Question

3). Let g(x) = x^4 - 6x^3 + 13x^2 - 24x + 36. Find all complex zeros and multiplicities, if 2i is a complexzero. Apply the.

Answer 1

We will determine the zeroes of the expression as follows:

$\begin{gathered} x^4-6x^3+13x^2-24x+36=0\Rightarrow(x-3)^2(x^2+4)=0 \\ \\ \Rightarrow x-3=0\Rightarrow x=3 \\ \Rightarrow x^2+4=0\Rightarrow x^2=-4 \\ \\ \Rightarrow x=3 \\ \Rightarrow x=2i \\ \Rightarrow x=-2i \end{gathered}$

So, the zeros of the expression are at x = 3, x = 2i & x = -2i.

***Explanation***

First, we factor the expression, that is:

$x^4-6x^3+13x^2-24x+36=0\Rightarrow(x-3)^2(x^2+4)=0$

Now, after we factor we simply determine the values for which the component factors are 0, that is:

For (x - 3)^2 = 0:

$(x-3)^2=0\Rightarrow x-3=0\Rightarrow x=3$

So, the first zero is x = 3.

For x^2 + 4 = 0:

$x^2+4=0\Rightarrow x^2=-4$

Now, this will have two possible solutions:

$\begin{gathered} x=2i \\ x=-2i \end{gathered}$

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.