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

In this lesson, we will learn how to factor quadratic equations, where the coefficient of*x*^{2} is greater than 1, using the trial and error method. In this method, we will need to try out different possibilities to get the
right factors for the given quadratic equation. In this lesson, we will also learn how to factor quadratic equations by grouping

We also have another lesson that will show you how to factor quadratic equations without trial and error.

Related Topics:

Other Factoring Techniques

More Algebra Lessons

Factoring Out Common Factors (GCF).

Factoring Quadratic Equations using Perfect Square Trinomial (Square of a Sum or Square of a Difference) or Difference of Two Squares.

Factoring Quadratic Equations where the coefficient of*x*^{2 }is 1.

Factoring Quadratic Equations by Completing the Square

Factoring Quadratic Equations using the Quadratic Formula.

### If the Coefficient of *x*^{2} Is Greater Than 1

* x* = 2, *x* = 5

Example 2: Get the values of *x* for the equation 7*x*^{2} + 18*x* + 11 = 0

Example 3: Get the values of*x* for the equation 4*x*^{2} + 26*x* + 12 = 0

**How to solve quadratic equations by factoring?**

The method for solving quadratic equations by factoring is based on the zero-factor property of real numbers.

The zero factor property states that it two numbers a and b are multiplied together and the rsulting product is 0, then at least one of the numbers must be 0.

Examples:

Solve

1. x^{2} - 9x = 0

2. x^{2} - 3x - 10 = 0

3. x^{2} - 36 = 0

4. 3x^{2} + 4x - 7 = 0

**Solving Quadratic Equations by Factoring**

Examples:

Solve

a) x^{2} - 9x + 14 = 0

b) 2x^{2} - 5x = 3

### Factoring Quadratic Equations by Grouping

Factoring by Grouping - 3 complete examples of solving quadratic equations using factoring by grouping are shown.

Examples:

1. x(x + 1) - 5(x + 1)

2. 2x^{2} + 5x + 2 = 0

3. 7x^{2} + 16x + 4 = 0

4. 6x^{2} - 17x + 12 = 0

**Solving Quadratic Equations by Factoring Using the Grouping Method**

Examples:

25x^{2} - 20x + 4 = 0

**Factor and Solve a Quadratic Equation - Factor by Grouping**

Examples:

1. x(x - 2) + 7(x - 2) = 0

2. 3x^{2} + 6x + 4x + 8 = 0
**How to factor and solve a quadratic equation by using the factor by grouping method?**

Examples:

1. x(2x - 3) + 2x - 3 = 0

2. 3x^{2} - 3x - x + 1 = 0

In this lesson, we will learn how to factor quadratic equations, where the coefficient of

We also have another lesson that will show you how to factor quadratic equations without trial and error.

Related Topics:

Other Factoring Techniques

More Algebra Lessons

Factoring Out Common Factors (GCF).

Factoring Quadratic Equations using Perfect Square Trinomial (Square of a Sum or Square of a Difference) or Difference of Two Squares.

Factoring Quadratic Equations where the coefficient of

Factoring Quadratic Equations by Completing the Square

Factoring Quadratic Equations using the Quadratic Formula.

Sometimes the coefficient of *x* in quadratic equations may not be 1, but the expression can be simplified by first finding common factors.

When the coefficient of *x*^{2} is greater than 1 and we cannot simplify the quadratic equation by finding common factors, we would need to consider the factors of the coefficient of *x*^{2} and the factors of *c* in order to get the numbers whose sum is *b*. If there are many factors to consider you may want to use the quadratic formula instead.

Example 1: Get the values of *x* for the equation 2*x*^{2} – 14*x* + 20 = 0

Step 1: Find common factors if you can.

2

x^{2}– 14x+ 20 = 2(x^{2}– 7x+ 10)

Step 2: Find the factors of (*x*^{2} – 7*x* + 10)

List out the factors of 10:

We need to get the negative factors of 10 to get a negative sum.

–1 × –10, –2 × –5

Step 3: Find the factors whose sum is – 7:

1 + ( –10) ≠ –7

–2 + ( –5) = –7

Step 4: Write out the factors and check using the distributive property.

2(

x– 2) (x– 5) = 2(x^{2}– 5x– 2x+ 10)

= 2(x^{2}– 7x+ 10) = 2x^{2}– 14x+ 20

Step 5: Going back to the original equation

Answer:2

x^{2}– 14x+ 20 = 0 Factorize the left hand side of the equation

2(x– 2) (x– 5) = 0We get two values for

x

Step 1: List out the factors of 7 and 11

Factors of 7:

1 × 7Factors of 11:

1 × 11Since 7 and 11 are prime numbers there are only two possibilities to try out.

Step 2: Write down the different combinations of the factors and perform the distributive property to check.

(7

x +1)(x+ 11) ≠ 7x^{2}+ 18x+ 11

(7x +11)(x+ 1) = 7x^{2}+ 18x+ 11

Step 3: Write out the factors and check using the distributive property.

(7

x +11)(x+ 1) = 7x^{2}+ 7x+ 11x+ 11 = 7x^{2}+ 18x+ 11

Step 4: Going back to the original equation

7

x^{2}+ 18x+ 11= 0 Factorize the left hand side of the equation

(7x +11)(x+ 1) = 0We get two values for

x

Answer:

Example 3: Get the values of

Step 1: Find common factors if you can.

4

x^{2}+ 26x+ 12 = 2(2x^{2}+ 13x+ 6)

Step 2: List out the factors of 2 & 6

Factors of 2:

1 × 2Factors of 6:

1 × 6, 2 × 3

Step 3: Write down the different combinations of the factors and perform the distributive property to check. When there are many factors to check, this becomes a tedious method to solve such quadratic equations, so you may want to try the quadratic formula instead.

(

x +1)(2x+ 6) ≠ (2x^{2}+ 13x+ 6)

(x +6)(2x+ 1) = (2x^{2}+ 13x+ 6)

Step 4: Going back to the original quadratic equation

4

x^{2}+ 26x+ 12 = 0 Factorize the left side of the equation

2(x +6)(2x+ 1) = 0We get two values for

x

Answer:

The method for solving quadratic equations by factoring is based on the zero-factor property of real numbers.

The zero factor property states that it two numbers a and b are multiplied together and the rsulting product is 0, then at least one of the numbers must be 0.

Examples:

Solve

1. x

2. x

3. x

4. 3x

Examples:

Solve

a) x

b) 2x

Examples:

1. x(x + 1) - 5(x + 1)

2. 2x

3. 7x

4. 6x

Examples:

25x

Examples:

1. x(x - 2) + 7(x - 2) = 0

2. 3x

Examples:

1. x(2x - 3) + 2x - 3 = 0

2. 3x

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