Question

-48x^2+3x^4
find the zeros of the polynomial

Answer 1

you equate this equation to zero
-48x^2+3x^4=0
3x^2(-16+x^2)=0
then from here
either 3x^2=0
or x^2-16=0
then x1=x2=0
and x^2=16 hence x=+4,+4
then the zeros of the equation are
0,0,-4,+4

Answer 2

iam sorry, x^2=16 hence x=-4,+4 not +4,+4