## How To Factorization Quadratic Equation

Equation | a | b | c |
---|---|---|---|

x^2 + 2x + 1 = 0 | 1 | 2 | 1 |

x^2 - 3x - 4 = 0 | 1 | -3 | -4 |

x^2 + x - 6 = 0 | 1 | 1 | -6 |

## Factorization Quadratic Equation Examples

here are some examples of factoring quadratic equations, along with the steps for each example:

Example 1: Factor the equation x^2 + 2x + 1 = 0

Write the quadratic equation in the form of ax^2 + bx + c = 0. In this case, a = 1, b = 2, and c = 1.

Determine whether the equation can be factored using the difference of squares or the sum/difference of cubes method. In this case, it can be factored using the difference of squares method.

Find two numbers whose product is equal to c and whose sum is equal to b. In this case, the product of the two numbers is 1 (the value of c), and their sum is 2 (the value of b). We can find two numbers that meet these criteria by using trial and error, or by using a list of factor pairs.

Factor the equation using the two numbers found in step 3. In this case, the two numbers are 1 and 1, so the equation can be factored as (x + 1)(x + 1) = 0.

Example 2: Factor the equation x^2 - 3x - 4 = 0

Write the quadratic equation in the form of ax^2 + bx + c = 0. In this case, a = 1, b = -3, and c = -4.

Determine whether the equation can be factored using the difference of squares or the sum/difference of cubes method. In this case, it can be factored using the difference of squares method.

Find two numbers whose product is equal to c and whose sum is equal to b. In this case, the product of the two numbers is -4 (the value of c), and their sum is -3 (the value of b). We can find two numbers that meet these criteria by using trial and error, or by using a list of factor pairs.

Factor the equation using the two numbers found in step 3. In this case, the two numbers are -2 and 2, so the equation can be factored as (x - 2)(x + 2) = 0.

Example 3: Factor the equation x^2 + x - 6 = 0

Write the quadratic equation in the form of ax^2 + bx + c = 0. In this case, a = 1, b = 1, and c = -6.

Determine whether the equation can be factored using the difference of squares or the sum/difference of cubes method. In this case, it can be factored using the sum/difference of cubes method.

Find two numbers whose product is equal to c and whose sum/difference is equal to a. In this case, the product of the two numbers is -6 (the value of c), and their sum/difference is 1 (the value of a). We can find two numbers that meet these criteria by using trial and error, or by using a list of factor pairs.

Factor the equation using the two numbers found in step 3. In this case, the two numbers are -3 and -2, so the equation can be factored as (x - 3)(x - 2) = 0.