To solve quadratic polynomials, we typically use one of three methods: factoring, completing the square, or using the quadratic formula. Let's review each method in detail and apply them to an example quadratic polynomial.

### General Form of a Quadratic Polynomial

A quadratic polynomial is generally given by:

\[ ax^2 + bx + c = 0 \]

where \( a \), \( b \), and \( c \) are constants, and \( a \neq 0 \).

### Solving by Factoring

1. **Find two numbers** that multiply to \( ac \) (the product of the coefficient of \( x^2 \) and the constant term) and add to \( b \) (the coefficient of \( x \)).
2. **Rewrite the middle term** using these two numbers.
3. **Factor by grouping.**

#### Example:

Solve \( x^2 + 5x + 6 = 0 \).

- \( ac = 1 \times 6 = 6 \)
- Two numbers that multiply to 6 and add to 5 are 2 and 3.
- Rewrite the equation: \( x^2 + 2x + 3x + 6 = 0 \).
- Factor by grouping: \( (x^2 + 2x) + (3x + 6) = 0 \).
- \( x(x + 2) + 3(x + 2) = 0 \).
- \( (x + 3)(x + 2) = 0 \).

Thus, the solutions are \( x = -3 \) and \( x = -2 \).

### Solving by Completing the Square

1. Move the constant term to the other side of the equation.
2. Divide the linear coefficient by 2, square it, and add it to both sides.
3. Rewrite the left side as a perfect square trinomial.
4. Solve for \( x \) by taking the square root of both sides.

#### Example:

Solve \( x^2 + 6x + 8 = 0 \).

- Move 8 to the other side: \( x^2 + 6x = -8 \).
- Take half of 6, square it, add to both sides: \( x^2 + 6x + 9 = 1 \).
- Rewrite: \( (x + 3)^2 = 1 \).
- Take the square root: \( x + 3 = \pm 1 \).
- Solve for \( x \): \( x = -3 \pm 1 \).

Thus, \( x = -2 \) or \( x = -4 \).

### Solving Using the Quadratic Formula

The quadratic formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

#### Example:

Solve \( 2x^2 + 4x - 6 = 0 \).

- Identify \( a = 2 \), \( b = 4 \), \( c = -6 \).
- Compute the discriminant: \( b^2 - 4ac = 4^2 - 4(2)(-6) = 16 + 48 = 64 \).
- Apply the quadratic formula:

  \[
  x = \frac{-4 \pm \sqrt{64}}{4} = \frac{-4 \pm 8}{4}
  \]

- The solutions are:

  \[
  x = \frac{-4 + 8}{4} = 1 \quad \text{and} \quad x = \frac{-4 - 8}{4} = -3
  \]

### Summary

Each method can be useful depending on the specific polynomial:

- **Factoring** is quick when the polynomial is easily factorable.
- **Completing the square** provides insight into the structure of the quadratic and is helpful for deriving the quadratic formula.
- **The quadratic formula** is a reliable method for finding roots when other methods are difficult to apply.

If you have specific quadratic equations to solve or need further clarification, feel free to ask!

### Related Questions:

1. What are the conditions under which a quadratic can be factored easily?
2. How can you determine if a quadratic has real or complex roots?
3. Why does the discriminant \( b^2 - 4ac \) determine the nature of the roots?
4. What are the benefits of completing the square in solving quadratics?
5. How does the quadratic formula derive from completing the square?
6. How do you solve quadratics with irrational roots using these methods?
7. What are some applications of quadratic equations in real-world problems?
8. How do you handle quadratics with fractional coefficients or complex numbers?

### Tip

When using the quadratic formula, pay close attention to the discriminant, \( b^2 - 4ac \), as it provides key information about the nature of the roots: if positive, there are two distinct real roots; if zero, one repeated real root; if negative, two complex roots.

Explore different methods to solve quadratic polynomials, including factoring, completing the square, and using the quadratic formula. Detailed examples and step-by-step solutions provided. Suitable for students in Grades 9-12.

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