Question

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

Answer 1

`(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 2

`(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.