Final answer:
To find the number of solutions or zeros, solve the polynomial equation $8q^3 - 4q^3 + 7q = 0$ for q. The equation has two possible solutions: q = 0 and $4q^2 + 7 = 0$. However, the equation $4q^2 + 7 = 0$ has no real solutions for q.
Step-by-step explanation:
This is a polynomial equation of the form $8q^3 - 4q^3 + 7q = 0$. To find the number of solutions or zeros, we need to solve for q. First, combine like terms: $8q^3 - 4q^3 = 4q^3$. Now we have the equation $4q^3 + 7q = 0$. To solve for q, set the equation equal to zero and factor out the common factor of q: $q(4q^2 + 7) = 0$. This equation has two possible solutions: q = 0 and $4q^2 + 7 = 0$. To find the solutions of $4q^2 + 7 = 0$, subtract 7 from both sides to get $4q^2 = -7$, then divide by 4 to get $q^2 = -rac{7}{4}$. Since we can't take the square root of a negative number, there are no real solutions for q in this equation.