1. The problem asks us to find which factorization correctly identifies the real zeros of the quadratic function $$f(x) = -20x^2 + 23x - 6$$. 2. To find the real zeros, we need to factor the quadratic expression or use the quadratic formula. The zeros are the values of $x$ for which $f(x) = 0$. 3. Let's check each factorization by expanding and comparing it to the original function. 4. Option 1: $(-10x + 2)(2x + 3)$ $$(-10x + 2)(2x + 3) = -20x^2 - 30x + 4x + 6 = -20x^2 - 26x + 6$$ This does not match $-20x^2 + 23x - 6$. 5. Option 2: $-(10x + 2)(2x - 3)$ $-(10x + 2)(2x - 3) = - (20x^2 - 30x + 4x - 6) = -20x^2 + 26x - 6$ This is close but the middle term is $26x$ instead of $23x$. 6. Option 3: $-(4x - 3)(5x + 2)$ $-(4x - 3)(5x + 2) = - (20x^2 + 8x - 15x - 6) = -20x^2 - 8x + 15x + 6 = -20x^2 + 7x + 6$ This does not match. 7. Option 4: $-(4x - 3)(5x - 2)$ $-(4x - 3)(5x - 2) = - (20x^2 - 8x - 15x + 6) = -20x^2 + 8x + 15x - 6 = -20x^2 + 23x - 6$ This matches the original function exactly. 8. Therefore, the correct factorization to identify the real zeros is $-(4x - 3)(5x - 2)$. 9. To find the zeros, set each factor equal to zero: $$4x - 3 = 0 \Rightarrow x = \frac{3}{4}$$ $$5x - 2 = 0 \Rightarrow x = \frac{2}{5}$$ 10. These are the real zeros of the function. Final answer: The correct factorization is $-(4x - 3)(5x - 2)$, and the real zeros are $x = \frac{3}{4}$ and $x = \frac{2}{5}$.