#3y^2+21y-36#

First extract the (obvious) constant factor of #3#
#3(y^2+7y-12)#

Unfortunately #(y^2+7y-12)# has no obvious rational roots.

We can apply the quadratic formula for roots:
#x=(-b+-sqrt(b^2-4ac))/(2a)#
to get
#x= (-7 +-sqrt(97))/2#

So with irrational factors we could continue to
#3(x+(7+sqrt(97))/2)(x+(7-sqrt(97))/2))#

Answer 2 · 2015-05-16T12:44:55

First notice that all the coefficients are divisible by 3

#3y^2 + 21y -36 = 3(y^2 + 7y - 12)#

To factor #y^2+7y-12#, notice that it is of the form #ay^2+by + c#, with #a=1#, #b=7# and #c=-12#.

By the general quadratic solution, this is zero for

#y = (-b +-sqrt(b^2 - 4ac))/(2a)#

#= (-7 +=sqrt(7^2-4*1*(-12)))/(2*1)#

#= (-7 +- sqrt(49+48)) / 2#

#= (-7 +-sqrt(97))/2#

#97# is not a perfect square, so the only factorization into linear factors has irrational coefficients, viz.

#y^2+7y-12#

#= (y-((-7 + sqrt(97))/2))(y-((-7 - sqrt(97))/2))#

So in summary:

#3y^2 + 21y -36#

#= 3(y-((-7 + sqrt(97))/2))(y-((-7 - sqrt(97))/2))#

The factorization would have been much simpler if we were finding the factors of #3y^2 + 21y + 36# or #3y^2 - 21y + 36#.

In those cases, we have:

#3y^2 + 21y + 36 = 3(y+3)(y+4)#

#3y^2 - 21y + 36 = 3(y-3)(y-4)#