The expression \(6x^2 + 13x + 5\) is a quadratic polynomial in the standard form \(ax^2 + bx + c\), where \(a = 6\), \(b = 13\), and \(c = 5\). We can analyze this polynomial in several ways, including factoring, finding the roots, or analyzing its graph. Here, let's start by factoring the polynomial.

### Factoring the Quadratic Polynomial

To factor the quadratic polynomial \(6x^2 + 13x + 5\), we look for two numbers that multiply to the product of \(a\) and \(c\) (which is \(6 \times 5 = 30\)) and add up to \(b\) (which is 13).

1. **Find two numbers that multiply to 30 and add to 13.**

   The numbers 10 and 3 work because:

   \[
   10 \times 3 = 30 \quad \text{and} \quad 10 + 3 = 13
   \]

2. **Rewrite the middle term using these numbers:**

   Replace \(13x\) with \(10x + 3x\):

   \[
   6x^2 + 10x + 3x + 5
   \]

3. **Factor by grouping:**

   \[
   6x^2 + 10x + 3x + 5 = (6x^2 + 10x) + (3x + 5)
   \]

   Factor out the greatest common factor in each group:

   \[
   2x(3x + 5) + 1(3x + 5)
   \]

4. **Factor out the common binomial factor:**

   \[
   (2x + 1)(3x + 5)
   \]

So, the factorization of \(6x^2 + 13x + 5\) is \((2x + 1)(3x + 5)\).

### Verification

To verify the factorization, expand \((2x + 1)(3x + 5)\):

\[
\begin{align*}
(2x + 1)(3x + 5) &= 2x \cdot 3x + 2x \cdot 5 + 1 \cdot 3x + 1 \cdot 5 \\
&= 6x^2 + 10x + 3x + 5 \\
&= 6x^2 + 13x + 5
\end{align*}
\]

The factorization is correct.

### Solving the Equation

To find the roots of the equation \(6x^2 + 13x + 5 = 0\), we use the factored form \((2x + 1)(3x + 5) = 0\).

1. **Solve each factor for zero:**

   \[
   2x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{2}
   \]

   \[
   3x + 5 = 0 \quad \Rightarrow \quad x = -\frac{5}{3}
   \]

Thus, the roots of the equation are \(x = -\frac{1}{2}\) and \(x = -\frac{5}{3}\).

### Graph of the Quadratic

The graph of the quadratic equation \(y = 6x^2 + 13x + 5\) is a parabola that opens upwards (since the coefficient of \(x^2\) is positive).

- **Vertex:** The vertex of the parabola can be found using the formula \(x = -\frac{b}{2a}\):

  \[
  x = -\frac{13}{2 \times 6} = -\frac{13}{12}
  \]

  Substitute back into the equation to find \(y\):

  \[
  y = 6\left(-\frac{13}{12}\right)^2 + 13\left(-\frac{13}{12}\right) + 5
  \]

  Calculating this gives the vertex coordinates.

- **Y-intercept:** The y-intercept is \(c = 5\), so the parabola crosses the y-axis at (0, 5).

Would you like further details on any of these steps, or do you have any other questions?

Here are some additional related questions you might find helpful:

1. How do you find the vertex of a quadratic equation using the vertex formula?
2. What is the significance of the discriminant in a quadratic equation?
3. How do you determine if a quadratic equation has real or complex roots?
4. What are the properties of a parabola derived from a quadratic equation?
5. How can you find the axis of symmetry for a quadratic function?
6. What are some methods to solve a quadratic equation besides factoring?
7. How does completing the square help in solving quadratic equations?
8. What are some real-life applications of quadratic equations?

**Tip:** When factoring a quadratic equation, always check your factorization by expanding it to ensure it matches the original equation.

Learn how to factor and solve the quadratic polynomial 6x^2 + 13x + 5 step-by-step with detailed explanations. This problem covers key concepts including factoring quadratic polynomials, solving quadratic equations, and graphing parabolas. Suitable for high school students in Grades 10-12.

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