# Thread: Cubic equation for pre-calc

1. ## Cubic equation for pre-calc

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.

$P(x) = -2x^3 + 6x^2 - 6x + 2 = 0$

$P(x) = x^3 - 3x^2 + 3x - 1 = 0$.

Notice that if you let $x = 1$, you get

$P(1) = 1^3 - 3(1)^2 + 3(1) - 1$

$= 1 - 3 + 3 - 1$

$= 0$.

By the factor and remainder theorems, this means that $(x - 1)$ is a factor.

Now long divide, and you should find

$x^3 - 3x^2 + 3x - 1 = 0$

$(x - 1)(x^2 - 2x + 1) = 0$

$(x - 1)(x-1)^2 = 0$

$(x - 1)^3 = 0$

$x - 1 = 0$

$x = 1$.

So the only solution to this equation is $x =1$.

P.S. If you're really good with the Binomial Theorem you can see that

$(x - 1)^2 = x^3 - 3x^2 + 3x - 1$ by observation...

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.
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

4. Originally Posted by CaptainBlack
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
You're better off taking out common factors by observation first...

### calac. cubic eq. root formula

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