How do you factor completely #-n^4 - 3n^2 - 2n^3#?

1 Answer
Jim G.
Sep 5, 2017

#-n^2(n+1-isqrt2)(n+1+isqrt2)#

Explanation:

#-n^2" is a "color(blue)"common factor"" in all 3 terms"#

#rArr-n^2(n^2+3+2n)#

#=-n^2(n^2+2n+3)#

#"check the "color(blue)"discriminant"" of "n^2+2n+3#

#"with "a=1,b=2,c=3#

#Delta=b^2-4ac=4-12=-8#

#"since "Delta<0" then the roots are not real"#

#"we can factorise by finding the roots of "n^2+2n+3#

#"using the "color(blue)"quadratic formula"#

#n=(-2+-sqrt(4-12))/2=(-2+-sqrt(-8))/2#

#color(white)(n)=-1+-isqrt2larrcolor(red)" complex roots"#

#rArr-n^4-3n^2-2n^3#

#=-n^2(n+1-isqrt2)(n+1+isqrt2)#

Related questions
See all questions in Factoring Completely
Impact of this question
1576 views around the world
You can reuse this answer
Creative Commons License