# Cubic Equation

• Oct 22nd 2008, 09:23 PM
Kaynight
Cubic Equation
I know cubic equations must have the form \$\displaystyle Ax^3 + Bx^2 + Cx + D = 0,\$ and I know how to solve that type of equation by doing long division etc:

However I got this cubic question,

\$\displaystyle x^3-3x^2+4\$

It doesn't have the same form as above and it's put me right off knowing what to do. I tried to simplify it \$\displaystyle ( x(x^2+3x) +4)\$ but I got no where fast. So any starters of where I should start to factorize would be very helpful!
• Oct 22nd 2008, 10:30 PM
Laurent
Quote:

Originally Posted by Kaynight
I know cubic equations must have the form \$\displaystyle Ax^3 + Bx^2 + Cx + D = 0,\$ and I know how to solve that type of equation by doing long division etc:

However I got this cubic question,

\$\displaystyle x^3-3x^2+4\$

It doesn't have the same form as above and it's put me right off knowing what to do. I tried to simplify it \$\displaystyle ( x(x^2+3x) +4)\$ but I got no where fast. So any starters of where I should start to factorize would be very helpful!

It has the same form as above, with \$\displaystyle C=0\$.
You can notice that \$\displaystyle -1\$ is a solution, hence \$\displaystyle x+1\$ divides your polynomial. It remains to compute \$\displaystyle Q(x)\$ such that \$\displaystyle x^3-3x^2+4=(x+1)Q(x)\$.
• Oct 22nd 2008, 11:00 PM
Kaynight
Quote:

Originally Posted by Laurent
It has the same form as above, with \$\displaystyle C=0\$.
You can notice that \$\displaystyle -1\$ is a solution, hence \$\displaystyle x+1\$ divides your polynomial. It remains to compute \$\displaystyle Q(x)\$ such that \$\displaystyle x^3-3x^2+4=(x+1)Q(x)\$.

Many Thanks! :D