Question

Let f(x)=ax^3+bx^2+cx+d and suppose that r is a root of the equation f(x)=0.
show that -r is a root of the equation f(-x)=0

Answer 1

you know that:
ar^3+br^2+cr+d=0

now look at
f(-x) = -ax^3+bx^2-cx+d
plug x=-r
-a*(-r)^3+b*(-r)^2-c*(-r)+d =
ar^3+br^2+cr+d
and this is 0 according to the second line :)

Answer 2

my "proof" is f(r)=0 and if -x=-r then f(-x)=f(-r)=0

Answer 3

does that work

Answer 4

how did u get f(-x) = f(-r) = 0 ? @muzzammil.raza
you need to prove it using coolsector work first right ?

Answer 5

ok thanks

Answer 6

:)