Solve this Quartic Equation $x^4+3x^3+4x^2+2x+1=0$. I have tried with various possible solution methods. But I can't find the answer.
[Math] Solving 4th Degree Polynomial Equation
polynomials
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Great question I believe! You probably know this but this is said in MacTutor's Quadratic, cubic and quartic equation.
"After Tartaglia had shown Cardan how to solve cubics, Cardan encouraged his own student, Lodovico Ferrari, to examine quartic equations. Ferrari managed to solve the quartic with perhaps the most elegant of all the methods that were found to solve this type of problem. Cardan published all 20 cases of quartic equations in Ars Magna. Here, again in modern notation, is Ferrari's solution of the case: x4 + px2 + qx + r = 0. First complete the square to obtain
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In the years after Cardan's Ars Magna many mathematicians contributed to the solution of cubic and quartic equations. Viète, Harriot, Tschirnhaus, Euler, Bezout and Descartes all devised methods. Tschirnhaus's methods were extended by the Swedish mathematician E S Bring near the end of the 18th Century."
I just wanted to put it here to emphasize how "non-trivial" the problem of solving quartic equations can be. (I believe that you are aware of methods of finding rational roots, Uspensky's Theory of Equations is a great reference for that; but the general problem is definitely non trivial).
Here are many of this methods explained. I also found some of them here.
The Rational Root Test shows that the only possible rational solutions are $\pm 1$. Substituting gives that $x = -1$ is one (but $x = 1$ is not), so polynomial long division gives $p(x) = -(x + 1) q(x)$ for some quintic $q$. Substituting $x = -1$ gives that $-1$ is not a root of $q$, so if $q$ factors over $\Bbb Q$, it does so into an irreducible quadratic and an irreducible cubic. One can find such a factorization without too much effort (this is made easier by the fact that the leading and constant coefficients are both $1$): We get $$p(x) = -(x + 1)\underbrace{(x^2 - x + 1)(x^3 - x^2 - 2 x + 1)}_{q(x)} .$$ The discriminant of the quadratic is $-3 < 0$, so the real root you've identified must be a factor of the cubic; since the cubic has no rational roots, one needs to use Cardano's Formula or the equivalent to extract it.
Best Answer
Consider the function and its derivatives $$f(x)=x^4+3 x^3+4 x^2+2 x+1$$ $$f'(x)=4 x^3+9 x^2+8 x+2$$ $$f''(x)=12 x^2+18 x+8$$ The second derivative does not show any real root and then it is always positive (so, at most, two real roots). This implies that the first derivative can only cancel once.
Using Cardano method, the first derivative cancels at $$x_*=\frac{1}{4} \left(-3-\frac{5^{2/3}}{\sqrt[3]{3 \left(9+4 \sqrt{6}\right)}}+\frac{\sqrt[3]{5 \left(9+4 \sqrt{6}\right)}}{3^{2/3}}\right)\approx -0.39417$$ and $$f( -0.39417)\approx 0.673553$$ which is the minimum value of the function.
So, no real root to $f(x)=0$.