That is precisely how I would do it.
Show that is the only real root of cubic equation:
Can I do it this way?
(1) Fill for x=-5 and show that it equals zero.
(2) Divide factor (x+5) into the given cubic.This will give me a quadratic.
(3) Then show that which means no other real roots
Would this be a correct approach?
Yup it is.. but i have a cooler method
In any general equation..
here,ax^3 +bx^2 + cx + d = 0
sum of roots taken 1 at a time is -b/a
sum of roots taken 2 at a time is c/a
sum of roots taken 3 at a time is -d/a and so on..
let m,n,o be the roots...
we're given m=-5
so,m+n+o=-3
=>n+o=-3+5=2 (1)
and mno=-40
=>no=-40/-5=8 (2)
(1)^2=(n+o)^2
=n^2 + o^2 + 2on=2^2=4
putting on=8
n^2 + o^2=4-16=-12
sum of two squares is nececarily positive.. thus,n,o are both imaginary/complex.
Cooler,right?
you can use this for equations with degree higher than 3 also.. u see??