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For the quartic equation:

$$x^4 - x^3 + 4x^2 + 3x + 5 = 0$$

I tried Ferrari so far and a few others but I just can't get its complex solutions. I know it has no real solutions.

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4 Answers

Note that your polynomial is equal to $(x^2-2x+5)(x^2+x+1)$.

Remark: Either the polynomial factors nicely over the integers, or we have kind of a mess. If it factors nicely, then the constant terms of the factors are $1$ and $5$, or $-1$ and $-5$.

Let's be positive and test $1$ and $5$ first. Let the coefficients of $x$ in the factors be $a$ and $b$ respectively. The coefficient of $x^3$ in the product is then $a+b$. This must be $-1$. The coefficient of $x$ in the product is $5a+b$. This must be $3$. Solve. If we are lucky, the coefficient of $x^2$ in the product will turn out to be $4$. It is.

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Yeah but how would I spot that, on a conceptual level. –  Tool Jun 7 '12 at 17:46
    
Either the polynomial factors pleasantly as a product of simple quadratics, or we have a real mess. If pleasantly, constant terms are $1$ and $5$, or $-1$ and $-5$. Play a bit, one bumps into it. –  André Nicolas Jun 7 '12 at 17:49
    
@Tool, every polynomial with real coefficients factors over the real numbers into quadratics and linear, and this may be useful if you are lucky, i.e. a human being made up the problem. For example, $$ x^4 + 4 = (x^2 + 2 x + 2) (x^2 - 2 x + 2) $$ –  Will Jagy Jun 7 '12 at 18:02
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$$x^4 - x^3 + 4x^2 + \underbrace{3x}_{4x-x} + \overbrace{5}^{4+1} = \\\color{red}{x^4-x^3}+4x^2+4x+4\color{red}{-x+1}\\={x^4-x^3}-x+1+4(x^2+x+1)\\={x^3(x-1)}-(x-1)+4(x^2+x+1)\\=(x-1)(x^3-1)+4(x^2+x+1)\\=(x-1)(x-1)(x^2+x+1)+4(x^2+x+1)\\=(x^2+x+1)(x-1)^2+4)=(x^2+x+1)(x^2-2x+5)$$

As for the roots, I assume you could solve those two quadratic equations and you could find the results on Wolfram|Alpha.

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+1 Nice regrouping of terms. –  Jyrki Lahtonen Jun 7 '12 at 18:09
    
+1 Elegant answer –  B. S. Jun 8 '12 at 9:47
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Let $f(x) = x^4 - x^3 +4x^2 +3x+5$. Once you know that $f(x)$ doesn't have real solutions, try some easy complex numbers like $i$, $\omega$, other $n^{th}$ roots of unity etc. Note that if we plug in $x = \omega$, where $\omega$ is the complex cube-root of unit, we get that $$f(\omega) = \omega^4 - \omega^3 +4 \omega^2 + 3 \omega + 5 = \omega - 1 + 4 \omega^2 + 3 \omega +5 = 4(\omega^2 + \omega + 1) = 0.$$ Hence, $x^2 + x + 1$ divides $f(x)$. Hence, $f(x) = (x^2 + x + 1) (x^2 + ax + b)$. This gives us that $$f(x) = x^4 + (a+1)x^3 + (1 + a + b)x^2 +(a+b)x + b$$ Hence, $a=-2$ and $b=5$. Hence, $$f(x) = (x^2 + x + 1) (x^2 - 2x + 5)$$ Hence, the roots are $$x = \omega, \omega^2, 1 \pm 2i$$

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@downvoter: Would you care to explain why you down voted? –  user17762 Jun 7 '12 at 18:18
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I was in a bad mood, and see no intuition behind your magical guessing the roots of the equation. No I see, and downvote removed. –  Norbert Jun 23 '12 at 11:21
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Use the magic quartic formula!

$$a = -1, b = 4, c = 3, d = 5$$

$$u = -\frac{29}{12}, \Delta_0 = 85, \Delta_1 = -826, Q = \sqrt[3]{666i-541}$$ The discriminant is 65712, meaning that there are either four real or four complex roots.

$$v = \frac{(\sqrt[3]{666i-541})^2 + 85}{3\sqrt[3]{666i-541}} = \frac{(666i-541)\sqrt[3]{666i-541} + 85(\sqrt[3]{666i-541})^2}{3(666i-541)}$$

$$=\frac{736237\sqrt[3]{666i-541} - 85(666i+541)(\sqrt[3]{666i-541})^2}{2208711}$$

And you should be able to take it from there.

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