Respuesta :
We will determine the zeroes of the expression as follows:
[tex]\begin{gathered} x^4-6x^3+13x^2-24x+36=0\Rightarrow(x-3)^2(x^2+4)=0 \\ \\ \Rightarrow x-3=0\Rightarrow x=3 \\ \Rightarrow x^2+4=0\Rightarrow x^2=-4 \\ \\ \Rightarrow x=3 \\ \Rightarrow x=2i \\ \Rightarrow x=-2i \end{gathered}[/tex]So, the zeros of the expression are at x = 3, x = 2i & x = -2i.
***Explanation***
First, we factor the expression, that is:
[tex]x^4-6x^3+13x^2-24x+36=0\Rightarrow(x-3)^2(x^2+4)=0[/tex]Now, after we factor we simply determine the values for which the component factors are 0, that is:
For (x - 3)^2 = 0:
[tex](x-3)^2=0\Rightarrow x-3=0\Rightarrow x=3[/tex]So, the first zero is x = 3.
For x^2 + 4 = 0:
[tex]x^2+4=0\Rightarrow x^2=-4[/tex]Now, this will have two possible solutions:
[tex]\begin{gathered} x=2i \\ x=-2i \end{gathered}[/tex]So, the zeros for this factor are x = 2i and x = -2i.
Finally, we will have that all the zeros of the expression are x = -2i, x = 2i and x = 3.
