solve the equation 4s^3+15s^2+16s+6=0

Unfortunately, this equation cannot be solved using simple factorization or the quadratic formula. It is a cubic equation, meaning it is of the form ax^3 + bx^2 + cx + d = 0.

One possible way to solve this equation is to use the rational root theorem, which states that any rational root of the equation must be of the form p/q, where p is a factor of the constant term d (in this case 6) and q is a factor of the leading coefficient a (in this case 4).

Using this theorem, we can quickly find that possible rational roots of the equation are ±1, ±2, ±3, and ±6. We can then test each of these values by substituting them into the equation and seeing if they make the equation true.

After some trial and error, we find that s = -1/2 is a root of the equation. This means that we can factor the equation as:

(2s + 1)(2s^2 + 11s + 6) = 0

The first factor gives us the root s = -1/2, and the second factor is a quadratic equation that can be solved using the quadratic formula or factoring. The quadratic factor can be factored as:

(2s + 1)(s + 6)(s + 1/2) = 0

This gives us three roots: s = -1/2 (from the first factor), s = -1/2 (from the third factor), and s = -6 (from the second factor).

Therefore, the solutions to the original equation are s = -1/2 and s = -6