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Knowing that $x_1 = 1-i$ is one of the roots of $P(x) = x^3 - (1-i)x^2 + x + (-1+i)$, find all the roots of P(x).

I only ever solved such problems for polynomials with real coefficients - then I could apply the complex conjugate root theorem and I also could do long division after finding successive roots to simplify the main polynomial. I don't know how to approach this problem.

Can I do long division of $P(x)$ by $(x-(1-i))$ somehow?

I'm looking more for universal solutions for approaching this kind of problems rather than tricks like factoring the polynomial because it may not work for similar polynomials.

The answer is: $x_1 = 1-i$, $x_2 = i$, $x_3 = -i$.

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  • $\begingroup$ Not an answer to your question, but alternatively you could spot from the coefficients of $P(x)$ that the sum of the roots is $1-i$, and the products of the roots is $-(-1+i)=1-i$. Knowing tat $1-i$ is a root, the sum of the remaining two roots is $0$ and their product is $1$. $\endgroup$ – Servaes Jan 5 '18 at 12:50
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Can I do long division of $P(x)$ by $(x-(1-i))$ somehow?

Yes! The same way you do long division of polynomial with real coefficients: Horner's algorithm. It gives you

$$P(x)=(x-(1-i))(x^2+1)$$ which you can also get through a sharp eye as seen by @mathlove.

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  • $\begingroup$ Thank you, I tried both Horner's algorithm and long division and got correct result. $\endgroup$ – Matt Jan 5 '18 at 13:19
  • $\begingroup$ Horner's algorithm is long division... $\endgroup$ – 5xum Jan 5 '18 at 13:19
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We have $$\color{red}{x^3-(1-i)x^2}+x+(-1+i)=\color{red}{x^2(x-1+i)}+(x-1+i)=(x-1+i)(x^2+1)$$

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$$x^3-(1-i)x^2-1+i=x^3-x^2+x-1+(x^2+1)i=$$ $$=x^2(x-1)+(x-1)+(x^2+1)i=(x^2+1)(x-1)+(x^2+1)i=(x^2+1)(x-1+i).$$

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