# Factoring Quadratic Equations - Trial and Error

In these lessons, we will learn how to factor quadratic equations, where the coefficient of x2 is 1, using the trial and error method (or guess and check method).

There are several techniques that can be used to factor quadratic equations.

In this trial and error method, we will need to try out different possibilities to get the right factors for the given quadratic equation.

### If the Coefficient Of x2 Is 1

To factorize quadratic equations of the form: x2 + bx + c, you will need to find two numbers whose product is c and whose sum is b. Example 1: (b and c are both positive)
Solve the quadratic equation: x2 + 7x + 10 = 0

Solution:
Step 1: List out the factors of 10:

1 × 10, 2 × 5

Step 2: Find the factors whose sum is 7:

1 + 10 ≠ 7
2 + 5 = 7 Step 3: Write out the factors and check using the distributive property.

(x + 2)(x + 5) = x2 + 5x + 2x + 10 = x2 + 7x + 10
The factors are (x + 2)(x + 5)

Step 4: Going back to the original quadratic equation

x2 + 7x + 10 = 0       Factorize the left side of the quadratic equation
(x + 2)(x + 5) = 0

We get two values for x. Answer:   x = – 2, x = – 5

Example 2: (b is positive and c is negative)

Get the values of x for the equation: x2 + 4x – 5 = 0

Solution:
Step 1: List out the factors of – 5:

1 × –5, –1 × 5

Step 2: Find the factors whose sum is 4:

1 – 5 ≠ 4
–1 + 5 = 4 Step 3: Write out the factors and check using the distributive property.

(x – 1)(x + 5)= x2 + 5xx – 5 = x2 + 4x – 5

Step 4: Going back to the original quadratic equation.

x2 + 4x – 5 = 0       Factorize the left hand side of the equation
(x – 1)(x + 5) = 0

We get two values for x. Answer:   x = 1, x = – 5

Example 3: (b and c are both negative)

Get the values of x for the equation: x2 – 5x – 6

Solution:
Step 1: List out the factors of – 6:

1 × –6, –1 × 6, 2 × –3, –2 × 3

Step 2: Find the factors whose sum is –5:

1 + ( –6) = –5 Step 3: Write out the factors and check using the distributive property.

(x + 1)(x – 6) = x2 – 6 x + x – 6 = x2 – 5x – 6

Step 4: Going back to the original quadratic equation

x2 – 5x – 6 = 0       Factorize the left hand side of the equation
(x + 1)(x – 6) = 0

We get two values for x. Answer: x = –1, x = 6

Example 4: (b is negative and c is positive)

Get the values of x for the equation: x2 – 6x + 8 = 0

Solution:
Step 1: List out the factors of 8:

We need to get the negative factors of 8 to get a negative sum.
–1 × – 8, –2 × –4

Step 2: Find the factors whose sum is –6:

–1 + ( –8) ≠ –6
–2 + ( –4) = –6 Step 3: Write out the factors and check using the distributive property.

(x – 2)(x – 4) = x2 – 4 x – 2x + 8 = x2 – 6x + 8

Step 4: Going back to the original quadratic equation

x2 – 6x + 8 = 0       Factorize the left hand side of the equation
(x – 2)(x – 4) = 0

We get two values for x. Answer: x = 2, x = 4

Examples:

1. Solve x2 + 6x + 8 = 0
2. Solve 2x2 + 20x + 50 = 0
3. Solve x2 - x - 30 = 0

coefficient of x2 = −1

Examples:

1. Factor x2 + 10x + 9
2. Factor x2 - 11x + 24
3. Factor x2 - x - 56
4. Factor -x2 - 5x + 24
5. Factor -x2 + 18x - 72

Example:
Solve x2 - 3x = 4

Examples:

1. x2 - 3x - 18
2. x2 - 6x - 16

Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. 