f (x) = 2 (x-3) (x+2)^2 + 2 (x+2) (x-3)^3
solve f(x) = 0
if we factorise, we get
f (x) = 2 (x-3) (x+2) (x^2 – 5x + 11)
0 = 2 (x-3) (x+2) (x^2 – 5x + 11)
is it meaning that x = 3, and x = 2 ?? or ?
One of the things I am sure yuo have already learned is that if abc= 0 then at least one of a, b, or c must be 0.
Yes, you can factor .
So you know immediately that x+2= 0, or x- 3= 0, or is 0. If x- 3= 0, then x= 3 so x= 3 is a root. If x+ 2= 0 then x= -2 (NOT 2) so x= -2 is a root. But there are two other roots- numbers that satisfy . That does not have any real number roots but you can find the complex roots by using the quadratic formula.