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Sagot :
Notice that Q(x) has a complex root; therefore, in order to get a polynomial with integer coefficients (real coefficients), we need to consider another root to be the complex conjugate of the given complex root.
In general, the conjugate of a complex number is
[tex]\begin{gathered} a+ib\rightarrow a-ib \\ \\ \\ \\ \end{gathered}[/tex]Thus, in our case,
[tex]1+i\rightarrow1-i[/tex]Then, Q(x) is
[tex]\Rightarrow Q(x)=(x+9)(x-1-i)(x-1+i)[/tex]Expanding Q(x),
[tex]\begin{gathered} Q(x)=(x+9)(x-(1+i))(x-(1-i)) \\ \Rightarrow Q(x)=(x+9)(x^2-((1+i)+(1-i))x+(1+i)(1-i) \\ \Rightarrow Q(x)=(x+9)(x^2-2x+2) \\ \end{gathered}[/tex][tex]\Rightarrow Q(x)=x^3+7x^2-16x+18[/tex]Thus, the answer is Q(x)=x^3+7x^2-16x+18
Calculations in more detail
[tex]\begin{gathered} (x-i-1)(x-1+i) \\ Set \\ z=i+1\text{ for simplicity} \\ \Rightarrow z^*=1-i \end{gathered}[/tex]Thus,
[tex]\begin{gathered} \Rightarrow(x-i-1)(x-1+i)=(x-z)(x-z^*)=x^2-zx-z^*x+zz^* \\ =x^2-(z+z^*)x+zz^* \end{gathered}[/tex]Then,
[tex]\begin{gathered} zz^*=(1+i)(1-i)=1+i-i-i^2=1-i^2=1-(-1)=2 \\ and \\ z+z^*=(1+i)+(1-i)=1+i+1-i=2 \end{gathered}[/tex]Therefore,
[tex]\Rightarrow(x-i-1)(x-1+i)=x^2-(2)x+2=x^2-2x+2[/tex]Then, we can express Q(x) as
[tex]\Rightarrow Q(x)=(x+9)(x^2-2x+2)[/tex]And we only need to multiply as shown above, but now all the numbers are integers; we do not need to deal with complex numbers anymore.
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