The expression \(5x^2 + 6x + 1\) is a quadratic polynomial. Here are several ways you can work with it:  

### 1. **Factoring the Quadratic**  
Let's factor this expression, if possible.

We need to find two numbers whose **product** is \(5 \cdot 1 = 5\) and whose **sum** is 6.  
- The pair \((5, 1)\) works because \(5 + 1 = 6\).

So, we rewrite the middle term as:
\[
5x^2 + 5x + x + 1
\]

Now, we group:
\[
(5x^2 + 5x) + (x + 1)
\]

Factor out common terms from each group:
\[
5x(x + 1) + 1(x + 1)
\]

Now factor out \((x + 1)\):
\[
(5x + 1)(x + 1)
\]

So, the **factored form** of the quadratic is:
\[
5x^2 + 6x + 1 = (5x + 1)(x + 1)
\]

---

### 2. **Roots of the Quadratic (Using the Quadratic Formula)**  
The **quadratic formula** is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For the equation \(5x^2 + 6x + 1 = 0\):
- \(a = 5\), \(b = 6\), \(c = 1\)

Now, plug in the values:
\[
x = \frac{-6 \pm \sqrt{6^2 - 4(5)(1)}}{2(5)}
\]
\[
x = \frac{-6 \pm \sqrt{36 - 20}}{10}
\]
\[
x = \frac{-6 \pm \sqrt{16}}{10}
\]
\[
x = \frac{-6 \pm 4}{10}
\]

Now, simplify the two roots:
1. \(x = \frac{-6 + 4}{10} = \frac{-2}{10} = -\frac{1}{5}\)
2. \(x = \frac{-6 - 4}{10} = \frac{-10}{10} = -1\)

So, the roots are:
\[
x = -1 \quad \text{and} \quad x = -\frac{1}{5}
\]

---

### 3. **Vertex of the Parabola**  
The vertex of a quadratic \(ax^2 + bx + c\) occurs at:
\[
x = \frac{-b}{2a}
\]
For \(5x^2 + 6x + 1\), we get:
\[
x = \frac{-6}{2(5)} = \frac{-6}{10} = -\frac{3}{5}
\]
Now, substitute \(x = -\frac{3}{5}\) back into the quadratic to find the **y-coordinate**:
\[
y = 5\left(-\frac{3}{5}\right)^2 + 6\left(-\frac{3}{5}\right) + 1
\]
\[
y = 5 \cdot \frac{9}{25} + 6 \cdot \left(-\frac{3}{5}\right) + 1
\]
\[
y = \frac{45}{25} - \frac{18}{5} + 1
\]
\[
y = \frac{9}{5} - \frac{18}{5} + \frac{5}{5}
\]
\[
y = \frac{-4}{5}
\]

So, the vertex is:
\[
\left(-\frac{3}{5}, -\frac{4}{5}\right)
\]

---

### 4. **Axis of Symmetry**  
The **axis of symmetry** is the vertical line passing through the vertex.  
Thus, the axis of symmetry is:
\[
x = -\frac{3}{5}
\]

---

### 5. **Graph Behavior**  
Since the leading coefficient \(a = 5\) is positive, the parabola opens **upwards**. The graph has a **minimum point** at the vertex \(\left(-\frac{3}{5}, -\frac{4}{5}\right)\).

---

Do you want details on any part or another operation with this polynomial?  

### Related Questions  
1. How do you use the discriminant to determine the nature of the roots?  
2. What is the general method for completing the square on \(5x^2 + 6x + 1\)?  
3. How do you graph a quadratic function step-by-step?  
4. What are the applications of quadratic functions in real life?  
5. How do you derive the quadratic formula from \(ax^2 + bx + c = 0\)?  

**Tip:** Always check whether a quadratic is factorable before using the quadratic formula, as it might save time.

Learn how to solve the quadratic equation 5x^2 + 6x + 1 with step-by-step factoring, using the quadratic formula, and determining the vertex and axis of symmetry. Ideal for high school students (Grades 9-12), this problem covers essential algebraic techniques for solving quadratics.

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