How do you factor completely -n^4 - 3n^2 - 2n^3?
1 Answer
Sep 5, 2017
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)
