math

x^3 - 3x^x2 +12x - 10  = 0

1 is a root

Using synthetic division, we can find the other two roots

1     1    - 3     12     -10

                1     -2      10

       1      -2    10        0

The remaining polynomial is

x^2 - 2x  + 10       this has no real roots

The other two roots are:

1- 3i    and   1 + 3i

what would the roots be for the equation x*x*x-3*x*x+12x-10

\(\begin{array}{lrcll} & x^3-3\cdot x^2+12\cdot x-10 &=& 0 \\ & 1^3-3\cdot 1^2-12\cdot 1-10 &=& 0 \\ \Rightarrow x_1=1 \end{array}\)

If we have \(x_1\), then we can calcutate \(x_2\) and \(x_3\)

\(\begin{array}{lrcll} x_2 &=& u+v\cdot i \\ x_3 &=& u-v\cdot i \\ \end{array}\)

We can calculate \(u\) and \(v\)

\(\begin{array}{rcll} a\cdot x^3+b\cdot x^2+c\cdot x+d &=& 0 \\ u &=& -\frac12 \cdot \left( x_1 + \frac{b}{a} \right) \\ v &=& \left( \sqrt{ \frac{d}{a\cdot x_1} + u^2 } \right) \cdot i \\ \end{array}\)

\(\begin{array}{rcll} x^3-3\cdot x^2+12\cdot x-10 &=& 0 \\\\ a\cdot x^3+b\cdot x^2+c\cdot x+d &=& 0 \quad & a = 1 \qquad b = -3 \qquad c = 12 \qquad d = -10 \qquad x_1 = 1\\ u &=& -\frac12 \cdot \left( 1 + \frac{-3}{1} \right) \\ &=& -\frac12 \cdot ( 1 -3 ) \\ &=& -\frac12 \cdot ( -2 ) \\ \mathbf{u} &\mathbf{=}& \mathbf{1} \\\\ v &=& \sqrt{ \frac{-10}{1\cdot 1} + 1^2 } \cdot i \\ &=& \sqrt{ \frac{-10}{1} + 1 } \cdot i \\ &=& \sqrt{ -10 + 1 } \cdot i \\ &=& \sqrt{ -9 } \cdot i \\ &=& \sqrt{ 9 } \cdot i \cdot i \\ &=& 3 \cdot i^2 \qquad & i^2 = -1\\ &=& 3 \cdot (-1)\\ \mathbf{v} &\mathbf{=}& \mathbf{-3} \end{array}\)

\(\begin{array}{lrcll} x_2 &=& u+v\cdot i \\ \mathbf{x_2} &\mathbf{=}& \mathbf{1-3\cdot i } \\\\ x_3 &=& u-v\cdot i \\ x_3 &=& 1-(-3)\cdot i \\ \mathbf{x_3} &\mathbf{=}& \mathbf{1+3\cdot i} \end{array}\)