Setting x = 0 in the equation gives y = -5, so the y-intercept is -5.

'a' represents the first term of the harmonic progression in the formula for the nth term.

The nth term of a harmonic progression can be expressed as the harmonic mean of n and m, which is 1/(1/n + 1/m).

The first term corresponds to n=1, which gives 1/1 = 1.

Evaluating f(1) gives 1^3 - 3(1^2) + 4 = 1 - 3 + 4 = 2.

The polynomial can be factored as (x - 1)(x - 2)(x - 3), so x - 1 is one of its factors.

Substituting x = 1 into the polynomial gives f(1) = 1 - 4 + 6 - 4 + 1 = 0.

The polynomial can be rewritten in a form that shows symmetry about the line x = 1.

A polynomial has a double root when its discriminant is equal to zero.

Using Vieta's formulas, the sum of the roots (-2 + 3) = 1, which means b = -1.

Using Vieta's formulas, the sum of the roots (3 + (-2)) = 1, hence b = -1.

According to Vieta's formulas, the sum of the roots r1 + r2 of the polynomial x^2 - 5x + 6 is equal to the coefficient of x (which is -(-5)) = 5.

Factoring the polynomial gives (x - 2)(x - 3), so the roots are 2 and 3.

Factoring the polynomial P(x) gives (x - 2)(x - 3), so the roots are 2 and 3.

Calculating P(1) gives 1^3 - 3(1^2) + 4 = 1 - 3 + 4 = 2.

According to Vieta's formulas, the sum of the roots p and q is -b/a and the product is c/a.

For real roots, the discriminant must be non-negative: k^2 - 4*1*16 >= 0, leading to k^2 >= 64.

For equal roots, the discriminant must be zero: (-4)^2 - 4*1*k = 0, leading to k = 4.

For the equation to have one real root, the discriminant must be zero. Thus, k must equal 4.

Factoring the equation gives (x - 2)(x - 3) = 0, thus the roots are 2 and 3.

According to Vieta's formulas, the sum of the roots p + q is equal to -(-5) = 5.

According to Vieta's formulas, the sum of the roots p + q = -b/a and the product pq = c/a.

For both roots to be negative, the sum (p) must be positive and the product (q) must also be positive.

Let the first term be a. The second term is a * r = a * 3 = 12, thus a = 12/3 = 4.

Let the first term be a. The second term is a * r = a * 2 = 12, thus a = 12/2 = 6.

Let the first term be a. The second term is a * r = a * 3 = 12, thus a = 12/3 = 4.

Let the first term be a. The second term is a * r = a * 3 = 6. Thus, a = 6/3 = 2.

Let the first term be a and the common ratio be r. Then, 8 = ar and 32 = ar^3. Dividing these gives r^2 = 4, so r = 2.

Substituting x = 2 into the equation gives 4(2) + 5y = 20, leading to y = 0.

The sum S of an infinite GP is given by S = a / (1 - r). Here, 20 = a / (1 - 1/4) = a / (3/4). Thus, a = 20 * (3/4) = 15.