The problem presents three equations: 1. $6x^3 + x^2 - 19x + 6 = 0$

1. Problem Description

The problem presents three equations:

1. $6x^3 + x^2 - 19x + 6 = 0$

2. $9x^3 + 10x^2 - 17x = 2$

3. $x^4 - 9x^2 - 4x + 12 = 0$

We need to find the solutions for each equation.

2. Solution Steps

Let's solve each equation separately.

Equation 1: $6x^3 + x^2 - 19x + 6 = 0$

We can try to find a rational root using the Rational Root Theorem. The possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6, \pm \frac{1}{2}, \pm \frac{3}{2}, \pm \frac{1}{3}, \pm \frac{2}{3}, \pm \frac{1}{6}$.

Let's try $x = \frac{1}{2}$:

$6(\frac{1}{2})^3 + (\frac{1}{2})^2 - 19(\frac{1}{2}) + 6 = 6(\frac{1}{8}) + \frac{1}{4} - \frac{19}{2} + 6 = \frac{3}{4} + \frac{1}{4} - \frac{38}{4} + \frac{24}{4} = \frac{3+1-38+24}{4} = \frac{-10}{4} \neq 0$

Let's try $x = \frac{2}{3}$:

$6(\frac{2}{3})^3 + (\frac{2}{3})^2 - 19(\frac{2}{3}) + 6 = 6(\frac{8}{27}) + \frac{4}{9} - \frac{38}{3} + 6 = \frac{16}{9} + \frac{4}{9} - \frac{114}{9} + \frac{54}{9} = \frac{16+4-114+54}{9} = \frac{-40}{9} \neq 0$

Let's try $x = -3$:

$6(-3)^3 + (-3)^2 - 19(-3) + 6 = 6(-27) + 9 + 57 + 6 = -162 + 9 + 57 + 6 = -162 + 72 = -90 \neq 0$

Let's try $x = 2$:

$6(2)^3 + (2)^2 - 19(2) + 6 = 6(8) + 4 - 38 + 6 = 48 + 4 - 38 + 6 = 58 - 38 = 20 \neq 0$

Let's try $x = 3/2$.

$6(3/2)^3 + (3/2)^2 - 19(3/2) + 6 = 6(27/8) + (9/4) - 57/2 + 6 = 81/4 + 9/4 - 114/4 + 24/4 = (81+9-114+24)/4 = 0/4 = 0$. So $x = 3/2$ is a root.

Now we can divide the polynomial by $(x - 3/2)$ or $(2x - 3)$.

$(6x^3 + x^2 - 19x + 6) / (2x - 3) = 3x^2 + 5x - 2$

Now we solve the quadratic equation $3x^2 + 5x - 2 = 0$.

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4(3)(-2)}}{2(3)} = \frac{-5 \pm \sqrt{25 + 24}}{6} = \frac{-5 \pm \sqrt{49}}{6} = \frac{-5 \pm 7}{6}$

So $x = \frac{-5+7}{6} = \frac{2}{6} = \frac{1}{3}$ and $x = \frac{-5-7}{6} = \frac{-12}{6} = -2$.

Therefore, the solutions to the first equation are $x = 3/2, x = 1/3, x = -2$.

Equation 2: $9x^3 + 10x^2 - 17x = 2 \implies 9x^3 + 10x^2 - 17x - 2 = 0$

Let's try $x = -2$: $9(-2)^3 + 10(-2)^2 - 17(-2) - 2 = 9(-8) + 10(4) + 34 - 2 = -72 + 40 + 34 - 2 = -74 + 74 = 0$

So $x = -2$ is a root.

Divide the polynomial by $(x + 2)$:

$(9x^3 + 10x^2 - 17x - 2) / (x + 2) = 9x^2 - 8x - 1$

Now we solve the quadratic equation $9x^2 - 8x - 1 = 0$.

Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4(9)(-1)}}{2(9)} = \frac{8 \pm \sqrt{64 + 36}}{18} = \frac{8 \pm \sqrt{100}}{18} = \frac{8 \pm 10}{18}$

So $x = \frac{8+10}{18} = \frac{18}{18} = 1$ and $x = \frac{8-10}{18} = \frac{-2}{18} = -\frac{1}{9}$.

Therefore, the solutions to the second equation are $x = -2, x = 1, x = -1/9$.

Equation 3: $x^4 - 9x^2 - 4x + 12 = 0$

Let's try integer roots.

Let's try $x = 1$: $1^4 - 9(1)^2 - 4(1) + 12 = 1 - 9 - 4 + 12 = 13 - 13 = 0$

Let's try $x = -1$: $(-1)^4 - 9(-1)^2 - 4(-1) + 12 = 1 - 9 + 4 + 12 = 17 - 9 = 8 \neq 0$

Let's try $x = 2$: $2^4 - 9(2)^2 - 4(2) + 12 = 16 - 36 - 8 + 12 = 28 - 44 = -16 \neq 0$

Let's try $x = -2$: $(-2)^4 - 9(-2)^2 - 4(-2) + 12 = 16 - 36 + 8 + 12 = 36 - 36 = 0$

So $x = 1$ and $x = -2$ are roots.

Thus $(x - 1)$ and $(x + 2)$ are factors.

$(x - 1)(x + 2) = x^2 + x - 2$

Now divide $x^4 - 9x^2 - 4x + 12$ by $x^2 + x - 2$:

$(x^4 - 9x^2 - 4x + 12) / (x^2 + x - 2) = x^2 - x - 6$

Now solve the quadratic equation $x^2 - x - 6 = 0$.

$(x - 3)(x + 2) = 0$

So $x = 3$ and $x = -2$.

Therefore, the solutions to the third equation are $x = 1, x = -2, x = 3, x = -2$.

The solutions are $x = 1, x = -2, x = 3$.

3. Final Answer

Equation 1: $x = 3/2, 1/3, -2$

Equation 2: $x = -2, 1, -1/9$

Equation 3: $x = 1, -2, 3$