How do you factor $x^3 - 6x^2 + 3x - 10$ ?

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Answer 1 · 2016-05-09T20:22:48

See explanation...

Use Cardano's method.

First a simple Tschirnhaus transformation, to eliminate the term of degree $2$ ...

$0 = x^3-6x^2+3x-10=(x-2)^3-9(x-2)-20$

Let $t=x-2$ to get the simplified cubic equation:

$t^3-9t-20 = 0$

Next substitute $t = u + v$ to get:

$u^3+v^3+3(uv-3)(u+v)-20 = 0$

Add the constraint $v = 3/u$ to eliminate the term in $(u+v)$ and get:

$u^3+(3/u)^3-20=0$

Multiply through by $u^3$ to get:

$(u^3)^2-20(u^3)+27 = 0$

$u^3 = (20+-sqrt(20^2-(4*1*27)))/(2*1)$

$=10+-sqrt(400-108)/2$

$=10+-sqrt(292)/2$

$=10+-sqrt(4*73)/2$

$=10+-sqrt(73)$

The derivation was symmetric in $u$ and $v$ , hence (noting $x = t+2$ ) we can deduce that the Real zero of the original cubic is:

$x_1 = 2+root(3)(10-sqrt(73))+root(3)(10+sqrt(73))$

and Complex zeros:

$x_2 = 2+omega root(3)(10-sqrt(73))+omega^2 root(3)(10+sqrt(73))$

$x_3 = 2+omega^2 root(3)(10-sqrt(73))+omega root(3)(10+sqrt(73))$

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

Then:

$x^3-6x^2+3x-10 = (x-x_1)(x-x_2)(x-x_3)$

How do you factor x^3 - 6x^2 + 3x - 10x^3 - 6x^2 + 3x - 10?