Real-coefficient polynomial with roots 2, 1 ± i; write the monic cubic.
Solution
Correct Answer: Option D
We are given roots \(2,\;1+i,\;1-i\) and asked for the monic cubic polynomial with real coefficients.
Since complex roots come in conjugate pairs for real-coefficient polynomials, the quadratic factor from \(1\pm i\) is
\[
(x-(1+i))(x-(1-i))=((x-1)-i)((x-1)+i)=(x-1)^2+1 = x^2-2x+2.
\]
Multiplying by the linear factor for the root \(2\) gives the cubic:
\[
(x-2)(x^2-2x+2).
\]
Expand:
\[
\begin{aligned}
(x-2)(x^2-2x+2) &= x^3-2x^2+2x -2x^2+4x-4\\
&= x^3-4x^2+6x-4.
\end{aligned}
\]
So the monic cubic is \(x^3-4x^2+6x-4\), which matches Option 1.
Since complex roots come in conjugate pairs for real-coefficient polynomials, the quadratic factor from \(1\pm i\) is
\[
(x-(1+i))(x-(1-i))=((x-1)-i)((x-1)+i)=(x-1)^2+1 = x^2-2x+2.
\]
Multiplying by the linear factor for the root \(2\) gives the cubic:
\[
(x-2)(x^2-2x+2).
\]
Expand:
\[
\begin{aligned}
(x-2)(x^2-2x+2) &= x^3-2x^2+2x -2x^2+4x-4\\
&= x^3-4x^2+6x-4.
\end{aligned}
\]
So the monic cubic is \(x^3-4x^2+6x-4\), which matches Option 1.
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