# How do you factor #-2x^3+6x^2-3#?

##### 1 Answer

#### Answer:

where:

#x_n = 1+2cos(1/3 cos^(-1)(1/4)+(2npi)/3)#

#### Explanation:

Given:

#f(x) = -2x^3+6x^2-3#

**Rational roots theorem**

By the rational roots theorem, any rational zeros of

That means that the only possible rational zeros are:

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

Trying each in ascending order, we find:

#f(-3) = 105#

#f(-3/2) = 69/4#

#f(-1) = 5#

#f(-1/2) = -5/4#

#f(1/2) = -7/4#

#f(1) = 1#

#f(3/2) = 15/4#

#f(3) = -3#

So

**Tschirnhaus transformation**

To make the cubic easier to solve, use a linear substitution called a Tschirnhaus transformation as follows:

#(-4)f(x) = 8x^3-24x^2+12#

#color(white)((-4)f(x)) = (2x-2)^3-12(2x-2)-4#

#color(white)((-4)f(x)) = t^3-12t-4#

where

**Trigonometric substitution**

Next we make a substitution:

#t = k cos theta#

where

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

Let

Then:

#0 = t^3-12t-4#

#color(white)(0) = (k cos theta)^3-12(k cos theta)-4#

#color(white)(0) = k(k^2 cos^3 theta-12 cos theta)-4#

#color(white)(0) = 4(16 cos^3 theta-12 cos theta)-4#

#color(white)(0) = 16(4 cos^3 theta-3 cos theta)-4#

#color(white)(0) = 16cos 3 theta-4#

Hence:

#cos 3 theta = 1/4#

So:

#3 theta = +-cos^(-1)(1/4)+2npi" "# for some integer#n#

So:

#theta = +-1/3cos^(-1)(1/4)+(2npi)/3#

So:

#cos theta = cos(+-1/3 cos^(-1)(1/4) + (2npi)/3)#

So:

#t_n = k cos theta = 4cos(1/3 cos^(-1)(1/4)+(2npi)/3)#

(( we can discard the

This last formula provides

Then

Hence zeros of our original cubic:

#x_n = 1+2cos(1/3 cos^(-1)(1/4)+(2npi)/3)" "# for#n = 0, 1, 2#

#x_0 ~~ 2.8100379#

#x_1 ~~ -0.6417835#

#x_2 ~~ 0.8317456#

Then

#-2x^3+6x^2-3 = -2(x-x_0)(x-x_1)(x-x_2)#