# How do you use the rational root theorem to list all possible roots for #x^3+5x^2-2x-15=0#?

##### 1 Answer

#### Answer:

The "possible" rational roots are

The actual roots are:

#x_k = 1/3(2sqrt(31) cos(1/3 cos^(-1)((65sqrt(31)) / 1922)+(2kpi)/3)-5)#

#### Explanation:

Given:

#f(x) = x^3+5x^2-2x-15#

**Rational root theorem**

The rational root theorem will only tell you about possible *rational* roots - not irrational or Complex ones.

By the rational root theorem, any *rational* zeros of

That means that the only possible *rational* roots are:

#+-1, +-3, +-5, +-15#

However, note that all of these are odd, resulting in all of the terms of

#f(x) = "odd" + "odd" + "even" + "odd" = "odd" != 0#

So

**Bonus**

Let us see what else we can find out about

**Discriminant**

The discriminant

#Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd#

In our example,

#Delta = 100+32+7500-6075+2700 = 4257#

Since

**Descartes's Rule of Signs**

The signs of the coefficients of

**Tschirnhaus transformation**

To make the task of solving the cubic simpler, we make the cubic simpler using a linear substitution known as a Tschirnhaus transformation.

#0=27f(x)=27x^3+135x^2-54x-405#

#=(3x+5)^3-93(3x+5)-65#

#=t^3-93t-65#

where

**Trigonometric substitution**

Look for a substitution of the form:

#t = k cos theta#

where

#4 cos^3 theta - 3 cos theta = cos 3 theta#

Let

Then:

#0 = t^3-93t-65#

#color(white)(0) = k^3cos^3 theta - 93k cos theta - 65#

#color(white)(0) = 2sqrt(31)(124 cos^3 theta - 93 cos theta) - 65#

#color(white)(0) = 62sqrt(31)(4 cos^3 theta - 3 cos theta) - 65#

#color(white)(0) = 62sqrt(31)cos 3theta - 65#

Hence:

#cos 3 theta = 65/(62 sqrt(31)) = (65sqrt(31)) / 1922#

So:

#3 theta = +-cos^(-1) ((65sqrt(31)) / 1922) + 2kpi#

#t = 2sqrt(31) cos(1/3 cos^(-1)((65sqrt(31)) / 1922)+(2kpi)/3)#

Then

#x_k = 1/3(2sqrt(31) cos(1/3 cos^(-1)((65sqrt(31)) / 1922)+(2kpi)/3)-5)#