Answer:
The real roots are [tex]x_1 = 2[/tex], [tex]x_2=1-\sqrt{7}\approx -1.65[tex] and [tex]x_3=1+\sqrt{7}\approx 3.65[/tex].
Step-by-step explanation:
In order to find the real solutions we should factor the given polynomial. Using the Ruffini-Horner rule we can find that
[tex] x^3-4x^2-2x+12 = (x-2)(x^2-2x+6).[/tex]
From here it is easy to affirm that [tex]x_1=2[/tex] is one real root of the equation. For the other two we use the general formula for second degree equations:
[tex]x_{2,3} = \frac{2\pm \sqrt{(-2)^2-4\cdot 1\cdot(-6)}}{2} = \frac{2\pm \sqrt{28}}{2} = \frac{2+2\sqrt{7}}{2} = 1\pm\sqrt{7}.[/tex]
In the chain of equalities above we used that [tex]\sqrt{28}=\sqrt{4\cdot 7} = \sqrt{4}\sqrt{7} = 2\sqrt{7}[/tex].
Then, the other two roots are [tex]x_2 = 1-\sqrt{7}[/tex] and [tex] x_3 = 1+\sqrt{7}[/tex].
