# How do you factor #3x^3 - 10x^2 + 2 = 0#?

##### 1 Answer

#### Answer:

where:

#x_k = sqrt(15)/10 sec((2kpi)/3+1/3arccos(-(9sqrt(15))/100))# for#k=0,1,2#

#{ (x_0 ~~ 0.48368), (x_1 ~~ -0.42137), (x_2 ~~ 3.271026) :}#

#### Explanation:

Given:

#f(x) = 3x^3-10x^2+2#

By the rational root theorem, any rational zeros of

That means that the only possible rational zeros are:

#+-1/3# ,#+-2/3# ,#+-1# ,#+-2#

None of these are zeros of

Note that:

#{ (f(-1) = -3-10+2 = -11 < 0), (f(0) = 0+0+2 = 2 > 0), (f(1) = 3-10+2 = -5 < 0), (f(4) = 192-160+2 = 34 > 0) :}#

So

I will use a trigonometric method to derive expressions for the three zeros:

Let

Then the zeros of

#2t^3-10t+3 = 0#

Let

The multiplier here is chosen to result in a match for the formula for

Then:

#0 = 2t^3-10t+3#

#=2((2sqrt(15))/3 cos theta)^3 - 10((2sqrt(15))/3 cos theta) + 3#

#=(20sqrt(15))/9(4 cos^3 theta - 3 cos theta)+3#

#=(20sqrt(15))/9cos 3 theta+3#

So:

#cos 3 theta = -27/(20sqrt(15)) = -(9sqrt(15))/100#

So

#3 theta = 2kpi +- arccos(-(9sqrt(15))/100)#

So:

#t = (2sqrt(15))/3 cos theta = (2sqrt(15))/3 cos((2kpi)/3+-1/3 arccos(-(9sqrt(15))/100))#

This gives us three distinct roots:

#t = (2sqrt(15))/3 cos((2kpi)/3+1/3 arccos(-(9sqrt(15))/100))# for#k=0, 1, 2#

Then:

#x=1/t = 3/(2sqrt(15)) sec((2kpi)/3+1/3 arccos(-(9sqrt(15))/100))# for#k=0, 1, 2#

#=sqrt(15)/10 sec((2kpi)/3+1/3 arccos(-(9sqrt(15))/100))# for#k=0, 1, 2#

Call these three roots

Then

**Footnote**

Why did I use a trigonometric method to solve this cubic?

When a cubic has

Since we are trying to describe Real roots, expressions involving explicit Complex arithmetic seem somewhat out of place.

In addition, note that taking cube roots relates in an essential way to trisecting angles, so trigonometry is appropriate. The trigonometric solution brings this out.