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

Question

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

1 Answer

Impact of this question

4273 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-09T20:22:48

See explanation...

Explanation:

Use Cardano's method.

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

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

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

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

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

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

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

$u^{3} + {(\frac{3}{u})}^{3} - 20 = 0$ $u^3+(3/u)^3-20=0$

Multiply through by $u^{3}$ $u^3$ to get:

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

Use the quadratic formula to find:

$u^{3} = \frac{20 \pm \sqrt{20^{2} - (4 \cdot 1 \cdot 27)}}{2 \cdot 1}$ $u^3 = (20+-sqrt(20^2-(4*1*27)))/(2*1)$

$= 10 \pm \frac{\sqrt{400 - 108}}{2}$ $=10+-sqrt(400-108)/2$

$= 10 \pm \frac{\sqrt{292}}{2}$ $=10+-sqrt(292)/2$

$= 10 \pm \frac{\sqrt{4 \cdot 73}}{2}$ $=10+-sqrt(4*73)/2$

$= 10 \pm \sqrt{73}$ $=10+-sqrt(73)$

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

$x_{1} = 2 + \sqrt[3]{10 - \sqrt{73}} + \sqrt[3]{10 + \sqrt{73}}$ $x_1 = 2+root(3)(10-sqrt(73))+root(3)(10+sqrt(73))$

and Complex zeros:

$x_{2} = 2 + ω \sqrt[3]{10 - \sqrt{73}} + ω^{2} \sqrt[3]{10 + \sqrt{73}}$ $x_2 = 2+omega root(3)(10-sqrt(73))+omega^2 root(3)(10+sqrt(73))$

$x_{3} = 2 + ω^{2} \sqrt[3]{10 - \sqrt{73}} + ω \sqrt[3]{10 + \sqrt{73}}$ $x_3 = 2+omega^2 root(3)(10-sqrt(73))+omega root(3)(10+sqrt(73))$

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

Then:

$x^{3} - 6 x^{2} + 3 x - 10 = (x - x_{1}) (x - x_{2}) (x - x_{3})$ $x^3-6x^2+3x-10 = (x-x_1)(x-x_2)(x-x_3)$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor x^3 - 6x^2 + 3x - 10x3−6x2+3x−10x^3 - 6x^2 + 3x - 10?

1 Answer

Explanation:

Related questions

Impact of this question

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