Solve the equation x^4 + 2x^3 - 16x^2 - 22x + 7 = 0, given that it has a root 2 + sqrt(3).

Given the equation:

x^4 + 2x^3 - 16x^2 - 22x + 7 = 0

and that one of its roots is 2 + √3, we know that if a polynomial with real coefficients has a root of the form 2 + √3, its conjugate, 2 - √3, must also be a root of the polynomial. So, the two roots 2 + √3 and 2 - √3 are part of the solution.

Step 1: Express the quadratic factor
The product of the two conjugate roots (2 + √3) and (2 - √3) is a quadratic expression. Using the difference of squares formula:

(2 + √3)(2 - √3) = 2^2 - (√3)^2 = 4 - 3 = 1

Therefore, the quadratic factor corresponding to these two roots is:

(x - (2 + √3))(x - (2 - √3)) = (x - 2 - √3)(x - 2 + √3)

This simplifies as:

(x - 2)^2 - (√3)^2 = (x - 2)^2 - 3

Now expand (x - 2)^2:

(x - 2)^2 = x^2 - 4x + 4

So, the quadratic factor becomes:

x^2 - 4x + 4 - 3 = x^2 - 4x + 1

Step 2: Divide the original polynomial by the quadratic factor
Now, we will divide the original polynomial x^4 + 2x^3 - 16x^2 - 22x + 7 by the quadratic factor x^2 - 4x + 1.

We perform polynomial long division:

Divide the first term of the dividend (x^4) by the first term of the divisor (x^2), which gives x^2.

Multiply x^2 by (x^2 - 4x + 1), resulting in x^4 - 4x^3 + x^2.

Subtract (x^4 - 4x^3 + x^2) from the original polynomial:

(x^4 + 2x^3 - 16x^2 - 22x + 7) - (x^4 - 4x^3 + x^2) = 6x^3 - 17x^2 - 22x + 7

Divide the first term of the new polynomial (6x^3) by the first term of the divisor (x^2), which gives 6x.

Multiply 6x by (x^2 - 4x + 1), resulting in 6x^3 - 24x^2 + 6x.

Subtract (6x^3 - 24x^2 + 6x) from (6x^3 - 17x^2 - 22x + 7):

(6x^3 - 17x^2 - 22x + 7) - (6x^3 - 24x^2 + 6x) = 7x^2 - 28x + 7

Divide the first term of the new polynomial (7x^2) by the first term of the divisor (x^2), which gives 7.

Multiply 7 by (x^2 - 4x + 1), resulting in 7x^2 - 28x + 7.

Subtract (7x^2 - 28x + 7) from (7x^2 - 28x + 7), which gives 0.

Thus, the quotient is:

x^2 + 6x + 7

Step 3: Write the complete factorization
The original polynomial is factored as:

(x^2 - 4x + 1)(x^2 + 6x + 7) = 0

Step 4: Solve for the roots
Now, solve each quadratic factor:

For x^2 - 4x + 1 = 0, use the quadratic formula:

x = [-(-4) ± √((-4)^2 - 4(1)(1))] / 2(1) x = [4 ± √(16 - 4)] / 2 x = [4 ± √12] / 2 x = [4 ± 2√3] / 2 x = 2 ± √3

So, the roots are x = 2 + √3 and x = 2 - √3.

For x^2 + 6x + 7 = 0, use the quadratic formula:

x = [-(6) ± √((6)^2 - 4(1)(7))] / 2(1) x = [-6 ± √(36 - 28)] / 2 x = [-6 ± √8] / 2 x = [-6 ± 2√2] / 2 x = -3 ± √2

So, the roots are x = -3 + √2 and x = -3 - √2.

Final Answer:
The roots of the equation are:

x = 2 + √3, x = 2 - √3, x = -3 + √2, and x = -3 - √2.