# How do you use the rational roots theorem to find all possible zeros of #f(x) = -2x^3 + 3x^2 -4x + 8#?

##### 1 Answer

#### Answer:

It has one positive Real zero and a Complex conjugate pair of non-Real zeros.

#### Explanation:

#f(x) = -2x^3+3x^2-4x+8#

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

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

#+-1/2, +-1, +-2, +-4, +-8#

Note that

So the only possible *rational* zeros are:

#1/2, 1, 2, 4, 8#

When

When

When

So that only leaves as a possible *rational* zero:

#1/2#

We find:

#f(1/2) = -2(1/8)+3(1/4)-4(1/2)+8#

#= (-1+3-8+32)/4#

#= 13/2#

So

**Bonus**

What can we find out about the zeros of

The discriminant

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

In our example,

#Delta = 144-512-864-6912+3456=-4688#

Since

Let's simplify

#0 = -4f(x) = 8x^3-12x^2+16x-32#

#=(2x-1)^3+5(2x-1)-26#

#=t^3+5t-26#

where

Use Cardano's method to find the roots:

Let

Then we want to solve:

#u^3+v^3+(3uv+5)(u+v)-26 = 0#

Add the constraint

#u^3-125/(27u^3)-26 = 0#

Multiply through by

#27(u^3)^2-702(u^3)-125 = 0#

Use the quadratic formula to find:

#u^3 = (702+-sqrt(702^2+4(27)(125)))/(2*27)#

#= (702+-sqrt(492804+13500))/54#

#= (702+-sqrt(506304))/54#

#= (702+-24sqrt(879))/54#

#= (351+-12sqrt(879))/27#

Since

#t_1 = 1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879)))#

and related Complex roots:

#t_2 = 1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879)))#

#t_3 = 1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879)))#

where

Then

#x_1 = 1/2(1+1/3(root(3)(351+12sqrt(879))+root(3)(351-12sqrt(879))))#

and related Complex zeros:

#x_2 = 1/2(1+1/3(omega root(3)(351+12sqrt(879))+omega^2 root(3)(351-12sqrt(879))))#

#x_3 = 1/2(1+1/3(omega^2 root(3)(351+12sqrt(879))+omega root(3)(351-12sqrt(879))))#