For our class our prof has given us the cubic function
-2x^3+6x^2-6x+2=0
Is there a formula to solve a cubic function like this? I can't remember ever doing one before.
Thank you.
Help on the steps through it would be appreciated.
For our class our prof has given us the cubic function
-2x^3+6x^2-6x+2=0
Is there a formula to solve a cubic function like this? I can't remember ever doing one before.
Thank you.
Help on the steps through it would be appreciated.
There is a Cubic Formula, but it only works for particular cubics.
You're better off using the remainder and factor theorems to factorise and then using the Null Factor Law.
$\displaystyle P(x) = -2x^3 + 6x^2 - 6x + 2 = 0$
$\displaystyle P(x) = x^3 - 3x^2 + 3x - 1 = 0$.
Notice that if you let $\displaystyle x = 1$, you get
$\displaystyle P(1) = 1^3 - 3(1)^2 + 3(1) - 1$
$\displaystyle = 1 - 3 + 3 - 1$
$\displaystyle = 0$.
By the factor and remainder theorems, this means that $\displaystyle (x - 1)$ is a factor.
Now long divide, and you should find
$\displaystyle x^3 - 3x^2 + 3x - 1 = 0$
$\displaystyle (x - 1)(x^2 - 2x + 1) = 0$
$\displaystyle (x - 1)(x-1)^2 = 0$
$\displaystyle (x - 1)^3 = 0$
$\displaystyle x - 1 = 0$
$\displaystyle x = 1$.
So the only solution to this equation is $\displaystyle x =1$.
P.S. If you're really good with the Binomial Theorem you can see that
$\displaystyle (x - 1)^2 = x^3 - 3x^2 + 3x - 1$ by observation...
Use the rational root theorem to look for rational roots of the cubic.
The rational root theorem tells you that if this has any rational roots they are amoung +/-1, +/-2, +/-1/2. Now x=1 appears to be a root giving (by long division or whatever other method you like):
-2x^3+6x^2-6x+2=(x-1)(-2x^2+4x-2)=0
and you should be able to finish from there.
CB