The coefficient of x^5 in (3x - 4)^7 is C(7, 5) * (3)^5 * (-4)^2 = 21 * 243 * 16 = 68016.

Using the binomial theorem, the coefficient of x^2 in (2x - 3)^4 is given by 4C2 * (2)^2 * (-3)^2 = 6 * 4 * 9 = 216.

The coefficient of x^2 in (3x - 2)^5 is given by 5C2 * (3x)^2 * (-2)^3 = 10 * 9 * (-8) = -720.

The coefficient of x^2 is given by C(6, 2)(4)^4 = 15 * 256 = 3840.

The coefficient of x^3 in (3x - 4)^5 is given by 5C3 * (3)^3 * (-4)^2 = 10 * 27 * 16 = -720.

The coefficient of x^3 in (x + 1)^8 is given by 8C3 = 56.

The coefficient of x^4 is C(8, 4) = 70.

The coefficient of x^5 in (3x + 2)^6 is C(6, 5)(3)^5(2)^1 = 6 * 243 * 2 = 2916.

The conjugate of z = 2 - 5i is z* = 2 + 5i.

(1 + i)² = 1² + 2(1)(i) + i² = 1 + 2i - 1 = 2i.

Using the binomial theorem, (1 + 2)^6 = 3^6 = 729.

Using the binomial theorem, (a + b)^4 = C(4, 0)a^4b^0 + C(4, 1)a^3b^1 + C(4, 2)a^2b^2 + C(4, 3)a^1b^3 + C(4, 4)a^0b^4. Substituting a = 2 and b = 3 gives 16 + 4*6 + 6*9 + 4*27 + 81 = 81.

The discriminant must be less than zero: k^2 - 4*1*16 < 0 leads to k < -8.

The coefficient of x^2 in (x + k)^4 is C(4, 2) * k^2 = 6. Thus, 6k^2 = 6, giving k^2 = 1, so k = 1 or -1.

The term containing x^4 is C(6,4) * (2)^2 * x^4 = 15 * 4 * x^4 = 60x^4. Setting 60 = 240 gives k = 4.

C(7, 4) = 7! / (4! * (7-4)!) = 7! / (4! * 3!) = (7*6*5)/(3*2*1) = 35.

Using the binomial theorem, the coefficient of x^4 in (a + b)^n is given by nCk * a^(n-k) * b^k. Here, n=6, a=x, b=-2, and k=2. Thus, the coefficient is 6C2 * (1)^4 * (-2)^2 = 15 * 4 = 60.

By substituting x = 1 into the equation, we find that it satisfies the equation, hence 1 is a root.

The roots can be found using the Rational Root Theorem and synthetic division, confirming that all roots are real.

The discriminant is 2^2 - 4(1)(1) = 0, indicating that the roots are real and equal.

The discriminant must be non-negative: 4^2 - 4*1*k >= 0 leads to 16 - 4k >= 0, thus k <= 4.

The condition for no real roots is that the discriminant must be less than zero: 6^2 - 4*1*k < 0 => 36 < 4k => k > 9.

The equation can be factored as (x-1)^3 = 0, which has one real root (x = 1) with multiplicity 3.

By substituting x = 2 into the equation, we find that it satisfies the equation, thus x = 2 is a root.

The product of the roots of the cubic equation ax^3 + bx^2 + cx + d = 0 is given by -d/a. Here, d = -6 and a = 1, so the product is -(-6)/1 = 6.

By substituting x = 2 into the equation, we find that 2 is a root since 2^3 - 6(2^2) + 11(2) - 6 = 0.

The polynomial can be factored as (x-1)^3, indicating that it has one real root with multiplicity 3, hence all roots are real and equal.

The sum of the roots is given by -b/a = 3/1 = 3.

The polynomial can be factored as (x - 1)^3, indicating that all roots are real and equal.

The discriminant D = b^2 - 4ac = 4^2 - 4(2)(2) = 16 - 16 = 0.