xplanation:
Step 1. e are given the following zeros of a polynomial function.
- 1 with a multiplicity of two, which means that is a zero two times:
[tex]\begin{gathered} x_1=1 \\ x_2=1 \end{gathered}[/tex]
and the other two zeros are:
[tex]\begin{gathered} x_3=2+\sqrt{2} \\ x_4=2-\sqrt{2} \end{gathered}[/tex]
And we need to find the equation that has these four zeros.
tep 2. 2R
[tex]f(x)=(x-x_1)(x-x_2)(x-x_3)(x-x_4)[/tex]
tep 3. S
[tex]f(x)=(x-1)(x-1)(x-(2+\sqrt{2}))(x-(2-\sqrt{2}))[/tex]
tep 4. S
[tex]f(x)=(x-1)(x-1)(x-2-\sqrt{2})(x-2+\sqrt{2})[/tex]
tep 5. T
[tex]f(x)=(x^2-2x+1)(x-2-\sqrt{2})(x-2+\sqrt{2})[/tex]
Then, let's multiply the second and third parentheses:
[tex]f(x)=(x^2-2x+1)(x^2-2x+x\sqrt{2}-2x+4-2\sqrt{2}-x\sqrt{2}+2\sqrt{2}-2)[/tex]
Multiple terms cancel and the result is:
[tex]f(x)=(x^2-2x+1)(x^2-4x+2)[/tex]
tep 6. TT
[tex]f(x)=x^4-4x^3+2x^2-2x^3+8x^2-4x+x^2-4x+2[/tex]
Combining like terms:
[tex]f(x)=x^4-6x^3+11x^2-8x+2[/tex]
This is shown in option D.
Answer:
D
[tex]f(x)=x^4-6x^3+11x^2-8x+2[/tex]
