Question

How do you factor `6y^6-5y^3-4` ?

Answer 1

`(2y^3+1)(3y^3-4)`

Explanation:

`"using a substitution reduces the expression to a"`
`"usual quadratic"`

`"let "u=y^3`

`rArr6y^6-5y^3-4=6u^2-5u-4`

`"using the a-c method for factoring"`

`"the factors of the product "6xx-4=-24`

`"which sum to - 5 are + 3 and - 8"`

`"split the middle term using these factors"`

`6u^2+3u-8u-4larrcolor(blue)"factor by grouping"`

`=color(red)(3u)(2u+1)color(red)(-4)(2u+1)`

`"take out the "color(blue)"common factor "(2u+1)`

`=(2u+1)(color(red)(3u-4))`

`"change the substitution back into terms in y"`

`rArr6y^6-5y^3-4=(2y^3+1)(3y^3-4)`