How do you factor #5x^3 -3x^2 +16x-6#?

1 Answer
Jan 25, 2017

Answer:

Use Cardano's method to find:

#5x^3-3x^2+16x-6 = 5(x-x_1)(x-x_2)(x-x_3)#

where:

#x_n = 1/15(3+omega^(n-1)root(3)(972+15sqrt(58983))+omega^(1-n)root(3)(972-15sqrt(58983)))#

Explanation:

Given:

#f(x) = 5x^3-3x^2+16x-6#

We can find the zeros #x_1#, #x_2# and #x_3#

Then:

#f(x) = 5(x-x_1)(x-x_2)(x-x_3)#

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Discriminant

The discriminant #Delta# of a cubic polynomial in the form #ax^3+bx^2+cx+d# is given by the formula:

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

In our example, #a=5#, #b=-3#, #c=16# and #d=-6#, so we find:

#Delta = 2304-81920-648-24300+25920 = -78644#

Since #Delta < 0# this cubic has #1# Real zero and #2# non-Real Complex zeros, which are Complex conjugates of one another.

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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=25f(x)=125x^3-75x^2+400x-150#

#=(5x-1)^3+77(5x-1)-72#

#=t^3+77t-72#

where #t=(5x-1)#

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Cardano's method

We want to solve:

#t^3+77t-72=0#

Let #t=u+v#.

Then:

#u^3+v^3+(3uv+77)(u+v)-72=0#

Add the constraint #v=-77/(3u)# to eliminate the #(u+v)# term and get:

#u^3-456533/(27u^3)-72=0#

Multiply through by #27u^3# and rearrange slightly to get:

#27(u^3)^2-1944(u^3)-456533=0#

Use the quadratic formula to find:

#u^3=(1944+-sqrt((-1944)^2-4(27)(-456533)))/(2*27)#

#=(1944+-sqrt(3779136+49305564))/54#

#=(1944+-sqrt(53084700))/54#

#=(1944+-sqrt(30^2*58983))/54#

#=(972+-15sqrt(58983))/3^3#

Since this is Real and the derivation is symmetric in #u# and #v#, we can use one of these roots for #u^3# and the other for #v^3# to find Real root:

#t_1=1/3(root(3)(972+15sqrt(58983))+root(3)(972-15sqrt(58983)))#

and related Complex roots:

#t_2=1/3(omega root(3)(972+15sqrt(58983))+omega^2 root(3)(972-15sqrt(58983)))#

#t_3=1/3(omega^2 root(3)(972+15sqrt(58983))+omega root(3)(972-15sqrt(58983)))#

where #omega=-1/2+sqrt(3)/2i# is the primitive Complex cube root of #1#.

Now #x=1/5(1+t)#. So the roots of our original cubic are:

#x_n = 1/15(3+omega^(n-1)root(3)(972+15sqrt(58983))+omega^(1-n)root(3)(972-15sqrt(58983)))#