the issue I'm having is this problem:
If $x=2$ is one of the solutions to $x^3-6x^2+13x=10$, then find the other solutions.
It says the other solutions are:
$x = 2+i$ and $x = 2-i$.
Can anyone show me how those are the solutions? This is really bothersome and confusing, thank you!
$\endgroup$ 33 Answers
$\begingroup$Firstly, if $x=a$ is a solution, then $(x-a)$ is a factor of the polynomial so you can divide it out. In your case $x=2$ is a solution so you should divide the polynomial by a factor of $(x-2)$.
$$ \frac{x^3-6x^2+13x-10}{x-2} = 0 $$ which gives you $$ x^2-4x+5=0 \quad\text{for}\, x\neq2$$ use the quadratic formula $$ \frac{4 \pm \sqrt{-4}}{2} = 2 \,\pm\,\sqrt{-1} $$ $$ x = 2 \pm i $$
$\endgroup$ $\begingroup$Here I have showed you how to get the answer with full detailed working.
$\begin{array}{lllllll} &&&+x^2&-4x&+5\\ \\ x&-2&&+x^3&-6x^2&+13x&-10\\ &&&+x^3&-2x^2\\ \\ &&&&-4x^2&+13x\\ &&&&-4x^2&+8x\\ \\ &&&&&+5x&-10\\ &&&&&+5x&-10 \end{array}$
The quotient $x^2-4x+5$
is the LHS of the equation you need to solve. 0 is the RHS.
