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So I'm asked to find the complex roots for the polynomial $f=x^4+3x-2$, knowing that it shares one of its roots with the polynomial $g=x^4+3x^3-3x+1$

My problem, however, is that when I try to find the GCD of f and g I get $(f:g)=1$, which would mean that f and g have no common roots. Am I wrong to assume this? Because else the GCD would be a grade 1 polynomial (if they share a root)

Maybe I made a mistake when calculating the GCD? (I did this by hand and they were many divisions through Euclid's Algorithm, but I can't find a mistake)

Thanks in advance

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    $\begingroup$ Hint: $x^4+3x-2 = (x^2 + x - 1) (x^2 - x + 2)$ and $x^4+3x^3-3x +1 = (x^2+2x-1)(x^2+x-1)$ $\endgroup$ – gammatester Oct 12 '17 at 20:35
  • $\begingroup$ Thanks, I'll try again later then $\endgroup$ – Francisco José Letterio Oct 12 '17 at 20:37
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You made a mistake. Their GCD is $x^2 + x - 1$.

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Let $a$ be the common root. Therefore, $f(a)=g(a)=0$. But $f(a)=g(a)$ implies $$a^4+3a-2 = a^4+3a^3-3a+1 \implies a^3-2a+1 = 0 \implies (a-1)(a^2+a-1)=0.$$ Since $a=1$ is not the root of $f(x)$ and $g(x)$, we solve $$a^2+a-1=0 \implies a = \frac{-1\pm\sqrt{5}}{2}.$$ Indeed, $$f\left(\frac{-1\pm\sqrt{5}}{2}\right)=g\left(\frac{-1\pm\sqrt{5}}{2}\right)=0.$$ Can you find the other roots of $f(x)$?

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  • $\begingroup$ Ohhh, thanks. This is much easier than running the algorithm. Thanks! I'll try to get the other roots when I get home $\endgroup$ – Francisco José Letterio Oct 12 '17 at 20:39
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The steps of Euclid:

$$\begin{cases}x^4+3x-2,\\x^4+3x^3-3x+1.\end{cases}$$

$$\begin{cases}x^4+3x-2,\\x^3-2x+1.\end{cases}$$

$$\begin{cases}x^2+x-1,\\x^3-2x+1.\end{cases}$$

$$\begin{cases}x^2+x-1,\\x^2+x-1.\end{cases}$$

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