Solve equations of the second degree by factoring could be several times the most simple and fastest solve the quadratic equation as long as you know how factor.

I urge you to study or review the following important unit: factoring

**Example # 1: Solving the equations of the second degree by factoring**

Solve x 2 + 3 x + 2 = 0.

First, you need to factor x 2 + 3 x + 2

Given that the coefficient of x 2 is 1 (x 2 = **1**x 2), you can factor in looking for factors of the last Parliament (latter term is 2) which adds to the coefficient of the second term (3 x, factor is 3)

2 = 1 × 2

-2 =-1 × 2

Given that 1 + 2 = 3, and 3 is the coefficient of the second term, x 2 + 3 x + 2 = (x + 2) × (x + 1)

x 2 + 3 x + 2 = 0 gives:

(x + 2) × (x + 1) = 0

(x + 2) × (x + 1) = 0 when x + 2 = 0 or x + 1 = 0.

x + 2 = 0 when x = – 2

x + 1 = 0 when x = – 1

Let us now check x =-2 and x =-1 are solutions of 2 x + 3 x + 2 = 0

(-2) 2 + 3 ×-2 + 2 = 4 + 6 + 2 = 3 + 6 = 0

(-1) 2 + 3 ×-1 + 2 = 1 + 3 + 2 = 3 + 3 = 0

If instead you were resolution x-2 + 3 x + 2 = 0, you’ll:

x-2 + 3 x + 2 = 0.

( x + -2) × ( x + -1) = 0

-(x + 2) × (x +-1) = 0 when x +-2 = 0 or x +-1 = 0.

-x + 2 = 0 when x = 2

-x + 1 = 0 when x = 1

Make sure that they are indeed solutions

**Example # 2: Resolution of the equations of the second degree by factoring**

Solve x 2 + x-30 = 0.

First, you need to factor x 2 + x – 30

-30 = 30 ×-1

-30-15 × 2 =

-30 = 6 ×-5

-30 =-30 × 1

-30-15 × 2 =

-30 =-6 × 5

Since only 6 +-5 = 1 and 1 are the coefficient of the second term (x = **1**x), x 2 + x – 30 = (x + 6) × (x – 5)

x 2 + x-30 = 0 gives:

(x + 6) × (x – 5) = 0

(x + 6) × (x – 5) = 0 when x + 6 = 0 or x – 5 = 0.

x + 6 = 0 when x = – 6

x – 5 = 0 when x = 5

Let us now check x =-6 and x = 5 are indeed x 2 + x-30 = 0 solutions

(-6) 2 +-6 – 30 = 36-6-30 = 30-30 = 0

2 (5) + 5 + 5-30-30 = 25 = 30-30 = 0.

If instead you were resolution x 2 + x – 30 = 0, you’ll:

x 2 + – x – 30 = 0.

(x – 6) × (x + 5) = 0

(x – 6) × (x + 5) = 0 when x – 6 = 0 or x + 5 = 0.

x – 6 = 0 when x = 6

x + 5 = 0 when x = – 5

Make sure that they are indeed solutions

Solve equations of the second degree by factoring can become very difficult. See below:

**Example # 3: Solving the equations of the second degree by factoring**

6 x 2 + 27 x + 30 = 0.

First factor 6 x 2 + 27 x + 30

6 x 2 + x + 30 = 27 (3 x +?) × (2 x +?) or (x 6 +?) × (x +?)

Now, the last factor term 30

30 = 30 × 1

30 = 15 × 2

30 = 6 × 5

To obtain 27 x, you should try to multiply cross below. There are 6 of them. **Cross multiply and add!**

6 x x

30 1

x 6 + 30 x = x 36

**6 x x15 2**

x 12 + 15 x = x 27

6 x x

6 5

x 30 + 6 x = x 36

3 x 2 x

30 1

3 x + x 60 = x 63

3 x 2 x

15 2

x 6 + 30 x = x 36

**3 x 2 x6 5**

15 x + 12 x = x 27

You have two choices in bold!

6 x 2 + 27 x + 30 = 0.

(6 x + 15) × (x + 2) = 0

x 6 + 15 = 0.

x 6 = – 15

x = – 15/6

x + 2 = 0.

x = – 2