Question

Find the complex zeros of each polynomial function f(x). Write f in factored form.

$f\left(x\right)=x^4-4x^3+9x^2-20x+20$

Short answer

The factored form is$f\left(x\right)={(x-2)}^2(x+\sqrt{5}i)(x-\sqrt{5}i)$

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=x^4-4x^3+9x^2-20x+20$

Step 2. Explanation  

The given function has a degree 4. So, there will be a maximum of 4 complex zeroes.

We can see from the function that it has three sign variations. So, there will be three or one positive real zero.

$$\begin{gathered} f(-x)={(-x)}^4-4{(-x)}^3+9{(-x)}^2-20(-x)+20 \\ =x^4+4x^3+9x^2+20x+20 \end{gathered}$$

Since there is no sign change in $f(-x)$, there is no negative real zero.

Step 3. Explanation  

We will list the integers which are factors of the constant term 20. $p=\pm 1,\pm 2,\pm 4.\pm 5,\pm 10$

Now, we will list the integers which are factors of the leading coefficient 1.

$q=\pm 1$

Next, list the possible potential rational zeroes which is the ratio$\frac{p}{q}$

$\frac{p}{q}=\pm 1,\pm 2,\pm 4,\pm 5,\pm 10$

Thus, the potential rational zero of function are 1,2,4,5 and 10.

Step 4. Calculation  

Using the synthetic division check whether 2 is a zero of function or not.

Thus, the depressed equation of function is $x^3-2x^2+5x-10=0$

Now, factorize the above equation by grouping.

$$\begin{gathered} (x^3-2x^2)+(5x-10)=0 \\ {x}^2(x-2)+5(x-2)=0 \\ (x-2)(x^2+5)=0 \\ x-2=0\Rightarrow x=2 \\ {x}^2+5=0\Rightarrow x=\pm \sqrt{5}i \end{gathered}$$

Thus, four complex zeroes of function are$\left\{2(multiplicity2),\sqrt{5}i,-\sqrt{5}i\right\}$