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I need to decompose (in $\Bbb{C}[x]$) the function

$$ f(x) = x^4 + 4x^3 - 4x^2 + 24x + 15 $$

in its simplest form, knowing that $1 - 2i$ is one of its roots. Any ideas?

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15  
Since the coefficients are real, $1+2i$ is another of its roots. Dividing $f$ by $(x-(1+2i))(x-(1-2i))$ will reduce it to a quadratic, which you can solve easily. –  Nick Thompson Jun 15 at 17:51
2  
If $\alpha +i\beta$ is a root, so is $\alpha -i\beta$. Thus $\underbrace{(x-(\alpha +i\beta))(x-(\alpha -i\beta))}_{\in \mathbb Q[x]}$ is a factor of the given polynomial. Proceed with polynomial division to get the other factor of degree $2$. –  Git Gud Jun 15 at 17:51
    
You know a second root. $1+2i$ multiply together the two linear factors with these roots and divide into your polynomial. –  Rene Schipperus Jun 15 at 17:52
    
After diving diving f by (x-(1+2i))(x-(1-2i)) = x^2 -2x + 5 I got a remainder of -10. But i'm pretty sure the long division is correct. Did I do something wrong? I got the polynomial x^2+6x+3 –  user157246 Jun 15 at 18:03
    
The polynomial $x^2+6x+3$ is right. –  André Nicolas Jun 15 at 18:04

2 Answers 2

Hint: Another root is $1+2i$. The sum and product of the two roots we know are $2$ and $5$.

The sum of all the roots is $-4$, and the product of all the roots is $15$. So the sum of the missing roots is $-6$, and the product is $3$.

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The answer has already been given in the comments, so I'll make a community answer out of it.

So we have the function $$f(x) = x^4 + 4x^3 - 4x^2 + 24x + 15$$ One root has been found at $x = 1-2i$.

Since the coefficients are real, there must be another root at $x = 1+2i$. We can now reduce this equation to a quadratic one, by dividing those out.

$$\begin{align} & \frac{x^4 + 4x^3 - 4x^2 + 24x + 15}{(x-(1-2i))(x-(1+2i))} \\ = \ & \frac{x^4 + 4x^3 - 4x^2 + 24x + 15}{(x-1+2i)(x-1-2i)} \\ = \ & \frac{x^4 + 4x^3 - 4x^2 + 24x + 15}{x^2-2x+5} \\ = \ & x^2+6x+3 \end{align}$$

The roots for that quadratic equation can be found at $x = -3-\sqrt{6}$ and $x = -3 + \sqrt{6}$.

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