Need help solving $x^4-3x^3-11x^2+3x+10=0$ Solve $x^4-3x^3-11x^2+3x+10=0$
I have tried to solve this equation using 'general formula from roots' from https://en.wikipedia.org/wiki/Quartic_function.
$$ax^4+bx^3+cx^2+dx+e=0$$
$$x_{1,2}=-\frac b{4a}-S \pm 0.5\sqrt{-4S^2-2P+ \frac q S}$$
$$x_{3,4}=-\frac b{4a} + S \pm 0.5\sqrt{-4S^2-2P-\frac q S}$$
$$p=\frac{8ac-3b^2}{8a^2}$$
$$q= \frac{b^3-4abc+8a^2d}{8a^3}$$
$$S=0.5\sqrt{-\frac{2p} 3 + \frac{Q+\Delta_0} Q}{3a}$$
$$Q=\left(\Delta_1+\sqrt{\Delta_1^2-4\Delta_0^3} \cdot 0.5\right)^{1/3}$$
$$\Delta_0=c^2-3bd+12ae$$
$$\Delta_1=2c^3-9bcd+27eb^2+27ad^2-72ace$$
Those formula does not work. I ended up with complex roots.
Please help me to use those formulas to solve quartic equation. Any help will be very much appreciated.
 A: One can check that $1$ and $-1$ are both roots of this polynomial. After factoring out $x-1$ and $x+1$, you're left with a quadratic polynomial.
As a general rule, the rational root theorem is a good place to start for questions like these.
A: We have
$$x^4-3x^3-11x^2+3x+10$$
Using rational root test:
$$(x-1)(x^3-2x^2-13x-10)$$
And rational root test again:
$$(x-1)(x+1)(x^2-3x-10)$$
Factoring quadratic:
$$(x-1)(x+1)(x-5)(x+2)$$
Thus, roots are $x=\pm1,\; x=-2,\; x=5$.
A: I coded up the formula in the Wikipedia page and got the right results. Perhaps you made a mistake in your program.  The intermediate values, using Wikipedia's variable names, are:
$p = -14.375$
$q = -16.875$
$\Delta_0 = 268$
$\Delta_1 = 7040$
$\sqrt{\Delta_1^2 - 4\Delta_0^3} = 5237.721642i$
$Q^3 = 3520 + 2618.860821i$
$Q = 16 + 3.464102i$
$S = 2.25$
$x = \{ -1, -2, 5, 1 \}$
Of course, in complex numbers there are three possible cube roots for $Q^3$. I tried all three and it gave the same roots $x_n$ in a different order .. surprising but pleasing :)
Taking the other square root in the above list results in $Q$ being conjugated, and $S$ stays the same.
A: $$x^4 - 3x^3 - 11x^2 + 3x + 10 = 0$$
$$x^4 - 3x^3 - 10x^2 - x^2 + 3x + 10 = 0$$
$$x^2(x^2 - 3x - 10) - (x^2 - 3x - 10) = 0$$
$$(x^2 - 1)(x^2 - 3x - 10) = 0$$
$$(x - 1)(x + 1)(x - 5)(x + 2) = 0$$
$$x = \{1,-1,5,2\}$$
