So it's been quite a while since I've worked with third degree polynomials, and I'm pretty rusty ...any help would be greatly appreciated. This is the equation:
"Find all solutions to the equation correct to two decimal places."
x^3 - 9x^2 - 4 = 0
I tried synthetic division but it didn't work out (I think).
Using Descartes' rule of signs you will find that this has one positive root and no negative roots.
It is also evident that the root is less than (Cauchy's bound tells us this).
The cubic has different signs at and , so the bisection method can be used to localise the root.
Another "hand-made" way to get a close answer is...
where "y" is the decimal part.
This leads to a cubic in "y"
for which the 81y term is dominant.
is very close to 4.
Hence the only real solution is very near x=9.05