Factorize #7 y^3 (9 y-4) + 9 y - 4# ?

2 Answers
Nov 9, 2016

Answer:

#(9y-4)(y+7^(-1/3))(7^(1/3) - 7^(2/3) y + 7 y^2)#

Explanation:

#7 y^3 (9 y-4) + 9 y - 4 = (9y-4)(1 + 7 y^3)#

but any odd order polynomial has at least one real root. In this case is

#y = -7^(-1/3)#

So

#1 + 7 y^3 = (y+7^(-1/3))(a y^2+b y+c)#

equating for all #y# we get the conditions

#{(1 - c/7^(1/3)=0),(b/7^(1/3) +c=0),(a/7^(1/3) + b=0),(7 - a=0):}# and finally

#7 y^3 (9 y-4) + 9 y - 4=(9y-4)(y+7^(-1/3))(7^(1/3) - 7^(2/3) y + 7 y^2)#

Answer:

#(9y-4)(7y^3+1)#

Explanation:

We have:

#7y^3(9y-4)+9y-4#

I'm going to rewrite this to group together the final two terms:

#7y^3(9y-4)+(9y-4)#

and now factor out the #(9y-4)#:

#(9y-4)(7y^3+1)#

We can't do anything with either term that will make this neater or easier to understand, so I'll leave it here.