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

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Answer 1 · 2016-11-09T03:11:37

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

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

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 · 2016-11-09T03:17:12

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

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.

Factorize 7 y^3 (9 y-4) + 9 y - 47 y^3 (9 y-4) + 9 y - 4 ?