# Thread: Cubic equations / real roots etc.

1. ## Cubic equations / real roots etc.

Show that $x=-5$ is the only real root of cubic equation:

$x^3+3x^2-2x+40=0$

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 $b^2-4ac<0$ which means no other real roots

Would this be a correct approach?

2. That is precisely how I would do it.

3. 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??