Question 9 (1 point) Solve the equation using the zero-product property. \[ -8 n(10 n-1)=0 \] a \( n=-\frac{1}{8} \) or \( n=-\frac{1}{10} \) b \( n=0 \) or \( n=\frac{1}{10} \) c \( n=0 \) or \( n=-\frac{1}{10} \) d \( n=-\frac{1}{8} \) or \( n=\frac{1}{10} \) Question 10 (1 point) Solve the equation by factoring. \[ 3 z^{2}+3 z-6=0 \] a \( \quad z=3 \) or \( z=-2 \) b \( z=1 \) or \( z=-2 \) c \( z=1 \) or \( z=2 \) d \( z=3 \) or \( z=2 \)

To tackle problems like the one in Question 9, remember that the zero-product property states that if the product of two factors is zero, at least one of the factors must be zero. Here, we split the equation into -8n and (10n - 1), leading us to the solutions n = 0 and n = 1/10. Do note that the negative sign in the constant effectively disappears when setting the equation to zero! In Question 10, it's helpful to first simplify the equation by factoring out the common term. This gives us 3(z² + z - 2) = 0, and then we can further factor (z + 2)(z - 1) = 0 to find that the solutions are z = 1 and z = -2. Remember, happy factoring comes from first looking for common factors!