How many distinct $x$-intercepts does the graph of $y = (x-5)(x^2+5x+6)$ have?
An $x$-intercept occurs when $y=0$.  So, the $x$-intercepts are the solutions to the equation $0 = (x-5)(x^2+5x+6)$.  From this equation, we see that solutions occur when $x-5=0$ and when $x^2+5x+6=0$.  Now, $x^2+5x+6$ factors to $(x+3)(x+2)$.  So, the solutions are $5, -2, -3$, which comes to $\boxed{3}$ intercepts.
3
Level 3
Algebra
Problem: How many distinct $x$-intercepts does the graph of $y = (x-5)(x^2+5x+6)$ have?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of distinct $x$-intercepts of the function $y = (x-5)(x^2+5x+6)$, we need to find the values of $x$ for which $y = 0$. This requires solving the equation $(x-5)(x^2+5x+6) = 0$. 

The equation is satisfied if either factor is equal to zero. First, we set the linear factor equal to zero: 

1. $x - 5 = 0$ leads to $x = 5$. This gives us one distinct $x$-intercept.

Next, we analyze the quadratic factor $x^2 + 5x + 6$. To find its roots, we can use the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = 5$, and $c = 6$.

2. Calculate the discriminant: $b^2 - 4ac = 5^2 - 4(1)(6) = 25 - 24 = 1$. Since the discriminant is positive, the quadratic has two distinct real roots.

3. Now, apply the quadratic formula: 
   - $x = \frac{-5 + \sqrt{1}}{2(1)} = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$.
   - $x = \frac{-5 - \sqrt{1}}{2(1)} = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$.

These calculations yield two additional distinct $x$-intercepts at $x = -2$ and $x = -3$.

In total, we have found three distinct $x$-intercepts: $x = 5$, $x = -2$, and $x = -3$. Therefore, the final answer is $3$.<|END_OF_LATENT|> The final answer is $3$.