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:
[tex]\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}[/tex]
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:
[tex]f(x)=(x-2)(2x^2-22)[/tex]
To find the other two zeros we have to solve:
[tex]2x^2-22=0[/tex]
Add 22 to both sides of the equation:
[tex]\begin{gathered} 2x^2-22+22=0+22 \\ 2x^2=22 \end{gathered}[/tex]
Divide both sides by 2:
[tex]\begin{gathered} \frac{2x^2}{2}=\frac{22}{2} \\ x^2=11 \end{gathered}[/tex]
And take square root. Remember that the square root has a positive and a negative result:
[tex]\begin{gathered} \sqrt[]{x^2}=\pm\sqrt[]{11} \\ x=\pm\sqrt[]{11} \end{gathered}[/tex]
So the other two zeros are √11 and -√11.
The three real zeros of f are 2, √11 and -√11
