Third degree polynomials like this one are not easily solved, but this one has a root at x = 0. The let us factorize this polynomial as x times a second degree polynomial:

$\begin{gathered} x^3-6x^2+6x=0 \\ x(x^2-6x+6)=0 \end{gathered}$

Now we can find the roots of the quadratic polynomial as:

$\begin{gathered} x=\frac{-(-6)\pm\sqrt[]{(-6)^2-4\cdot1\cdot6}}{2\cdot1} \\ x=\frac{6\pm\sqrt[]{36-24}}{2} \\ x=\frac{6\pm\sqrt[]{12}}{2} \\ x=\frac{6\pm\sqrt[]{4\cdot3}}{2} \\ x=\frac{6\pm2\sqrt[]{3}}{2} \\ x=3\pm\sqrt[]{3} \\ x_1=3-\sqrt[]{3} \\ x_2=3+\sqrt[]{3} \end{gathered}$

Then, the solutions to the equation are:

x = 0

x = 3 - √3

x = 3 + √3

4 0

1 year ago

\frac{5}{2}\mathrm{g}-\frac{1}{2}\mathrm{g}= 2 5 g− 2 1 g simlily

antoniya [11.8K]

Answer:

Step-by-step explanation:

7 0

3 years ago

Your class has raised $760 for a hunger relief organization. The organization provides farm animals that people can use to produ

melomori [17]

Answer:

760 for hunder relif oragniz

5 0

2 years ago