Zeros of Polynomials Quick Check

To find a possible sketch of the polynomial function \( f(x) = x^3 + 5x^2 + 6x \), we can start by analyzing its zeros and general behavior.

1. **Factoring**:
The polynomial can be factored by first taking out a common factor of \( x \):
\[
f(x) = x(x^2 + 5x + 6)
\]
Now, we can further factor the quadratic \( x^2 + 5x + 6 \):
\[
x^2 + 5x + 6 = (x + 2)(x + 3)
\]
Therefore, the complete factorization of \( f(x) \) is:
\[
f(x) = x(x + 2)(x + 3)
\]

2. **Finding zeros**:
The zeros of the polynomial are given by setting it to zero:
\[
x = 0, \, x = -2, \, x = -3
\]
This indicates that the function crosses the x-axis at these values.

3. **Behavior at infinity**:
Since the leading term is \( x^3 \), the polynomial will go to \( -\infty \) as \( x \) approaches \( -\infty \) and to \( +\infty \) as \( x \) approaches \( +\infty \).

4. **Sign Analysis**:
- For \( x < -3 \), the function is positive (since all factors are negative).
- For \( -3 < x < -2 \), the function is negative (since \( x + 2 > 0 \)).
- For \( -2 < x < 0 \), the function is positive again (since both \( x + 2 > 0 \) and \( x + 3 > 0 \)).
- For \( x > 0 \), the function remains positive.

Given this analysis:
- The graph should intersect the x-axis at \( x = 0 \), \( x = -2 \), and \( x = -3 \).
- The graph should change signs between these points as described and be an S-shaped curve, starting in the third quadrant, passing through the origin (the first zero), and continuing upward into the first quadrant.

Among the provided options, we look for the sketch that shows these characteristics:
- A graph that passes through quadrants 3, then intersects the x-axis, and continues up into quadrants 1.

The option that matches this description would be the one that shows a curve passing through quadrants 3, 1, and possibly 4, as it rises after crossing the x-axis.

Without the ability to see the actual graphs, you should select the option that indicates this S-shaped behavior that meets the criteria above.