Need help solving $x^4-3x^3-11x^2+3x+10=0$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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 $x=\pm1,\; x=-2,\; x=5$.