Solve the equation $x^3 - 3x^2 - 9x + 11 = 0$ - Leaving Cert Mathematics - Question 2 - 2015

Since $x - 1$ is a factor, we can perform polynomial long division on $x^3 - 3x^2 - 9x + 11$ by $(x - 1)$:

1. Divide:

   • $x^3$ by $x$ gives $x^2$
   • Multiply $x^2$ by $(x - 1)$ results in $x^3 - x^2$. Subtract this from the original polynomial.

2. The new polynomial is:

   • $(-3x^2 + x^2) - 9x + 11 = -2x^2 - 9x + 11$

3. Repeat the steps:

   • $-2x^2/x = -2$
   • Multiply $-2$ by $(x - 1)$ gives $-2x + 2$. Subtract to find:

$(-9x + 2) - (-2x) + 11 = -7x + 9$

Thus, we have:

$(x - 1)(x^2 + Ax + B) = f(x)$

where $A = -2$ and $B = -11$.

Now we write the factorization:

$x^3 - 3x^2 - 9x + 11 = (x - 1)(x^2 - 2x - 11)$

Next, we solve the quadratic equation $x^2 - 2x - 11 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-11)}}{2(1)}$

This simplifies to:

$x = \frac{2 \pm \sqrt{4 + 44}}{2} = \frac{2 \pm \sqrt{48}}{2} = \frac{2 \pm 4 \sqrt{3}}{2} = 1 \pm 2\sqrt{3}$

The irrational solutions are:

$x = 1 + 2\sqrt{3}, \quad x = 1 - 2\sqrt{3}$

Thus, we can express the solutions in the form $a + b \sqrt{c}$ where $a = 1$, $b = 2$, and $c = 3$.