Question

Use the Rational Zeros Theorem to find all the real zeros of each polynomial function. Use the zeros to factor f over the real numbers.

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

Short answer

The real zeroes of the function is 2 with multiplicity 2 and the factored form is$f\left(x\right)=(x-2)(x-2)(x^2+5)$

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

Since the degree of the function is 4 so it has atmost 4 real zeroes.

By descarte's rule of signs, there are 4, 2 or 0 positive zeroes as there are 4 variations in sign of the function.

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

So, there are no negative zeroes as there are no variations in sign in$f(-x)$

Step 3. Calculation

As the leading coefficient of the function, $a_4=1$and the constant term is 20, the potential rational zeroes are 1,2,4,5,10 and 20.

The negative integers are neglected because there are no negative zeroes for the polynomial function.

Test the potential zero 1, using the synthetic division

Since, the remainder is 6, 1 is not a zero of the function.

Next, take the value 2 and perform the synthetic division.

The remainder of this is 0. So, 2 is a zero and$x-2$is a factor of the function.

Step 4. Calculation

Using the entries in the bottom row of the above synthetic division, the function can be factored as follows,

$$\begin{gathered} f\left(x\right)=x^4-4x^3+9x^2-20x+20 \\ =(x-2)(x^3-2x^2+5x-10) \end{gathered}$$

Now, the polynomial $x^3-2x^2+5x-10$is again factored using synthetic division by taking the value 2.

The remainder of$f\left(2\right)=0$. So, 2 is a repeated zero and$x-2$is a repeated factor of the function.

Step 5. Calculation

Using the above entries in the bottom row of the synthetic division, the function can be factored.

$$\begin{gathered} f\left(x\right)=(x-2)(x^3-2x^2+5x-10) \\ =(x-2)(x-2)(x^2+5) \end{gathered}$$

Use the zero product property to check the remaining zeroes satisfy the new equation.

$$\begin{gathered} x^2+5=0 \\ {x}^2=-5 \\ x=\sqrt{-5} \end{gathered}$$

Since, the obtained number is an imaginary number, it is not a real zero of the function.