Question #f8d98

1 Answer
A08
Jun 13, 2016

Factors are #2y(y-9)(y+3)#
Equation as following solutions
#y=0,9,-3#

Explanation:

The question could be how do you factor #2y^3-12y^2-54y=0# and solve the equation.

In inspection we see that #2y# is a common factor. Therefore taking that out we have
#2y^3-12y^2-54y=0#
#=>2y(y^2-6y-27)=0#
Now the factor within the bracket can be further factorized by using the split the middle term.
We make two parts of middle term so that product of these two parts is equal to product of first and third terms. We have two parts as #-9y and 3y#. The equation becomes
#2y(y^2-9y+3y-27)=0#
Now paring the first two and last two we get
#2y((y^2-9y)+(3y-27))=0#

Taking the common factor out of both the pairs, #y# from the first and #3# from the second pair we get
#2y(y(y-9)+3(y-9))=0#
Taking out #(y-9)# common factor out of both pairs we get
#2y(y-9)(y+3)=0#
We have the factors as above.
Now solving for #y#, we set each factor #=0#

  1. #2y=0#
    #=>y=0#
  2. #y-9=0#
    #=>y=9#
  3. #y+3=0#
    #=>y=-3#
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