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

1 Answer
May 7, 2018

Answer:

#(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)#