Question

Could you please tell me the answers and explain how to find them?

Answer 1

Answer:

$x=2,x=3+i\sqrt{2},x=3-i\sqrt{2}$

Explanation:

Given the function:

$y=x^3-8x^2+23x-22$

As seen on the graph, from the point (2,0) one of the roots of the function:

$x=2$

This means that x-2 is a factor of the polynomial y.

To get the other roots, first, divide y by x-2:

So, we have that:

$\begin{gathered} y=x^{3}-8x^{2}+23x-22 \\ =(x-2)(x^2-6x+11) \end{gathered}$

We then solve the quadratic equation for the other roots of y:

$\begin{gathered} x^2-6x+11=0 \\ x=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a},a=1,b=-6,c=11 \\ \\ x=\dfrac{-(-6)\pm\sqrt{(-6)^2-4(1)(11)}}{2\times1} \\ =\dfrac{6\pm\sqrt{36-44}}{2} \\ =\dfrac{6\pm\sqrt{-8}}{2} \\ =\frac{6}{2}\pm\frac{i\sqrt{8}}{2} \\ =3\pm\frac{i2\sqrt{2}}{2} \\ x=3\pm i\sqrt{2} \end{gathered}$

The other two roots of y are:

$\begin{gathered} x=3+i\sqrt{2} \\ \begin{equation*} x=3-\mathrm{i}\sqrt{2} \end{equation*} \end{gathered}$

Thus, the exact roots of y are:

$\begin{gathered} x=3+i\sqrt{2} \\ \begin{equation*} x=3-\mathrm{i}\sqrt{2} \end{equation*} \\ x=2 \end{gathered}$