The equation provided is:

$f(x) = x^4 - 3x^3 - 5x^2 + 9x - 2 = 0$ with one root given as $2 + \sqrt{3}$.

Since $2 + \sqrt{3}$ is a root, its conjugate $2 - \sqrt{3}$ will also be a root because the coefficients of the polynomial are real. This suggests that $(x - (2 + \sqrt{3}))(x - (2 - \sqrt{3}))$ is a factor of the polynomial.

Let's first expand the product of these factors:

$(x - (2 + \sqrt{3}))(x - (2 - \sqrt{3})) = (x - 2 - \sqrt{3})(x - 2 + \sqrt{3})$ This is a difference of squares:

$(x - 2)^2 - (\sqrt{3})^2 = (x - 2)^2 - 3 = x^2 - 4x + 1$

Now, we know that $x^2 - 4x + 1$ is a factor of the polynomial $f(x)$. To find the remaining factors, we will divide $f(x)$ by $x^2 - 4x + 1$.

I'll perform the division now.The polynomial division yields:

$f(x) = (x^2 - 4x + 1)(x^2 + x - 2)$

Since the remainder is 0, we know that $f(x)$ factors completely as:

$f(x) = (x^2 - 4x + 1)(x^2 + x - 2)$

Now, we solve the quadratic equation $x^2 + x - 2 = 0$ to find the other roots. Factoring this quadratic:

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

Thus, the complete factorization of $f(x)$ is:

$f(x) = (x - (2 + \sqrt{3}))(x - (2 - \sqrt{3}))(x - 1)(x + 2)$

The roots of the equation are:

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

Would you like more details or have any questions?

Related Questions:

1. How do you find roots of higher-degree polynomials?
2. What is the difference between rational and irrational roots?
3. How does the conjugate root theorem apply to polynomials with real coefficients?
4. How can polynomial division help in factorization?
5. What is the significance of the factor theorem in solving polynomials?

Tip:

Always check if the given polynomial has complex roots by applying the discriminant for quadratics when simplifying.