ANSWER
A. x = 2, √11, -√11
EXPLANATION
To find the zeros of f, which is a 3rd-degree polynomial, we can try some common values to see if any of them is a zero: 1, 2, -1, -2. In this case 2 is a zero of the function:
![\begin{gathered} f(2)=2\cdot2^3-4\cdot2^2-22\cdot2+44 \\ f(2)=2\cdot8-4\cdot4-44+44 \\ f(2)=16\cdot16 \\ f(2)=0 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20f%282%29%3D2%5Ccdot2%5E3-4%5Ccdot2%5E2-22%5Ccdot2%2B44%20%5C%5C%20f%282%29%3D2%5Ccdot8-4%5Ccdot4-44%2B44%20%5C%5C%20f%282%29%3D16%5Ccdot16%20%5C%5C%20f%282%29%3D0%20%5Cend%7Bgathered%7D)
Now we can reduce our polynomial by dividing it by the factor (x - 2) and obtain a 2nd-degree polynomial, whose zeros are easier to find:
So we can rewrite f(x) as:
![f(x)=(x-2)(2x^2-22)](https://tex.z-dn.net/?f=f%28x%29%3D%28x-2%29%282x%5E2-22%29)
To find the other two zeros we have to solve:
![2x^2-22=0](https://tex.z-dn.net/?f=2x%5E2-22%3D0)
Add 22 to both sides of the equation:
![\begin{gathered} 2x^2-22+22=0+22 \\ 2x^2=22 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%202x%5E2-22%2B22%3D0%2B22%20%5C%5C%202x%5E2%3D22%20%5Cend%7Bgathered%7D)
Divide both sides by 2:
![\begin{gathered} \frac{2x^2}{2}=\frac{22}{2} \\ x^2=11 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7B2x%5E2%7D%7B2%7D%3D%5Cfrac%7B22%7D%7B2%7D%20%5C%5C%20x%5E2%3D11%20%5Cend%7Bgathered%7D)
And take square root. Remember that the square root has a positive and a negative result:
![\begin{gathered} \sqrt[]{x^2}=\pm\sqrt[]{11} \\ x=\pm\sqrt[]{11} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Csqrt%5B%5D%7Bx%5E2%7D%3D%5Cpm%5Csqrt%5B%5D%7B11%7D%20%5C%5C%20x%3D%5Cpm%5Csqrt%5B%5D%7B11%7D%20%5Cend%7Bgathered%7D)
So the other two zeros are √11 and -√11.
The three real zeros of f are 2, √11 and -√11